I want to delete lines from the file using sed command in linux but problem is that
if , I want to match "abcd" it matches "abcd.efgh" also . I just want to match not "abcd.efgh"
. I have searched on google but didnt find any solution .
With awk you can do:
awk '/abcd/ && !/abcd\.efgh/' file
It will match lines that are abcd but not abcd.efgh
Perhaps this is simply what you want:
sed -i '/^abcd$/d' file
It would only delete lines that give an exact match of it.
perl -lne 'print if(/\babcd\b/)' your_file
This might work for you (GUN sed):
sed '/abcd\.efgh/b;/abcd/d' file
or to put another way:
sed '/abcd/{/abcd\.efgh/!d}' file
Related
I have a project directory with folders containing .html files. I want to find those files which have the pattern -
'btn-primary.*{.*Save'
And replace the
'btn-primary' word with 'btn-primary Save'
only in those lines.
What I have done:
grep -rl -e 'btn-primary.*{Save' . |xargs sed -i 's/btn-primary/btn-primary Save/g'
What this did:
This found all files that have that pattern, that's okay. Then, sed ran on all of those files and replaced 'btn-primary' with 'btn-primary save' wherever it got - which is not what I want
What I want: to replace on those lines where there is 'Save' somewhere after 'btn-primary'.
Any help will be very much appreciated.
Regards,
Rahul
Why are you using grep at all? Sed does pattern matching:
sed -e 's/btn-primary\(.*{.*Save\)/btn-primary Save\1/g'
or:
sed -e 's/\(btn-primary\)\(.*{.*Save\)/\1 Save\2/g'
If you are using grep to try to trim down the number of files that sed will operate on, you're fooling yourself if you believe that is more efficient. By doing that, you will read every file that doesn't match only once, but every file that does match will be read twice. If you only use sed, every file will be read only once.
I'm looking for an in place command to change all file lines which end with :.:.
From
chr01 1453173 . C T 655.85 PASS . GT:AD:DP:PGT:PID 0/1:25,29:54:.:.
To
chr01 1453173 . C T 655.85 PASS . GT:AD:DP 0/1:25,29:54
In words, I'm basically deleting :PGT:PID and :.:. from any line ending with :.:.
With GNU sed and Solaris sed:
sed '/:\.:\.$/{s///;s/PGT:PID//;}' file
If you want to edit your file with GNU sed "in place" use option -i.
With awk that'd be:
awk 'sub(/:\.:\.$/,""){sub(/:PGT:PID/,"")} 1' file
chr01 1453173 . C T 655.85 PASS . GT:AD:DP 0/1:25,29:54
and for inplace editing with gawk you could add the -i inplace option while with any awk you can just add > tmp && mv tmp file.
You are looking for something like:
sed -i.bak "/^.*:\.:\.$/ {s/:PGT:PID//g; s/:\.:\.//g;}" file
it does inplace replace with file and creates a backup as file.bak
the /^.*:\.:\.$/ restricts the s command to lines ending in :.:. the . need quoting because they are special characters for regexes
the s sommand replaces the strings with the empty string
As a one-liner:
sed -i -e '/:\.:\.$/!n' -e 's///' -e 's/:PGT:PID//g' "$file"
Expanded:
/:\.:\.$/!n # leave lines untouched unless they end in ":.:."
s/// # replace the matched ":.:." with nothing
s/:PGT:PID//g # replace ":PGT:PID" with nothing, everywhere
We use the -i flag to perform in-place edits. We pass each line as a separate -e expression for portability: some sed implementations allow several commands to be concatenated with ;, but that is not required by the POSIX standard.
Have
08-01-12|07-30-13|08-09-32|12-43-56|
Want
08-01-12|07-30-13|08-09-32|12-43-56
I want to remove just the last |.
quick and dirty
sed 's/.$//' YourFile
a bit secure
sed 's/[|]$//' YourFile
allowing space
sed 's/[|][[:space:]]*$//' YourFile
same for only last char of last line (thansk #amelie for this comment) :
add a $in front so on quick and dirty it gives sed '$ s/.$//' YourFile
Since you tagged with awk, find here two approaches, one for every interpretation of your question:
If you want to remove | if it is the last character:
awk '{sub(/\|$/,"")}1' file
Equivalent to sed s'/|$//' file, only that escaping | because it has a special meaning in regex content ("or").
If you want to remove the last character, no matter what it is:
awk '{sub(/.$/,"")}1' file
Equivalent to sed s'/.$//' file, since . matches any character.
Test
$ cat a
08-01-12|07-30-13|08-09-32|12-43-56|
rrr.
$ awk '{sub(/\|$/,"")}1' a
08-01-12|07-30-13|08-09-32|12-43-56
rrr. # . is kept
$ awk '{sub(/.$/,"")}1' a
08-01-12|07-30-13|08-09-32|12-43-56
rrr # . is also removed
I wan't to replace a string like Europe12 with Europe12_yesturday in a file. Without changing the Europe12-36 strings that also exists in the file.
I tried:
$basename=Europe12
sed -i 's/\b$basename\b/${basename}_yesterday/g' file.txt
but this also changed the Europe12-36 strings.
Require a space or end of line character:
sed 's/Europe12\([ ]|$\)/Europe12_yesturday\1/g' input
Manually construct the delimiter list you want instead of using \b, \W or \<. - is not part of the word characters (alphanumericals), so that's why this also matches your other string. So try something like this, expanding the list as needed: [-a-zA-Z0-9].
You can do it in 2 times:
sed -e 's/Europe12/Europe12_yesturday/g' -e 's/Europe12_yesturday-36/Europe12-36/g' file.txt
sed 's/\(Europe12[[:blank:]]\)/\1_yesturday/g;s/Europe12$/&_yesturday/' YourFile
[[:blank:]] could be completeted with any boundary you accept also like .,;:/]) etc (be carrefull of regex meaning of this char in this case)
It is little late to reply..
It can be achieved easily by "word boundary" notation (\<..\>)
sed -i 's/\<$basename\>/${basename}_yesterday/g' file.txt
I'm trying to do the opposite of this question, replacing Unix line endings with Windows line endings, so that I can use SQL Server bcp over samba to import the file. I have sed installed but not dos2unix. I tried reversing the examples but to no avail.
Here's the command I'm using.
sed -e 's/\n/\r\n/g' myfile
I executed this and then ran od -c myfile, expecting to see \r\n where there used to be \n. But there all still \n. (Or at least they appear to be. The output of od overflows my screen buffer, so I don't get to see the beginning of the file).
I haven't been able to figure out what I'm doing wrong. Any suggestions?
When faced with this, I use a simple perl one-liner:
perl -pi -e 's/\n/\r\n/' filename
because sed behavior varies, and I know this works.
What is the problem with getting dos2unix onto the machine?
What is the platform you are working with?
Do you have GNU sed or regular non-GNU sed?
On Solaris, /usr/bin/sed requires:
sed 's/$/^M/'
where I entered the '^M' by typing controlV controlM. The '$' matches at the end of the line, and replaces the end of line with the control-M. You can script that, too.
Mechanisms expecting sed to expand '\r' or '\\r' to control-M are going to be platform-specific, at best.
You don't need the -e option.
$ matches the endline character. This sed command will insert a \r character before the end of line:
sed 's/$/\r/' myfile
Just adding a \r (aka ^M, see Jonathan Leffler's answer) in front of \n is not safe because the file might have mixed mode EOL, so then you risk ending up with some lines becomming \r\r\n. The safe thing to do is first remove all '\r' characters, and then insert (a single) \r before \n.
#!/bin/sh
sed 's/^M//g' ${1+"$#"} | sed 's/$/^M/'
Updated to use ^M.
sed 's/\([^^M]\)$/\0^M/' your_file
This makes sure you only insert a \r when there is no \r before \n. This worked for me.
Try using:
echo " this is output" > input
sed 's/$/\r/g' input |od -c
Maybe if you try it this way
cat myfile | sed 's/\n/\r\n/g' > myfile.win
will work, from my understanding your just making the replacements to the console output, you need to redirect output to a file, in this case myfile.win, then you could just rename it to whatever you want. The whole script would be (running inside a directory full of this kind of files):
#!/bin/bash
for file in $(find . -type f -name '*')
do
cat $file | sed 's/\n/\r\n/g' > $file.new
mv -f $file.new $file
done