I have function that prints a header that needs to be applied across several files, but if I utilize a sed process substitution the lines prior to the last have a backslash \ on them.
E.g.
function print_header() {
cat << EOF
-------------------------------------------------------------------
$(date '+%B %d, %Y # ~ %r') ID:$(echo $RANDOM)
EOF
}
If I then take a file such as test.txt:
line 1
line 2
line 3
line 4
line 5
sed "1 i $(print_header | sed 's/$/\\/g')" test.txt
I get:
-------------------------------------------------------------------\
November 24, 2015 # ~ 11:18:28 AM ID:13187
line 1
line 2
line 3
line 4
line 5
Notice the troublesome backslash at the end of the first line, I'd like to not have that backslash appear. Any ideas?
I would use cat for that:
cat <(print_header) file > file_with_header
This behavior depends on the sed dialect. Unfortunately, it's one of the things which depends on which version you have.
To simplify debugging, try specifying verbatim text. Here's one from a Debian system.
vnix$ sed '1i\
> foo\
> bar' <<':'
> hello
> goodbye
> :
foo
bar
hello
goodbye
Your diagnostics appear to indicate that your sed dialect does not in fact require the backslash after the first i.
Since you are generating the contents of the header programmatically anyway, my recommended solution would be to refactor the code so that you can avoid this conundrum. If you don't want cat <<EOF test.txt then maybe experiment with sed 1r/dev/stdin' <<EOF test.txt (I could not get 1r- to work, but /dev/stdin should be portable to any Linux.)
Here is my kludgy fix, if you can find something more elegant I'll gladly credit you:
sed "1 i $(print_header | sed 's/$/\\/g;$s/$/\x01/')" test.txt | tr -d '\001'
This puts an unprintable SOH (\x01) ascii Start Of Header character after the inserted text, that precludes the backslashes and then I run it over tr to delete the SOH chars.
Related
I have a file called error_log for the apache and I want to see the first line and the last line of this file using sed command. Would you please help me how can I do that?
I know how to do that with head and tail commands, but I'm curious if it's possible in sed command too.
I have read the man sed and have googled a lot but nothing is found unfortunately.
This might work for you (GNU sed):
sed '1b;$b;d' file
All sed commands can be prefixed by either an address or a regexp. An address is either a line number or the $ which represents the last line. If neither an address or a regexp is present, the following command applies to all other lines.
The normal sed cycle, presents each line of input (less its newline) in the pattern space. The sed commands are then applied and the final act of the cycle is to re-attach the newline and print the result.
The b command controls command flow; if by itself it jumps out of the following sed commands to the final act of the cycle i.e. where the newline is re-attached and the result printed.
The d command deletes the pattern space and since there is nothing to be printed no further processing is executed (including re-attaching the newline and printing the result).
Thus the solution above prints the first line and the last and deletes the rest.
Sed has some command line options, one of which turns of the implicit printing of the result of the pattern space -n. The p command prints the current state of the pattern space. Thus the dual of the above solution is:
sed -n '1p;$p' file
N.B. If the input file is only one line the first solution will only print one line whereas the second solution will print the same line twice. Also if more than one file is input both solutions will print the first line of the first file and last line of the last file unless the -i option is in place, in which case each file will be amended. The -s option replicates this without amending each file but streams the results to stdout as if each file is treated separately.
This will work:
sed -n '1p ; $p' error_log
1p will print the first line and $p will print the last line.
As a suggestion, take a look at info sed, not only man sed. You can find the some examples about your question at the paragraph 2.1.
First line:
sed '2,$d' error_log
Last line:
sed '$!d' error_log
Based on your new requirement to output nothing if the input file is just 1 line (see How to find only the first and last line of a file using sed):
awk 'NR==1{first=$0} {last=$0} END{if (NR>1) print first ORS last}'
Original answer:
This is one of those things that you can, at face value, do easily enough in sed:
$ seq 3 7
3
4
5
6
7
$ seq 3 7 | sed -n '1p; $p'
3
7
but then how to handle edge cases like one line of input is non-obvious, e.g. is this REALLY the correct output:
$ printf 'foo\n' | sed -n '1p; $p'
foo
foo
or is the correct output just:
foo
and if the latter, how do you tweak that sed command to produce that output? #potong suggested a GNU sed command:
$ printf 'foo\n' | sed '1b;$b;d'
foo
which works but may be GNU-only (idk) and more importantly doesn't look much like the command we started with so the tiniest change in requirements meant a complete rewrite using different constructs.
Now, how about if you want to enhance it to, say, only print the first and last line if the file contained foo? I expect that'd be another challenging exercise with sed and probably involve non-portable constructs too.
It's just all pointless to learn how to do this with sed when you can use a different tool like awk and do whatever you like in a simple, consistent, portable syntax:
$ seq 3 7 |
awk 'NR==1{first=$0} {last=$0} END{print first ORS last}'
3
7
$ printf 'foo\n' |
awk 'NR==1{first=$0} {last=$0} END{print first ORS last}'
foo
foo
$ printf 'foo\n' |
awk 'NR==1{first=$0} {last=$0} END{print first (NR>1 ? ORS last : "")}'
foo
$ printf '3\nfoo\n7\n' |
awk 'NR==1{first=$0} /foo/{f=1} {last=$0} END{if (f) print first (NR>1 ? ORS last : "")}'
3
7
$ printf '3\nbar\n7\n' |
awk 'NR==1{first=$0} /foo/{f=1} {last=$0} END{if (f) print first (NR>1 ? ORS last : "")}'
$
Notice that:
Every command looks like every other command.
A minor change in requirements leads to a minor change in the code, not a complete rewrite.
Once you learn how to do any given thing A, how to do similar things B, C, D, etc. just builds on top of the syntax you already used, you don't have to learn a completely different syntax.
Each of those commands will work using any awk in any shell on every UNIX box.
Now, how about if you want to do that for multiple files such as would be created by the following commands?
$ seq 3 7 > file1
$ seq 12 25 > file2
With awk you can just store the lines in an array for printing in the END:
$ awk 'FNR==1{first[++cnt]=$0} {last[cnt]=$0}
END{for (i=1;i<=cnt;i++) print first[i] ORS last[i]}' file1 file2
3
7
12
25
or with GNU awk you can print them from ENDFILE:
$ awk 'FNR==1{first=$0} {last=$0} ENDFILE{print first ORS last}' file1 file2
3
7
12
25
With sed? An exercise left for the reader.
I want to get the first line of a file that is not commented out with an hash, then append a line of text just after that line just before that line.
I managed to get the number of the line:
sed -n '/^\s*#/!{=;q}' file // prints 2
and also to insert text (specifying the line manually):
sed '2 a extralinecontent' file
I can't get them working together as a one liner or in a batch.
I tried command substitution (with $(command) and also with backticks) but I get an error from bash:
sed '$(sed -n '/^\s*#/!{=;q}' file) a extralinecontent' file
-bash: !{=: event not found
and also tried many other combinations, but no luck.
I'm using gnu-sed (via brew) on macOS.
This might work for you (GNU sed):
sed -e '/^\s*#/b;a extra line content' -e ':a;n;ba' file
Bail out of any lines beginning with a comment at the beginning of the file, append an extra line following the first line that is not a comment and keep fetching/printing all the remaining lines of the file.
Here's a way to do it with GNU sed without reading the file twice
$ cat ip.txt
#comment
foo baz good
123 456 7889
$ sed -e '0,/^\s*[^#[:space:]]/ {// a XYZ' -e '}' ip.txt
#comment
foo baz good
XYZ
123 456 7889
GNU sed allows first address to be 0 if the other address is regex, that way this will work even if first line matches the condition
/^\s*[^#[:space:]]/ as sed doesn't support possessive quantifier, need to ensure that the first character being matched by the character class isn't either a # or a whitespace character
// is a handy shortcut to repeat the last regex
a XYZ your required line to be appended (note that your question mentiones insert, so if you want that, use i instead of a)
Have a file that has been created incorrectly. There are several space delimited fields in the file but one text field has some unwanted newlines. This is causing a big problem.
How can I remove these characters but not the wanted line ends?
file is:
'Number field' 'Text field' 'Number field'
1 Some text 999999
2 more
text 111111111
3 Even more text 8888888888
EOF
So there is a NL after the word "more".
I've tried sed:
sed 's/.$//g' test.txt > test.out
and
sed 's/\n//g' test.txt > test.out
But none of these work. The newlines do not get removed.
tr -d '\n' does too much - I need to remove ONLY the newlines that are preceded by a space.
How can I delete newlines that follow a space?
SunOS 5.10 Generic_144488-09 sun4u sparc SUNW,Sun-Fire-V440
A sed solution is
sed '/ $/{N;s/\n//}'
Explanation:
/ $/: whenever the line ends in space, then
N: append a newline and the next line of input, and
s/\n//: delete the newline.
It might be simplest with Perl:
perl -p0 -e 's/ \n/ /g'
The -0 flag makes Perl read the entire file as one line. Then we can substitute using s in the usual way. You can, of course, also add the -i option to edit the file in-place.
How can I delete newlines that follow a space?
If you want every occurrence of $' \n' in the original file to be replaced by a space ($' '), and if you know of a character (e.g. a control character) that does not appear in the file, then the task can be accomplished quite simply using sed and tr (as you requested). Let's suppose, for example, that control-A is a character that is not in the file. For the sake of simplicity, let's also assume we can use bash. Then the following script should do the job:
#!/bin/bash
A=$'\01'
tr '\n' "$A" | sed "s/ $A/ /g" | tr "$A" '\n'
We have a process which can use a file containing sed commands to alter piped input.
I need to replace a placeholder in the input with a variable value, e.g. in a single -e type of command I can run;
$ echo "Today is XX" | sed -e "s/XX/$(date +%F)/"
Today is 2012-10-11
However I can only specify the sed aspects in a file (and then point the process at the file), E.g. a file called replacements.sed might contain;
s/XX/Thursday/
So obviously;
$ echo "Today is XX" | sed -f replacements.sed
Today is Thursday
If I want to use an environment variable or shell value, though, I can't find a way to make it expand, e.g. if replacements.txt contains;
s/XX/$(date +%F)/
Then;
$ echo "Today is XX" | sed -f replacements.sed
Today is $(date +%F)
Including double quotes in the text of the file just prints the double quotes.
Does anyone know a way to be able to use variables in a sed file?
This might work for you (GNU sed):
cat <<\! > replacements.sed
/XX/{s//'"$(date +%F)"'/;s/.*/echo '&'/e}
!
echo "Today is XX" | sed -f replacements.sed
If you don't have GNU sed, try:
cat <<\! > replacements.sed
/XX/{
s//'"$(date +%F)"'/
s/.*/echo '&'/
}
!
echo "Today is XX" | sed -f replacements.sed | sh
AFAIK, it's not possible. Your best bet will be :
INPUT FILE
aaa
bbb
ccc
SH SCRIPT
#!/bin/sh
STRING="${1//\//\\/}" # using parameter expansion to prevent / collisions
shift
sed "
s/aaa/$STRING/
" "$#"
COMMAND LINE
./sed.sh "fo/obar" <file path>
OUTPUT
fo/obar
bbb
ccc
As others have said, you can't use variables in a sed script, but you might be able to "fake" it using extra leading input that gets added to your hold buffer. For example:
[ghoti#pc ~/tmp]$ cat scr.sed
1{;h;d;};/^--$/g
[ghoti#pc ~/tmp]$ sed -f scr.sed <(date '+%Y-%m-%d'; printf 'foo\n--\nbar\n')
foo
2012-10-10
bar
[ghoti#pc ~/tmp]$
In this example, I'm using process redirection to get input into sed. The "important" data is generated by printf. You could cat a file instead, or run some other program. The "variable" is produced by the date command, and becomes the first line of input to the script.
The sed script takes the first line, puts it in sed's hold buffer, then deletes the line. Then for any subsequent line, if it matches a double dash (our "macro replacement"), it substitutes the contents of the hold buffer. And prints, because that's sed's default action.
Hold buffers (g, G, h, H and x commands) represent "advanced" sed programming. But once you understand how they work, they open up new dimensions of sed fu.
Note: This solution only helps you replace entire lines. Replacing substrings within lines may be possible using the hold buffer, but I can't imagine a way to do it.
(Another note: I'm doing this in FreeBSD, which uses a different sed from what you'll find in Linux. This may work in GNU sed, or it may not; I haven't tested.)
I am in agreement with sputnick. I don't believe that sed would be able to complete that task.
However, you could generate that file on the fly.
You could change the date to a fixed string, like
__DAYOFWEEK__.
Create a temp file, use sed to replace __DAYOFWEEK__ with $(date +%Y).
Then parse your file with sed -f $TEMPFILE.
sed is great, but it might be time to use something like perl that can generate the date on the fly.
To add a newline in the replacement expression using a sed file, what finally worked for me is escaping a literal newline. Example: to append a newline after the string NewLineHere, then this worked for me:
#! /usr/bin/sed -f
s/NewLineHere/NewLineHere\
/g
Not sure it matters but I am on Solaris unix, so not GNU sed for sure.
how to remove comment lines (as # bal bla ) and empty lines (lines without charecters) from file with one sed command?
THX
lidia
If you're worried about starting two sed processes in a pipeline for performance reasons, you probably shouldn't be, it's still very efficient. But based on your comment that you want to do in-place editing, you can still do that with distinct commands (sed commands rather than invocations of sed itself).
You can either use multiple -e arguments or separate commands with a semicolon, something like (just one of these, not both):
sed -i 's/#.*$//' -e '/^$/d' fileName
sed -i 's/#.*$//;/^$/d' fileName
The following transcript shows this in action:
pax> printf 'Line # with a comment\n\n# Line with only a comment\n' >file
pax> cat file
Line # with a comment
# Line with only a comment
pax> cp file filex ; sed -i 's/#.*$//;/^$/d' filex ; cat filex
Line
pax> cp file filex ; sed -i -e 's/#.*$//' -e '/^$/d' filex ; cat filex
Line
Note how the file is modified in-place even with two -e options. You can see that both commands are executed on each line. The line with a comment first has the comment removed then all is removed because it's empty.
In addition, the original empty line is also removed.
#paxdiablo has a good answer but it can be improved.
(1) The '/^$/d' clause only matches 100% blank lines.
If you want to also match lines that are entirely whitespace (spaces, tabs etc.) use this instead:
'/^\s*$/d'
(2) The 's/#.*$//' clause only matches lines that start with the # character in column 0.
If you want to also match lines that have only whitespace before the first # use this instead:
'/^\s*#.*$/d'
The above criteria may not be universal (e.g. within a HEREDOC block, or in a Python multi-line string the different approaches could be significant), but in many cases the conventional definition of "blank" lines include whitespace-only, and "comment" lines include whitespace-then-#.
(3) Lastly, on OSX at least, the #paxdiablo solution in which the first clause turns comment lines into blank lines, and the second clause strips blank lines (including what were originally comments) doesn't work. It seems to be more portable to make both clauses /d delete actions as I've done.
The revised command incorporating the above is:
sed -e '/^\s*#.*$/d' -e '/^\s*$/d' inputFile
This tiny jewel removes all # comments, no matter where they begin in a line (see caution below):
sed -e 's/\s*#.*$//'
Example:
text="
this is a # test
#this is a test
#this is a #test
this is # another #test
"
$echo "$text" | sed -e 's/\s*#.*$//'
this is a
this is
Next this removes any resulting blank lines:
$echo "$text" | sed -e 's/\s*#.*$//' | sed -e '/^\s*$/d'
Caution: Depending on the syntax and/or interpretation of the lines your processing, this might not be an appropriate solution, as it just stupidly removes end of lines, even if the '#' is part of your data or code. However, for use cases where you'll never use a hash except for as an end of line comment then it works fine. So just as with all coding, context must be taken into consideration.
Alternative variant, using grep:
cat file.txt | grep -Ev '(#.*$)|(^$)'
you can use awk
awk 'NF{gsub(/^[ \t]*#/,"");print}' file
First example(paxdiablo) is very good except its not change file, just output result. If you want to change it inline:
sudo sed -i 's/#.*$//;/^$/d' inputFile
On (one of) my linux boxes, sed understands extended regular expressions with the -r option, so:
sed -r '/(^\s*#)|(^\s*$)/d' squid.conf.installed
is very useful for showing all non-blank, non comment lines.
The regex matches either start of line followed by zero or more spaces or tabs followed by either a hash or end of line, and deletes those matching lines from the input.