sed oneliner to add a line before another line - sed

I need to add a line with bar before each line with foo with sed.
I need to do this in a Makefile and so I cannot use i\ because it needs a newline in standard sed (not GNU sed, e.g., the one in Mac OS X) and this cannot be done in a Makefile (at least, not nicely).
The solution I found is:
sed '/foo/{h;s/.*/bar/;p;g;}' < in > out
This saves the line, replaces its contents with bar, prints the new line, restores the old line (and prints it by default).
Is there a simpler solution?

BSD sed
This will put bar before every line with foo:
sed $'/foo/{s/^/bar\\\n/;}' in >out
GNU sed
This will put bar before every line with foo:
sed '/foo/{s/^/bar\n/;}' in >out
How it works
/foo/
This selects lines that contain foo.
s/^/bar\n/
^ matches the beginning of the line. Thus, for the selected lines, this substitutes in bar\n at the beginning of the line. This effectively adds a new line to precede the one containing foo.
Under GNU, one can write \n and sed interprets it as a newline. This doesn't work under BSD sed. Hence the different version.

This might work for you (GNU sed):
sed '/foo/ibar' file
I'm not sure why you say you need a newline but just incase the inserted line needs to longer than one line you can employ bash so:
sed $'/foo/ibar\\\nbaz' file

Related

Sed - replace with variable first occurrence only [duplicate]

I would like to update a large number of C++ source files with an extra include directive before any existing #includes. For this sort of task, I normally use a small bash script with sed to re-write the file.
How do I get sed to replace just the first occurrence of a string in a file rather than replacing every occurrence?
If I use
sed s/#include/#include "newfile.h"\n#include/
it replaces all #includes.
Alternative suggestions to achieve the same thing are also welcome.
A sed script that will only replace the first occurrence of "Apple" by "Banana"
Example
Input: Output:
Apple Banana
Apple Apple
Orange Orange
Apple Apple
This is the simple script: Editor's note: works with GNU sed only.
sed '0,/Apple/{s/Apple/Banana/}' input_filename
The first two parameters 0 and /Apple/ are the range specifier. The s/Apple/Banana/ is what is executed within that range. So in this case "within the range of the beginning (0) up to the first instance of Apple, replace Apple with Banana. Only the first Apple will be replaced.
Background: In traditional sed the range specifier is also "begin here" and "end here" (inclusive). However the lowest "begin" is the first line (line 1), and if the "end here" is a regex, then it is only attempted to match against on the next line after "begin", so the earliest possible end is line 2. So since range is inclusive, smallest possible range is "2 lines" and smallest starting range is both lines 1 and 2 (i.e. if there's an occurrence on line 1, occurrences on line 2 will also be changed, not desired in this case). GNU sed adds its own extension of allowing specifying start as the "pseudo" line 0 so that the end of the range can be line 1, allowing it a range of "only the first line" if the regex matches the first line.
Or a simplified version (an empty RE like // means to re-use the one specified before it, so this is equivalent):
sed '0,/Apple/{s//Banana/}' input_filename
And the curly braces are optional for the s command, so this is also equivalent:
sed '0,/Apple/s//Banana/' input_filename
All of these work on GNU sed only.
You can also install GNU sed on OS X using homebrew brew install gnu-sed.
# sed script to change "foo" to "bar" only on the first occurrence
1{x;s/^/first/;x;}
1,/foo/{x;/first/s///;x;s/foo/bar/;}
#---end of script---
or, if you prefer: Editor's note: works with GNU sed only.
sed '0,/foo/s//bar/' file
Source
An overview of the many helpful existing answers, complemented with explanations:
The examples here use a simplified use case: replace the word 'foo' with 'bar' in the first matching line only.
Due to use of ANSI C-quoted strings ($'...') to provide the sample input lines, bash, ksh, or zsh is assumed as the shell.
GNU sed only:
Ben Hoffstein's anwswer shows us that GNU provides an extension to the POSIX specification for sed that allows the following 2-address form: 0,/re/ (re represents an arbitrary regular expression here).
0,/re/ allows the regex to match on the very first line also. In other words: such an address will create a range from the 1st line up to and including the line that matches re - whether re occurs on the 1st line or on any subsequent line.
Contrast this with the POSIX-compliant form 1,/re/, which creates a range that matches from the 1st line up to and including the line that matches re on subsequent lines; in other words: this will not detect the first occurrence of an re match if it happens to occur on the 1st line and also prevents the use of shorthand // for reuse of the most recently used regex (see next point).1
If you combine a 0,/re/ address with an s/.../.../ (substitution) call that uses the same regular expression, your command will effectively only perform the substitution on the first line that matches re.
sed provides a convenient shortcut for reusing the most recently applied regular expression: an empty delimiter pair, //.
$ sed '0,/foo/ s//bar/' <<<$'1st foo\nUnrelated\n2nd foo\n3rd foo'
1st bar # only 1st match of 'foo' replaced
Unrelated
2nd foo
3rd foo
A POSIX-features-only sed such as BSD (macOS) sed (will also work with GNU sed):
Since 0,/re/ cannot be used and the form 1,/re/ will not detect re if it happens to occur on the very first line (see above), special handling for the 1st line is required.
MikhailVS's answer mentions the technique, put into a concrete example here:
$ sed -e '1 s/foo/bar/; t' -e '1,// s//bar/' <<<$'1st foo\nUnrelated\n2nd foo\n3rd foo'
1st bar # only 1st match of 'foo' replaced
Unrelated
2nd foo
3rd foo
Note:
The empty regex // shortcut is employed twice here: once for the endpoint of the range, and once in the s call; in both cases, regex foo is implicitly reused, allowing us not to have to duplicate it, which makes both for shorter and more maintainable code.
POSIX sed needs actual newlines after certain functions, such as after the name of a label or even its omission, as is the case with t here; strategically splitting the script into multiple -e options is an alternative to using an actual newlines: end each -e script chunk where a newline would normally need to go.
1 s/foo/bar/ replaces foo on the 1st line only, if found there.
If so, t branches to the end of the script (skips remaining commands on the line). (The t function branches to a label only if the most recent s call performed an actual substitution; in the absence of a label, as is the case here, the end of the script is branched to).
When that happens, range address 1,//, which normally finds the first occurrence starting from line 2, will not match, and the range will not be processed, because the address is evaluated when the current line is already 2.
Conversely, if there's no match on the 1st line, 1,// will be entered, and will find the true first match.
The net effect is the same as with GNU sed's 0,/re/: only the first occurrence is replaced, whether it occurs on the 1st line or any other.
NON-range approaches
potong's answer demonstrates loop techniques that bypass the need for a range; since he uses GNU sed syntax, here are the POSIX-compliant equivalents:
Loop technique 1: On first match, perform the substitution, then enter a loop that simply prints the remaining lines as-is:
$ sed -e '/foo/ {s//bar/; ' -e ':a' -e '$!{n;ba' -e '};}' <<<$'1st foo\nUnrelated\n2nd foo\n3rd foo'
1st bar
Unrelated
2nd foo
3rd foo
Loop technique 2, for smallish files only: read the entire input into memory, then perform a single substitution on it.
$ sed -e ':a' -e '$!{N;ba' -e '}; s/foo/bar/' <<<$'1st foo\nUnrelated\n2nd foo\n3rd foo'
1st bar
Unrelated
2nd foo
3rd foo
1 1.61803 provides examples of what happens with 1,/re/, with and without a subsequent s//:
sed '1,/foo/ s/foo/bar/' <<<$'1foo\n2foo' yields $'1bar\n2bar'; i.e., both lines were updated, because line number 1 matches the 1st line, and regex /foo/ - the end of the range - is then only looked for starting on the next line. Therefore, both lines are selected in this case, and the s/foo/bar/ substitution is performed on both of them.
sed '1,/foo/ s//bar/' <<<$'1foo\n2foo\n3foo' fails: with sed: first RE may not be empty (BSD/macOS) and sed: -e expression #1, char 0: no previous regular expression (GNU), because, at the time the 1st line is being processed (due to line number 1 starting the range), no regex has been applied yet, so // doesn't refer to anything.
With the exception of GNU sed's special 0,/re/ syntax, any range that starts with a line number effectively precludes use of //.
sed '0,/pattern/s/pattern/replacement/' filename
this worked for me.
example
sed '0,/<Menu>/s/<Menu>/<Menu><Menu>Sub menu<\/Menu>/' try.txt > abc.txt
Editor's note: both work with GNU sed only.
You could use awk to do something similar..
awk '/#include/ && !done { print "#include \"newfile.h\""; done=1;}; 1;' file.c
Explanation:
/#include/ && !done
Runs the action statement between {} when the line matches "#include" and we haven't already processed it.
{print "#include \"newfile.h\""; done=1;}
This prints #include "newfile.h", we need to escape the quotes. Then we set the done variable to 1, so we don't add more includes.
1;
This means "print out the line" - an empty action defaults to print $0, which prints out the whole line. A one liner and easier to understand than sed IMO :-)
Quite a comprehensive collection of answers on linuxtopia sed FAQ. It also highlights that some answers people provided won't work with non-GNU version of sed, eg
sed '0,/RE/s//to_that/' file
in non-GNU version will have to be
sed -e '1s/RE/to_that/;t' -e '1,/RE/s//to_that/'
However, this version won't work with gnu sed.
Here's a version that works with both:
-e '/RE/{s//to_that/;:a' -e '$!N;$!ba' -e '}'
ex:
sed -e '/Apple/{s//Banana/;:a' -e '$!N;$!ba' -e '}' filename
With GNU sed's -z option you could process the whole file as if it was only one line. That way a s/…/…/ would only replace the first match in the whole file. Remember: s/…/…/ only replaces the first match in each line, but with the -z option sed treats the whole file as a single line.
sed -z 's/#include/#include "newfile.h"\n#include'
In the general case you have to rewrite your sed expression since the pattern space now holds the whole file instead of just one line. Some examples:
s/text.*// can be rewritten as s/text[^\n]*//. [^\n] matches everything except the newline character. [^\n]* will match all symbols after text until a newline is reached.
s/^text// can be rewritten as s/(^|\n)text//.
s/text$// can be rewritten as s/text(\n|$)//.
#!/bin/sed -f
1,/^#include/ {
/^#include/i\
#include "newfile.h"
}
How this script works: For lines between 1 and the first #include (after line 1), if the line starts with #include, then prepend the specified line.
However, if the first #include is in line 1, then both line 1 and the next subsequent #include will have the line prepended. If you are using GNU sed, it has an extension where 0,/^#include/ (instead of 1,) will do the right thing.
Just add the number of occurrence at the end:
sed s/#include/#include "newfile.h"\n#include/1
A possible solution:
/#include/!{p;d;}
i\
#include "newfile.h"
:a
n
ba
Explanation:
read lines until we find the #include, print these lines then start new cycle
insert the new include line
enter a loop that just reads lines (by default sed will also print these lines), we won't get back to the first part of the script from here
I know this is an old post but I had a solution that I used to use:
grep -E -m 1 -n 'old' file | sed 's/:.*$//' - | sed 's/$/s\/old\/new\//' - | sed -f - file
Basically use grep to print the first occurrence and stop there. Additionally print line number ie 5:line. Pipe that into sed and remove the : and anything after so you are just left with a line number. Pipe that into sed which adds s/.*/replace to the end number, which results in a 1 line script which is piped into the last sed to run as a script on the file.
so if regex = #include and replace = blah and the first occurrence grep finds is on line 5 then the data piped to the last sed would be 5s/.*/blah/.
Works even if first occurrence is on the first line.
i would do this with an awk script:
BEGIN {i=0}
(i==0) && /#include/ {print "#include \"newfile.h\""; i=1}
{print $0}
END {}
then run it with awk:
awk -f awkscript headerfile.h > headerfilenew.h
might be sloppy, I'm new to this.
As an alternative suggestion you may want to look at the ed command.
man 1 ed
teststr='
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
'
# for in-place file editing use "ed -s file" and replace ",p" with "w"
# cf. http://wiki.bash-hackers.org/howto/edit-ed
cat <<-'EOF' | sed -e 's/^ *//' -e 's/ *$//' | ed -s <(echo "$teststr")
H
/# *include/i
#include "newfile.h"
.
,p
q
EOF
I finally got this to work in a Bash script used to insert a unique timestamp in each item in an RSS feed:
sed "1,/====RSSpermalink====/s/====RSSpermalink====/${nowms}/" \
production-feed2.xml.tmp2 > production-feed2.xml.tmp.$counter
It changes the first occurrence only.
${nowms} is the time in milliseconds set by a Perl script, $counter is a counter used for loop control within the script, \ allows the command to be continued on the next line.
The file is read in and stdout is redirected to a work file.
The way I understand it, 1,/====RSSpermalink====/ tells sed when to stop by setting a range limitation, and then s/====RSSpermalink====/${nowms}/ is the familiar sed command to replace the first string with the second.
In my case I put the command in double quotation marks becauase I am using it in a Bash script with variables.
Using FreeBSD ed and avoid ed's "no match" error in case there is no include statement in a file to be processed:
teststr='
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
'
# using FreeBSD ed
# to avoid ed's "no match" error, see
# *emphasized text*http://codesnippets.joyent.com/posts/show/11917
cat <<-'EOF' | sed -e 's/^ *//' -e 's/ *$//' | ed -s <(echo "$teststr")
H
,g/# *include/u\
u\
i\
#include "newfile.h"\
.
,p
q
EOF
This might work for you (GNU sed):
sed -si '/#include/{s//& "newfile.h\n&/;:a;$!{n;ba}}' file1 file2 file....
or if memory is not a problem:
sed -si ':a;$!{N;ba};s/#include/& "newfile.h\n&/' file1 file2 file...
If anyone came here to replace a character for the first occurrence in all lines (like myself), use this:
sed '/old/s/old/new/1' file
-bash-4.2$ cat file
123a456a789a
12a34a56
a12
-bash-4.2$ sed '/a/s/a/b/1' file
123b456a789a
12b34a56
b12
By changing 1 to 2 for example, you can replace all the second a's only instead.
The use case can perhaps be that your occurences are spread throughout your file, but you know your only concern is in the first 10, 20 or 100 lines.
Then simply adressing those lines fixes the issue - even if the wording of the OP regards first only.
sed '1,10s/#include/#include "newfile.h"\n#include/'
The following command removes the first occurrence of a string, within a file. It removes the empty line too. It is presented on an xml file, but it would work with any file.
Useful if you work with xml files and you want to remove a tag. In this example it removes the first occurrence of the "isTag" tag.
Command:
sed -e 0,/'<isTag>false<\/isTag>'/{s/'<isTag>false<\/isTag>'//} -e 's/ *$//' -e '/^$/d' source.txt > output.txt
Source file (source.txt)
<xml>
<testdata>
<canUseUpdate>true</canUseUpdate>
<isTag>false</isTag>
<moduleLocations>
<module>esa_jee6</module>
<isTag>false</isTag>
</moduleLocations>
<node>
<isTag>false</isTag>
</node>
</testdata>
</xml>
Result file (output.txt)
<xml>
<testdata>
<canUseUpdate>true</canUseUpdate>
<moduleLocations>
<module>esa_jee6</module>
<isTag>false</isTag>
</moduleLocations>
<node>
<isTag>false</isTag>
</node>
</testdata>
</xml>
ps: it didn't work for me on Solaris SunOS 5.10 (quite old), but it works on Linux 2.6, sed version 4.1.5
Nothing new but perhaps a little more concrete answer: sed -rn '0,/foo(bar).*/ s%%\1%p'
Example: xwininfo -name unity-launcher produces output like:
xwininfo: Window id: 0x2200003 "unity-launcher"
Absolute upper-left X: -2980
Absolute upper-left Y: -198
Relative upper-left X: 0
Relative upper-left Y: 0
Width: 2880
Height: 98
Depth: 24
Visual: 0x21
Visual Class: TrueColor
Border width: 0
Class: InputOutput
Colormap: 0x20 (installed)
Bit Gravity State: ForgetGravity
Window Gravity State: NorthWestGravity
Backing Store State: NotUseful
Save Under State: no
Map State: IsViewable
Override Redirect State: no
Corners: +-2980+-198 -2980+-198 -2980-1900 +-2980-1900
-geometry 2880x98+-2980+-198
Extracting window ID with xwininfo -name unity-launcher|sed -rn '0,/^xwininfo: Window id: (0x[0-9a-fA-F]+).*/ s%%\1%p' produces:
0x2200003
POSIXly (also valid in sed), Only one regex used, need memory only for one line (as usual):
sed '/\(#include\).*/!b;//{h;s//\1 "newfile.h"/;G};:1;n;b1'
Explained:
sed '
/\(#include\).*/!b # Only one regex used. On lines not matching
# the text `#include` **yet**,
# branch to end, cause the default print. Re-start.
//{ # On first line matching previous regex.
h # hold the line.
s//\1 "newfile.h"/ # append ` "newfile.h"` to the `#include` matched.
G # append a newline.
} # end of replacement.
:1 # Once **one** replacement got done (the first match)
n # Loop continually reading a line each time
b1 # and printing it by default.
' # end of sed script.
A possible solution here might be to tell the compiler to include the header without it being mentioned in the source files. IN GCC there are these options:
-include file
Process file as if "#include "file"" appeared as the first line of
the primary source file. However, the first directory searched for
file is the preprocessor's working directory instead of the
directory containing the main source file. If not found there, it
is searched for in the remainder of the "#include "..."" search
chain as normal.
If multiple -include options are given, the files are included in
the order they appear on the command line.
-imacros file
Exactly like -include, except that any output produced by scanning
file is thrown away. Macros it defines remain defined. This
allows you to acquire all the macros from a header without also
processing its declarations.
All files specified by -imacros are processed before all files
specified by -include.
Microsoft's compiler has the /FI (forced include) option.
This feature can be handy for some common header, like platform configuration. The Linux kernel's Makefile uses -include for this.
I needed a solution that would work both on GNU and BSD, and I also knew that the first line would never be the one I'd need to update:
sed -e "1,/pattern/s/pattern/replacement/"
Trying the // feature to not repeat the pattern did not work for me, hence needing to repeat it.
I will make a suggestion that is not exactly what the original question asks for, but for those who also want to specifically replace perhaps the second occurrence of a match, or any other specifically enumerated regular expression match. Use a python script, and a for loop, call it from a bash script if needed. Here's what it looked like for me, where I was replacing specific lines containing the string --project:
def replace_models(file_path, pixel_model, obj_model):
# find your file --project matches
pattern = re.compile(r'--project.*')
new_file = ""
with open(file_path, 'r') as f:
match = 1
for line in f:
# Remove line ending before we do replacement
line = line.strip()
# replace first --project line match with pixel
if match == 1:
result = re.sub(pattern, "--project='" + pixel_model + "'", line)
# replace second --project line match with object
elif match == 2:
result = re.sub(pattern, "--project='" + obj_model + "'", line)
else:
result = line
# Check that a substitution was actually made
if result is not line:
# Add a backslash to the replaced line
result += " \\"
print("\nReplaced ", line, " with ", result)
# Increment number of matches found
match += 1
# Add the potentially modified line to our new file
new_file = new_file + result + "\n"
# close file / save output
f.close()
fout = open(file_path, "w")
fout.write(new_file)
fout.close()
sed -e 's/pattern/REPLACEMENT/1' <INPUTFILE

Substring file name in Unix using sed command

I want to substring the File name in unix using sed command.
File name : Test_Test1_Test2_10082019_030013.csv.20191008-075740
I want the characters after the 3rd underscore or (all the characters after Test2 ) i need to be printed .
Can this be done using sed command?
I have tried this command
sed 's/^.*_\([^_]*\)$/\1/' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
but this is giving result as 030013.csv.20191008-075740
I need it from 10082019_030013.csv.20191008-075740
Thanks
Neha
To remove from the beginning up to including the 3rd underscore you can use
sed 's/^\([^_]*_\)\{3\}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
This removes the initial part that consists of 3 groups of (any number of non-underscore characters followed by an underscore). The result is
10082019_030013.csv.20191008-075740
If you use GNU sed you can switch it to extended regular expressions and omit the backslashes.
sed -r 's/^([^_]*_){3}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
Could you please try following.
sed 's/\([^_]*\)_\([^_]*\)_\([^_]*\)_\(.*\)/\4/' Input_file
Or as per Bodo's nice suggestion:
sed 's/[^_]*_[^_]*_[^_]_\(.*\)/\1/' Input_file
This might work for you (GNU sed):
sed 's/_/\n/3;s/.*\n//;t;s/Test2/\n/;s/.*\n//;t;d' file
Replace the third _ by a newline and then remove everything upto and including the first newline. If this succeeds, bail out and print the result. Otherwise, try the same method with Test2 and if this fails delete the entire line.

Why is my sed multiline find-and-replace not working as expected?

I have a simple sed command that I am using to replace everything between (and including) //thistest.com-- and --thistest.com with nothing (remove the block all together):
sudo sed -i "s#//thistest\.com--.*--thistest\.com##g" my.file
The contents of my.file are:
//thistest.com--
zone "awebsite.com" {
type master;
file "some.stuff.com.hosts";
};
//--thistest.com
As I am using # as my delimiter for the regex, I don't need to escape the / characters. I am also properly (I think) escaping the . in .com. So I don't see exactly what is failing.
Why isn't the entire block being replaced?
You have two problems:
Sed doesn't do multiline pattern matches—at least, not the way you're expecting it to. However, you can use multiline addresses as an alternative.
Depending on your version of sed, you may need to escape alternate delimiters, especially if you aren't using them solely as part of a substitution expression.
So, the following will work with your posted corpus in both GNU and BSD flavors:
sed '\#^//thistest\.com--#, \#^//--thistest\.com# d' /tmp/corpus
Note that in this version, we tell sed to match all lines between (and including) the two patterns. The opening delimiter of each address pattern is properly escaped. The command has also been changed to d for delete instead of s for substitute, and some whitespace was added for readability.
I've also chosen to anchor the address patterns to the start of each line. You may or may not find that helpful with this specific corpus, but it's generally wise to do so when you can, and doesn't seem to hurt your use case.
# separation by line with 1 s//
sed -n -e 'H;${x;s#^\(.\)\(.*\)\1//thistest.com--.*\1//--thistest.com#\2#;p}' YourFile
# separation by line with address pattern
sed -e '\#//thistest.com--#,\#//--thistest.com# d' YourFile
# separation only by char (could be CR, CR/LF, ";" or "oneline") with s//
sed -n -e '1h;1!H;${x;s#//thistest.com--.*\1//--thistest.com##;p}' YourFile
Note:
assuming there is only 1 section thistest per file (if not, it remove anything between the first opening until the last closing section) for the use of s//
does not suite for huge file (load entire file into memory) with s//
sed using addresses pattern cannot select section on the same line, it search 1st pattern to start, and a following line to stop but very efficient on big file and/or multisection

Use sed to take all lines containing regex and append to end of file

I'm trying to come up with a sed script to take all lines containing a pattern and move them to the end of the output. This is an exercise in learning hold vs pattern space and I'm struggling to come up with it (though I feel close).
I'm here:
$ echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed -E '/foo/H; //d; $G'
hi
bar
something
yo
foo1
foo2
But I want the output to be:
hi
bar
something
yo
foo1
foo2
I understand why this is happening. It is because the first time we find foo the hold space is empty so the H appends \n to the blank hold space and then the first foo, which I suppose is fine. But then the $G does it again, namely another append which appends \n plus what is in the hold space to the pattern space.
I tried a final delete command with /^$/d but that didn't remove the blank line (I think this is because this pattern is being matched not against the last line, but against the, now, multiline pattern space which has a \n\n in it.
I'm sure the sed gurus have a fix for me.
This might work for you (GNU sed):
sed '/foo/H;//!p;$!d;x;//s/.//p;d' file
If the line contains the required string append it to the hold space (HS) otherwise print it as normal. If it is not the last line delete it otherwise swap the HS for the pattern space (PS). If the required string(s) is now in the PS (what was the HS); since all such patterns were appended, the first character will be a newline, delete the first character and print. Delete whatever is left.
An alternative, using the -n flag:
sed -n '/foo/H;//!p;$!b;x;//s/.//p' file
N.B. When the d or b (without a parameter) command is performed no further sed commands are, a new line is read into the PS and the sed script begins with the first command i.e. the sed commands do not resume following the previous d command.
Why? Stuff like this is absolutely trivial in awk, awk is available everywhere that sed is, and the resulting awk script will be simpler, more portable, faster and better in almost every other way than a sed script to do the same task. All that hold space stuff was necessary in sed before the mid-1970s when awk was invented but there's absolutely no use for it now other than as a mental exercise.
$ echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" |
awk '/foo/{buf = buf $0 RS;next} {print} END{printf "%s",buf}'
hi
bar
something
yo
foo1
foo2
The above will work as-is in every awk on every UNIX installation and I bet you can figure out how it works very easily.
This feels like a hack and I think it should be possible to handle this situation more gracefully. The following works on GNU sed:
echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed -r '/foo/{H;d;}; $G; s/\n\n/\n/g'
However, on OSX/BSD sed, results in this odd output:
hi
bar
something
yonfoo1
foo2
Note the 2 consecutive newlines was replaced with the literal character n
The OSX/BSD vs GNU sed is explained in this article. And the following works (in GNU SED as well):
echo -e "hi\nfoo1\nbar\nsomething\nfoo2\nyo" | sed '/foo/{H;d;}; $G; s/\n\n/\'$'\n''/'
TL;DR; in BSD sed, it does not accept escaped characters in the RHS of the replacement expression and so you either have to put a true LF/newline in there at the command line, or do the above where you split the sed script string where you need the newline on the RHS and put a dollar sign in front of '\n' so the shell interprets it as a line feed.

Matching strings even if they start with white spaces in SED

I'm having issues matching strings even if they start with any number of white spaces. It's been very little time since I started using regular expressions, so I need some help
Here is an example. I have a file (file.txt) that contains two lines
#String1='Test One'
String1='Test Two'
Im trying to change the value for the second line, without affecting line 1 so I used this
sed -i "s|String1=.*$|String1='Test Three'|g"
This changes the values for both lines. How can I make sed change only the value of the second string?
Thank you
With gnu sed, you match spaces using \s, while other sed implementations usually work with the [[:space:]] character class. So, pick one of these:
sed 's/^\s*AWord/AnotherWord/'
sed 's/^[[:space:]]*AWord/AnotherWord/'
Since you're using -i, I assume GNU sed. Either way, you probably shouldn't retype your word, as that introduces the chance of a typo. I'd go with:
sed -i "s/^\(\s*String1=\).*/\1'New Value'/" file
Move the \s* outside of the parens if you don't want to preserve the leading whitespace.
There are a couple of solutions you could use to go about your problem
If you want to ignore lines that begin with a comment character such as '#' you could use something like this:
sed -i "/^\s*#/! s|String1=.*$|String1='Test Three'|g" file.txt
which will only operate on lines that do not match the regular expression /.../! that begins ^ with optional whiltespace\s* followed by an octothorp #
The other option is to include the characters before 'String' as part of the substitution. Doing it this way means you'll need to capture \(...\) the group to include it in the output with \1
sed -i "s|^\(\s*\)String1=.*$|\1String1='Test Four'|g" file.txt
With GNU sed, try:
sed -i "s|^\s*String1=.*$|String1='Test Three'|" file
or
sed -i "/^\s*String1=/s/=.*/='Test Three'/" file
Using awk you could do:
awk '/String1/ && f++ {$2="Test Three"}1' FS=\' OFS=\' file
#String1='Test One'
String1='Test Three'
It will ignore first hits of string1 since f is not true.