Insert line after match using sed - shell

For some reason I can't seem to find a straightforward answer to this and I'm on a bit of a time crunch at the moment. How would I go about inserting a choice line of text after the first line matching a specific string using the sed command. I have ...
CLIENTSCRIPT="foo"
CLIENTFILE="bar"
And I want insert a line after the CLIENTSCRIPT= line resulting in ...
CLIENTSCRIPT="foo"
CLIENTSCRIPT2="hello"
CLIENTFILE="bar"

Try doing this using GNU sed:
sed '/CLIENTSCRIPT="foo"/a CLIENTSCRIPT2="hello"' file
if you want to substitute in-place, use
sed -i '/CLIENTSCRIPT="foo"/a CLIENTSCRIPT2="hello"' file
Output
CLIENTSCRIPT="foo"
CLIENTSCRIPT2="hello"
CLIENTFILE="bar"
Doc
see sed doc and search \a (append)

Note the standard sed syntax (as in POSIX, so supported by all conforming sed implementations around (GNU, OS/X, BSD, Solaris...)):
sed '/CLIENTSCRIPT=/a\
CLIENTSCRIPT2="hello"' file
Or on one line:
sed -e '/CLIENTSCRIPT=/a\' -e 'CLIENTSCRIPT2="hello"' file
(-expressions (and the contents of -files) are joined with newlines to make up the sed script sed interprets).
The -i option for in-place editing is also a GNU extension, some other implementations (like FreeBSD's) support -i '' for that.
Alternatively, for portability, you can use perl instead:
perl -pi -e '$_ .= qq(CLIENTSCRIPT2="hello"\n) if /CLIENTSCRIPT=/' file
Or you could use ed or ex:
printf '%s\n' /CLIENTSCRIPT=/a 'CLIENTSCRIPT2="hello"' . w q | ex -s file

Sed command that works on MacOS (at least, OS 10) and Unix alike (ie. doesn't require gnu sed like Gilles' (currently accepted) one does):
sed -e '/CLIENTSCRIPT="foo"/a\'$'\n''CLIENTSCRIPT2="hello"' file
This works in bash and maybe other shells too that know the $'\n' evaluation quote style. Everything can be on one line and work in
older/POSIX sed commands. If there might be multiple lines matching the CLIENTSCRIPT="foo" (or your equivalent) and you wish to only add the extra line the first time, you can rework it as follows:
sed -e '/^ *CLIENTSCRIPT="foo"/b ins' -e b -e ':ins' -e 'a\'$'\n''CLIENTSCRIPT2="hello"' -e ': done' -e 'n;b done' file
(this creates a loop after the line insertion code that just cycles through the rest of the file, never getting back to the first sed command again).
You might notice I added a '^ *' to the matching pattern in case that line shows up in a comment, say, or is indented. Its not 100% perfect but covers some other situations likely to be common. Adjust as required...
These two solutions also get round the problem (for the generic solution to adding a line) that if your new inserted line contains unescaped backslashes or ampersands they will be interpreted by sed and likely not come out the same, just like the \n is - eg. \0 would be the first line matched. Especially handy if you're adding a line that comes from a variable where you'd otherwise have to escape everything first using ${var//} before, or another sed statement etc.
This solution is a little less messy in scripts (that quoting and \n is not easy to read though), when you don't want to put the replacement text for the a command at the start of a line if say, in a function with indented lines. I've taken advantage that $'\n' is evaluated to a newline by the shell, its not in regular '\n' single-quoted values.
Its getting long enough though that I think perl/even awk might win due to being more readable.

A POSIX compliant one using the s command:
sed '/CLIENTSCRIPT="foo"/s/.*/&\
CLIENTSCRIPT2="hello"/' file

Maybe a bit late to post an answer for this, but I found some of the above solutions a bit cumbersome.
I tried simple string replacement in sed and it worked:
sed 's/CLIENTSCRIPT="foo"/&\nCLIENTSCRIPT2="hello"/' file
& sign reflects the matched string, and then you add \n and the new line.
As mentioned, if you want to do it in-place:
sed -i 's/CLIENTSCRIPT="foo"/&\nCLIENTSCRIPT2="hello"/' file
Another thing. You can match using an expression:
sed -i 's/CLIENTSCRIPT=.*/&\nCLIENTSCRIPT2="hello"/' file
Hope this helps someone

The awk variant :
awk '1;/CLIENTSCRIPT=/{print "CLIENTSCRIPT2=\"hello\""}' file

I had a similar task, and was not able to get the above perl solution to work.
Here is my solution:
perl -i -pe "BEGIN{undef $/;} s/^\[mysqld\]$/[mysqld]\n\ncollation-server = utf8_unicode_ci\n/sgm" /etc/mysql/my.cnf
Explanation:
Uses a regular expression to search for a line in my /etc/mysql/my.cnf file that contained only [mysqld] and replaced it with
[mysqld]
collation-server = utf8_unicode_ci
effectively adding the collation-server = utf8_unicode_ci line after the line containing [mysqld].

I had to do this recently as well for both Mac and Linux OS's and after browsing through many posts and trying many things out, in my particular opinion I never got to where I wanted to which is: a simple enough to understand solution using well known and standard commands with simple patterns, one liner, portable, expandable to add in more constraints. Then I tried to looked at it with a different perspective, that's when I realized i could do without the "one liner" option if a "2-liner" met the rest of my criteria. At the end I came up with this solution I like that works in both Ubuntu and Mac which i wanted to share with everyone:
insertLine=$(( $(grep -n "foo" sample.txt | cut -f1 -d: | head -1) + 1 ))
sed -i -e "$insertLine"' i\'$'\n''bar'$'\n' sample.txt
In first command, grep looks for line numbers containing "foo", cut/head selects 1st occurrence, and the arithmetic op increments that first occurrence line number by 1 since I want to insert after the occurrence.
In second command, it's an in-place file edit, "i" for inserting: an ansi-c quoting new line, "bar", then another new line. The result is adding a new line containing "bar" after the "foo" line. Each of these 2 commands can be expanded to more complex operations and matching.

Related

Remove first two characters from a column in a text file excluding the headers

I want to remove the first two characters of a column in a text file.
I am using the below but this is also truncating the headers.
sed -i 's/^..//' file1.txt
Below is my file:
FileName,Age
./Acct_Bal_Tgt.txt,7229
./IDQ_HB1.txt,5367
./IDQ_HB_LOGC.txt,5367
./IDQ_HB.txt,5367
./IGC_IDQ.txt,5448
./JobSchedule.txt,3851
I want the ./ to be removed from each line in the file name.
Transferring comments to an answer, as requested.
Modify your script to:
sed -e '2,$s/^..//' file1.txt
The 2,$ prefix limits the change to lines 2 to the end of the file, leaving line 1 unchanged.
An alternative is to remove . and / as the first two characters on a line:
sed -e 's%^[.]/%%' file1.txt
I tend to use -e to specify that the script option follows; it isn't necessary unless you split the script over several arguments (so it isn't necessary here where there's just one argument for the script). You could use \. instead of [.]; I'm allergic to backslashes (as you would be if you ever spent time working out whether you needed 8 or 16 consecutive backslashes to get the right result in a troff document).
Advice: Don't use the -i option until you've got your script working correctly. It overwrites your file with the incorrect output just as happily as it will with the correct output. Consequently, if you're asking about how to write a sed script on SO, it isn't safe to be using the -i option. Also note that the -i option is non-standard and behaves differently with different versions of sed (when it is supported at all). Specifically, on macOS, the BSD sed requires a suffix specified; if you don't want a backup, you have to use two arguments: -i ''.
Use this Perl one-liner:
perl -pe 's{^[.]/}{}' file1.txt > output.txt
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
s{^[.]/}{} : Replace a literal dot ([.]) followed by a slash ('/'), found at the beginning of the line (^), with nothing (delete them). This does not modify the header since it does not match the regex.
If you prefer to modify the file in-place, you can use this:
perl -i.bak -pe 's{^[.]/}{}' file1.txt
This creates the backup file file1.txt.bak.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc perlrequick: Perl regular expressions quick start

sed substitute and show line number

I'm working in bash trying to use sed substitution on a file and show both the line number where the substitution occurred and the final version of the line. For a file with lines that contain foo, trying with
sed -n 's/foo/bar/gp' filename
will show me the lines where substitution occurred, but I can't figure out how to include the line number. If I try to use = as a flag to print the current line number like
sed -n 's/foo/bar/gp=' filename
I get
sed: -e expression #1, char 14: unknown option to `s'
I can accomplish the goal with awk like
awk '{if (sub("foo","bar",$0)){print NR $0}}' filename
but I'm curious if there's a way to do this with one line of sed. If possible I'd love to use a single sed statement without a pipe.
I can't think of a way to do it without listing the search pattern twice and using command grouping.
sed -n "/foo/{s/foo/bar/g;=;p;}" filename
EDIT: mklement0 helped me out there by mentioning that if the pattern space is empty, the default pattern space is the last one used, as mentioned in the manual. So you could get away with it like this:
sed -n "/foo/{s//bar/g;=;p;}" filename
Before that, I figured out a way not to repeat the pattern space, but it uses branches and labels. "In most cases," the docs specify, "use of these commands indicates that you are probably better off programming in something like awk or Perl. But occasionally one is committed to sticking with sed, and these commands can enable one to write quite convoluted scripts." [source]
sed -n "s/foo/bar/g;tp;b;:p;=;p" filename
This does the following:
s/foo/bar/g does your substitution.
tp will jump to :p iff a substitution happened.
b (branch with no label) will process the next line.
:p defines label p, which is the target for the tp command above.
= and p will print the line number and then the line.
End of script, so go back and process the next line.
See? Much less readable...and maybe a distant cousin of :(){ :|:& };:. :)
It cannot be done in any reasonable way with sed, here's how to really do it clearly and simply in awk:
awk 'sub(/foo/,"bar"){print NR, $0}' filename
sed is an excellent tool for simple substitutions on a single line, for anything else use awk.

How to replace all spaces in .txt file using SED in Cygwin

I have a huge .txt file that I want all spaces, line-breaks, indentations etc removed. It should literally be one long string.
I tried
sed -i 's/\ //g' test.txt
but nothing happens
sed -n "s/[[:blank:]]//g;H
$ {x;s/\n//g;p;}"
The H than $ are needed if you want to include New line due to fact that sed treat by default line by line (so no new line inside a line). The -n and p are needed to avoid double display with use of H
Seems to work ok for me:
[~/Desktop]
==> cat test.txt
the quick brown fox
[~/Desktop]
==> sed -i "s/\ //g" test.txt
[~/Desktop]
==> cat test.txt
thequickbrownfox
Sometimes using " " directly is hard and especially when you use double quotes (which involves that bash will interpret the string before passing it to sed).
sed -i -e 's/\s//g' file.txt
... should work (it works for me). "\s" means all whitespace characters, and with single quotes '', for bash not to interpret it before you passing it to sed.
While you use cygwin I think your OS is windows, then you don't need to use bash to implement your goal. Just open your txt file with the text editor, and replace the while space with nothing, then all of the whit space in you txt file will be removed.
This method can meet almost all kinds of removal. And also can apply in excel or word and so on.
Good luck!

Limiting SED to the first 10 characters of a line

I'm running sed as a part of a shell script to clean up bind logs for insertion into a database.
One of the sed commands is the following:
sed -i 's/-/:/g' $DPath/named.query.log
This turns out to be problematic as it disrupts any resource requests that also include a dash (I'm using : as a delimiter for an awk statement further down).
My question is how do I limit the sed command above to only the first ten characters of the line? I haven't seen a specific switch that does this, and I'm nowhere near good enough with RegEx to even start on developing one that works. I can't just use regex to match the preceding numbers because it's possible that the pattern could be part of a resource request. Heck, I can't even use pattern matching for ####-##-## because, again, it could be part of the resource.
Any ideas are much appreciated.
It's [almost always] simpler with awk:
awk '{target=substr($0,1,10); gsub(/-/,":",target); print target substr($0,11)}' file
I think the shortest solution, and perhaps the simplest, is provided by sed itself, rather than awk[ward]:
sed "h;s/-/:/g;G;s/\(..........\).*\n........../\1/"
Explanation:
(h) copy everything to the hold space
(s) do the substitution (to the entire pattern space)
(G) append the hold space, with a \n separator
(s) delete the characters up to the tenth after the \n, but keep the first ten.
Some test code:
echo "--------------------------------" > foo
sed -i "h;s/-/:/g;G;s/\(..........\).*\n........../\1/" foo
cat foo
::::::::::----------------------
I'm not sure how make sed do it per se, however, I do know that you can feed sed the first 10 characters then paste the rest back in, like so:
paste -d"\0" <(cut -c1-10 $DPath/named.query.log | sed 's/\-/:/g') <(cut -c11- $DPath/named.query.log)
You can do the following:
cut -c 1-10 $DPath/named.query.log | sed -i 's/-/:/g'
The cut statemnt takes only the first 10 chars of each line in that file. The output of that should be piped in a file. As of now it will just output to your terminal

How do I alter the n-th line in multiple files using SED?

I have a series of text files that I want to convert to markdown. I want to remove any leading spaces and add a hash sign to the first line in every file. If I run this:
sed -i.bak '1s/ *\(.*\)/\#\1/g' *.md
It alters the first line of the first file and processes them all, leaving the rest of the files unchanged.
What am I missing that will search and replace something on the n-th line of multiple files?
Using bash on OSX 10.7
The problem is that sed by default treats any number of files as a single stream, and thus line-number offsets are relative to the start of the first file.
For GNU sed, you can use the -s (--separate) flag to modify this behavior:
sed -s -i.bak '1s/^ */#/' *.md
...or, with non-GNU sed (including the one on Mac OS X), you can loop over the files and invoke once per each:
for f in *.md; do sed -i.bak '1s/^ */#/' "$f"; done
Note that the regex is a bit simplified here -- no need to match parts of the line that you aren't going to change.
XARgs will do the trick for you:
http://en.wikipedia.org/wiki/Xargs
Remove the *.md from the end of your sed command, then use XArgs to gather your files one at a time and send them to your sed command as a single entity, sorry I don't have time to work it out for you but the wikiPedia article should show you what you need to know.
sed -rsi.bak '1s/^/#/;s/^[ \t]+//' *.md
You don't need g(lobally) at the end of the command(s), because you wan't to replace something at the begin of line, and not multiple times.
You use two commands, one to modify line 1 (1s...), seperated from the second command for the leading blanks (and tabs? :=\t) with a semicolon. To remove blanks in the first line, switch the order:
sed -rsi.bak 's/^[ \t]+//;1s/^/#/' *.md
Remove the \t if you don't need it. Then you don't need a group either:
sed -rsi.bak 's/^ +//;1s/^/#/' *.md
-r is a flag to signal special treatment of regular expressions. You don't need to mask the plus in that case.

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