How to find structures matching a pattern
struct struct_name {
....
....
};
I'm using
sed -n -e '/struct{/,/}/p'
how to search for any struct_name
To extract all struct definitions (POSIX-compliant command):
sed -n '/struct [^ {]\{1,\} {/,/}/p' file
More robust with respect to whitespace variations (POSIX-compliant):
sed -n '/struct[[:blank:]]\{1,\}[^ {]\{1,\}[[:blank:]]*{/,/}/p' file
Alternative, using an extended regular expression (works with both GNU and BSD/macOS sed):
sed -E -n '/struct[[:blank:]]+[^ {]+[[:blank:]]*\{/,/\}/p' file
awk alternative (awk only uses extended regexes):
awk '/struct[[:blank:]]+[^ {]+[[:blank:]]*\{/,/\}/' file
The awk solution has the added advantage that a given struct definition will also be extracted correctly if it is all on a single line: awk looks for the end of a range on the same input line as the start of the range, whereas sed does not.
To extract a specific struct definition by name:
sed doesn't support variables, so your best bet is to splice in a shell variable that the shell expands up front.
name='struct_name' # define name to search for as shell var.
sed -n '/struct '"$name"' {/,/}/p' file # splice shell var. into sed script
Note that I've deliberately not used sed -n "/struct $name {/,/}/p" - a single, double-quoted string expanded by the shell as a whole - so as to make it clear which part of the sed script is expanded by the shell up front.
This works in this simple case, but is tricky business in general, because you must ensure that the expanded variable value contains no regex/sed metacharacters that break the command.
Here's an awk alternative that uses awk variables and literal substring matching to bypass the problem of potentially having to escape the variable value:
awk -v name='struct_name' 'index($0, "struct " name " {"),/}/' file
This solution has the added advantage that the struct definition will also be extracted correctly if it is all on a single line: awk looks for the end of a range on the same input line as the start of the range, whereas sed does not.
This will search a text file for struct_name. You can use the -E switch to use a regular expression.
grep -no struct_name test.txt
The -n switch causes the line number to be included, the -o means only the matching element of the line will be displayed.
Related
I am trying to create a filter command to reduce the lines from a log file, assume each line contains partition made of date,
/iamthepath01/20200301/file01.txt
/iamthepath02/20200302/file02.txt
....
/iamthepathxx/20210619/filexx.txt
then from thousands of lines I only want to keep the ones with two string in the path
/202106
/202105
and remove any other lines
I have tried following command
sed -i -e '\(/202105\|/202106\)!d' ~/log.txt
above command threw
sed: -e expression #1, char 24: unterminated address regex
You can use
sed -i '/\/20210[56]/!d' ~/log.txt
Or, if you need to use more specific alternatives and further enhance the pattern:
sed -i -E '/\/(202105|202106)/!d' ~/log.txt
Details:
-i - GNU sed option for inline file replacement
-E - option enabling POSIX ERE regex syntax
/\/20210[56]/ - regex that matches /20210 and then either 5 or 6
\/(202105|202106) - the POSIX ERE pattern that matches / and then either 202105 or 202106
!d - removes the lines not matching the pattern.
See the online demo:
#!/bin/bash
s='/iamthepath01/20200301/file01.txt
/iamthepath02/20200302/file02.txt
/iamthepathxx/20210619/filexx.txt'
sed '/\/20210[56]/!d' <<< "$s"
Output:
/iamthepathxx/20210619/filexx.txt
sed is the wrong tool for this. If you want a script that's as fragile as the sed one then use grep as it's the tool that exists solely to do a simple g/re/p (hence the name) like you're doing:
$ grep '/20210[56]' file
/iamthepathxx/20210619/filexx.txt
or if you want a more robust solution that focuses just on the part of the line you want to match and so will avoid false matches, then use awk:
$ awk -F '/' '$3 ~ /^20210[56]/' file
/iamthepathxx/20210619/filexx.txt
This might work for you (GNU sed):
sed -ni '\#/20210[56]#p' file
This uses seds -n grep-like option to turn off implicit printing and -i option to edit the file in place.
Normally sed uses the /.../ to match but other delimiters may be used if the first is escaped e.g. \#...#.
So the above solution will filter the existing file down to lines that contain either /202105 or /202106.
N.B. grep will almost certainly be faster in finding the above lines however the use of the -i option may be the ultimate reason for choosing sed (although the same outcome can be achieved by tacking on the > tmpFile && mv tmpFile file to a grep solution).
I need to comment out a line in a crontab file through a script, so it contains directories, spaces and symbols. This specific line is stored in a variable and I am starting to get mixed up on how to escape the variable. Since the line changes on a regular basis I dont want any escaping in there. I don't want to simply add # in front of it, since I also need to switch it around and replace the line again with the original without the #.
So the goal is to replace $line with #$line (comment) with the possibility to do it the other way around (uncomment).
So I have a variable:
line="* * * hello/this/line & /still/this/line"
This is a line that occurs in a file, file.txt. Wich needs to get comment out.
First try:
sed -i "s/^${line}/#${line}/" file.txt
Second try:
sed -i 's|'${line}'|'"#${line}"'|g' file.txt
choroba's helpful answer shows an effective solution using perl.
sed solution
If you want to use sed, you must use a separate sed command just to escape the $line variable value, because sed has no built-in way to escape strings for use as literals in a regex context:
lineEscaped=$(sed 's/[^^]/[&]/g; s/\^/\\^/g' <<<"$line") # escape $line for use in regex
sed -i "s/^$lineEscaped\$/#&/" file.txt # Note the \$ to escape the end-of-line anchor $
With BSD/macOS sed, use -i '' instead of just -i for in-place updating without backup.
And the reverse (un-commenting):
sed -i "s/^#\($lineEscaped\)\$/\1/" file.txt
See this answer of mine for an explanation of the sed command used for escaping, which should work with any input string.
Also note how variable $lineEscaped is only referenced once, in the regex portion of the s command, whereas the substitution-string portion simply references what the regex matched (which avoids the need to escape the variable again, using different rules):
& in the substitution string represents the entire match, and \1 the first capture group (parenthesized subexpression, \(...\)).
For simplicity, the second sed command uses double quotes in order to embed the value of shell variable $lineEscaped in the sed script, but it is generally preferable to use single-quoted scripts so as to avoid confusion between what the shell interprets up front vs. what sed ends up seeing.
For instance, $ is special to both the shell and sed, and in the above script the end-of-line anchor $ in the sed regex must therefore be escaped as \$ to prevent the shell from interpreting it.
One way to avoid confusion is to selectively splice double-quoted shell-variable references into the otherwise single-quoted script:
sed -i 's/^'"$lineEscaped"'$/#&/' file.txt
awk solution
awk offers literal string matching, which obviates the need for escaping:
awk -v line="$line" '$0 == line { $0 = "#" $0 } 1' file.txt > $$.tmp && mv $$.tmp file.txt
If you have GNU Awk v4.1+, you can use -i inplace for in-place updating.
And the reverse (un-commenting):
awk -v line="#$line" '$0 == line { $0 = substr($0, 2) } 1' file.txt > $$.tmp &&
mv $$.tmp file.txt
Perl has ways to do the quoting/escaping for you:
line=$line perl -i~ -pe '$regex = quotemeta $ENV{line}; s/^$regex/#$ENV{line}/' -- input.txt
This question already has answers here:
Using different delimiters in sed commands and range addresses
(3 answers)
Closed 1 year ago.
I have a Visual Studio project, which is developed locally. Code files have to be deployed to a remote server. The only problem is the URLs they contain, which are hard-coded.
The project contains URLs such as ?page=one. For the link to be valid on the server, it must be /page/one .
I've decided to replace all URLs in my code files with sed before deployment, but I'm stuck on slashes.
I know this is not a pretty solution, but it's simple and would save me a lot of time. The total number of strings I have to replace is fewer than 10. A total number of files which have to be checked is ~30.
An example describing my situation is below:
The command I'm using:
sed -f replace.txt < a.txt > b.txt
replace.txt which contains all the strings:
s/?page=one&/pageone/g
s/?page=two&/pagetwo/g
s/?page=three&/pagethree/g
a.txt:
?page=one&
?page=two&
?page=three&
Content of b.txt after I run my sed command:
pageone
pagetwo
pagethree
What I want b.txt to contain:
/page/one
/page/two
/page/three
The easiest way would be to use a different delimiter in your search/replace lines, e.g.:
s:?page=one&:pageone:g
You can use any character as a delimiter that's not part of either string. Or, you could escape it with a backslash:
s/\//foo/
Which would replace / with foo. You'd want to use the escaped backslash in cases where you don't know what characters might occur in the replacement strings (if they are shell variables, for example).
The s command can use any character as a delimiter; whatever character comes after the s is used. I was brought up to use a #. Like so:
s#?page=one&#/page/one#g
A very useful but lesser-known fact about sed is that the familiar s/foo/bar/ command can use any punctuation, not only slashes. A common alternative is s#foo#bar#, from which it becomes obvious how to solve your problem.
add \ before special characters:
s/\?page=one&/page\/one\//g
etc.
In a system I am developing, the string to be replaced by sed is input text from a user which is stored in a variable and passed to sed.
As noted earlier on this post, if the string contained within the sed command block contains the actual delimiter used by sed - then sed terminates on syntax error. Consider the following example:
This works:
$ VALUE=12345
$ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
MyVar=12345
This breaks:
$ VALUE=12345/6
$ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
sed: -e expression #1, char 21: unknown option to `s'
Replacing the default delimiter is not a robust solution in my case as I did not want to limit the user from entering specific characters used by sed as the delimiter (e.g. "/").
However, escaping any occurrences of the delimiter in the input string would solve the problem.
Consider the below solution of systematically escaping the delimiter character in the input string before having it parsed by sed.
Such escaping can be implemented as a replacement using sed itself, this replacement is safe even if the input string contains the delimiter - this is since the input string is not part of the sed command block:
$ VALUE=$(echo ${VALUE} | sed -e "s#/#\\\/#g")
$ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
MyVar=12345/6
I have converted this to a function to be used by various scripts:
escapeForwardSlashes() {
# Validate parameters
if [ -z "$1" ]
then
echo -e "Error - no parameter specified!"
return 1
fi
# Perform replacement
echo ${1} | sed -e "s#/#\\\/#g"
return 0
}
this line should work for your 3 examples:
sed -r 's#\?(page)=([^&]*)&#/\1/\2#g' a.txt
I used -r to save some escaping .
the line should be generic for your one, two three case. you don't have to do the sub 3 times
test with your example (a.txt):
kent$ echo "?page=one&
?page=two&
?page=three&"|sed -r 's#\?(page)=([^&]*)&#/\1/\2#g'
/page/one
/page/two
/page/three
replace.txt should be
s/?page=/\/page\//g
s/&//g
please see this article
http://netjunky.net/sed-replace-path-with-slash-separators/
Just using | instead of /
Great answer from Anonymous. \ solved my problem when I tried to escape quotes in HTML strings.
So if you use sed to return some HTML templates (on a server), use double backslash instead of single:
var htmlTemplate = "<div style=\\"color:green;\\"></div>";
A simplier alternative is using AWK as on this answer:
awk '$0="prefix"$0' file > new_file
You may use an alternative regex delimiter as a search pattern by backs lashing it:
sed '\,{some_path},d'
For the s command:
sed 's,{some_path},{other_path},'
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 can I make sed filter matching lines according to some expression, but ignore non-matching lines, instead of letting them print?
As a real example, I want to run scalac (the Scala compiler) on a set of files, and read from its -verbose output the .class files created. scalac -verbose outputs a bunch of messages, but we're only interested in those of the form [wrote some-class-name.class].
What I'm currently doing is this (|& is bash 4.0's way to pipe stderr to the next program):
$ scalac -verbose some-file.scala ... |& sed 's/^\[wrote \(.*\.class\)\]$/\1/'
This will extract the file names from the messages we're interested in, but will also let all other messages pass through unchanged! Of course we could do instead this:
$ scalac -verbose some-file.scala ... |& grep '^\[wrote .*\.class\]$' |
sed 's/^\[wrote \(.*\.class\)\]$/\1/'
which works but looks very much like going around the real problem, which is how to instruct sed to ignore non-matching lines from the input. So how do we do that?
If you don't want to print lines that don't match, you can use the combination of
-n option which tells sed not to print
p flag which tells sed to print what is matched
This gives:
sed -n 's/.../.../p'
Another way with plain sed:
sed -e 's/.../.../;t;d'
s/// is a substituion, t without any label conditionally skips all following commands, d deletes line.
No need for perl or grep.
(edited after Nicholas Riley's suggestion)
Rapsey raised a relevant point about multiple substitutions expressions.
First, quoting an Unix SE answer, you can "prefix most sed commands with an address to limit the lines to which they apply".
Second, you can group commands within curly braces {} (separated with a semi-colon ; or a new line)
Third, add the print flag p on the last substitution
Syntax:
sed -n -e '/^given_regexp/ {s/regexp1/replacement1/flags1;[...];s/regexp1/replacement1/flagsnp}'
Example (see Here document for more details):
Code:
sed -n -e '/^ha/ {s/h/k/g;s/a/e/gp}' <<SAMPLE
haha
hihi
SAMPLE
Result:
keke
sed -n '/.../!p'
There is no need for a substitution.