I currently have this sed command that replaces foo.us.param=value with foo.param=value:
sed -i -e 's/\.us\./\./' file.txt
I also need it to delete any lines that contain .eu. anywhere but leave all other lines untouched. Any help would save me a long time trying to figure this out alone and would be greatly appreciated.
sed -i -e 's/\.us\./\./' -e '/\.eu\./d' file.txt
instead of sed you could also use grep
grep -v '\.eu\.'
Related
I have a bunch of txt-files containing stuff like this:
text_i_need_to_remove{text_i_need_to_retain}
text_i need_to_remove{text_i_need_to_retain}
...
How do I remove text before curly braces (and curly braces themselves) and retain just only text_i_need_to_retain?
Deleting everything upto { or } at end of line
:%s/.*{\|}$//g
From bash shell, you can use text processing tools like sed and awk. Assume file is named ip.txt
1) With sed, which is pretty similar to regex we used inside vim. The -i flag allows to make change in place, i.e it modifies the input file itself.
$ sed -i 's/.*{\|}$//g' ip.txt
2) With awk, one can again use substitution or in this case, split the line on curly brackets and use only the second column.
$ awk -F'{|}' '{print $2}' ip.txt > tmp && mv tmp ip.txt
If you have GNU awk, there is -i inplace option for in place editing
$ gawk -i inplace -F'{|}' '{print $2}' ip.txt
To make changed to all files in current directory, use
sed -i 's/.*{\|}$//g' *
Or if they have common extension, say .txt, use
sed -i 's/.*{\|}$//g' *.txt
:%s/^.*{\(.*\)}$/\1/ or in bash, sed 's/^.*{\(.*\)}$/\1/ foo.txt
\(.*\) is a control group which feeds into \1 and looks like a lumbering zombie.
you can use this in vim;
:%s/^.*{// | %s/}$//
you can also use this script; first run this, if everythink is ok, uncomment sed with -i option as below;
#!/bin/bash
for item in $(ls /dir/where/my/files/are)
do
sed -i 's/^.*{//;s/}$//' /dir/where/my/files/are/$item
done
sed -i ; inplace replace
or
Only use as below;
sed -i 's/^.*{//;s/}$//' /dir/where/my/files/are/*
Perl can be used to do the substitution on all files:
perl -i -pe 's/.*{|}$//g' *.txt
My script gets this string for example:
/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file
let's say I don't know how long the string until the /importance.
I want a new variable that will keep only the /importance/lib1/lib2/lib3/file from the full string.
I tried to use sed 's/.*importance//' but it's giving me the path without the importance....
Here is the command in my code:
find <main_path> -name file | sed 's/.*importance//
I am not familiar with the regex, so I need your help please :)
Sorry my friends I have just wrong about my question,
I don't need the output /importance/lib1/lib2/lib3/file but /importance/lib1/lib2/lib3 with no /file in the output.
Can you help me?
I would use awk:
$ echo "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file" | awk -F"/importance/" '{print FS$2}'
importance/lib1/lib2/lib3/file
Which is the same as:
$ awk -F"/importance/" '{print FS$2}' <<< "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file"
importance/lib1/lib2/lib3/file
That is, we set the field separator to /importance/, so that the first field is what comes before it and the 2nd one is what comes after. To print /importance/ itself, we use FS!
All together, and to save it into a variable, use:
var=$(find <main_path> -name file | awk -F"/importance/" '{print FS$2}')
Update
I don't need the output /importance/lib1/lib2/lib3/file but
/importance/lib1/lib2/lib3 with no /file in the output.
Then you can use something like dirname to get the path without the name itself:
$ dirname $(awk -F"/importance/" '{print FS$2}' <<< "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file")
/importance/lib1/lib2/lib3
Instead of substituting all until importance with nothing, replace with /importance:
~$ echo $var
/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file
~$ sed 's:.*importance:/importance:' <<< $var
/importance/lib1/lib2/lib3/file
As noted by #lurker, if importance can be in some dir, you could add /s to be safe:
~$ sed 's:.*/importance/:/importance/:' <<< "/dir1/dirimportance/importancedir/..../importance/lib1/lib2/lib3/file"
/importance/lib1/lib2/lib3/file
With GNU sed:
echo '/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file' | sed -E 's#.*(/importance.*)#\1#'
Output:
/importance/lib1/lib2/lib3/file
pure bash
kent$ a="/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file"
kent$ echo ${a/*\/importance/\/importance}
/importance/lib1/lib2/lib3/file
external tool: grep
kent$ grep -o '/importance/.*' <<<$a
/importance/lib1/lib2/lib3/file
I tried to use sed 's/.*importance//' but it's giving me the path without the importance....
You were very close. All you had to do was substitute back in importance:
sed 's/.*importance/importance/'
However, I would use Bash's built in pattern expansion. It's much more efficient and faster.
The pattern expansion ${foo##pattern} says to take the shell variable ${foo} and remove the largest matching glob pattern from the left side of the shell variable:
file_name="/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file"
file_name=${file_name##*importance}
Removeing the /file at the end as you ask:
echo '<path>' | sed -r 's#.*(/importance.*)/[^/]*#\1#'
Input /dir1/dir2/dir3.../importance/lib1/lib2/lib3/file
Returns: /importance/lib1/lib2/lib3
See this "Match groups" tutorial.
I'd like to use sed to remove all characters between "foo=1&" and "bar=2&" for all occurrences in an xml file.
<url>http://example.com?addr=123&foo=1&s=alkjasldkffjskdk$bar=2&f=jkdng</url>
<url>http://example.com?addr=124&foo=1&k=d93ndkdisnskiisndjdjdj$bar=2&p=dnsks</url>
Here is my sed command:
sed -e '/foo=1&/,/bar=2&/d' sample.xml
When I run this, the file is unchanged.
The above is based on this example: Find "string1" and delete between that and "string2"
Use the substitution command instead of the delete command:
sed -e 's/\(foo=1&\).*\(bar=2&\)/\1\2/'
You should use
sed -i -e 's/\(foo=1&\).*\(bar=2&\)/\1\2/' your_html.xml
I'm trying to write a sed command to remove a specific string followed by two digits. So far I have:
sed -e 's/bizzbuzz\([0-9][0-9]\)//' file.txt
but I cant seem to get the syntax right. Any suggestions?
sed -re 's/bizzbuzz[0-9]{2}//' file.txt
and
sed -re 's/\bbizzbuzz[0-9]{2}\b//' file.txt
if the searched string have word boundary
sed -e 's/bizzbuzz[0-9]\{2\}//' file.txt
if you don't have GNU sed
Your current approach seems like it should work fine:
$ echo 'FOO bizzbuzz56 BAR' | sed -e 's/bizzbuzz\([0-9][0-9]\)//'
FOO BAR
As said in other answer, the syntax seems to be fine (with unnecesary parenthesis).
But may be you want to replace all the strings found in each line ? In that case, you should add a 'g' at the end of the 's' command:
sed -e 's/bizzbuzz\([0-9][0-9]\)//g' file.txt
I have a big apache log file and I need to filter that and leave only (in a new file) the log from a certain IP: 192.168.1.102
I try using this command:
sed -e "/^192.168.1.102/d" < input.txt > output.txt
But "/d" removes those entries, and I needt to leave them.
Thanks.
What about using grep?
cat input.txt | grep -e "^192.168.1.102" > output.txt
EDIT: As noted in the comments below, escaping the dots in the regex is necessary to make it correct. Escaping in the regex is done with backslashes:
cat input.txt | grep -e "^192\.168\.1\.102" > output.txt
sed -n 's/^192\.168\.1\.102/&/p'
sed is faster than grep on my machines
I think using grep is the best solution but if you want to use sed you can do it like this:
sed -e '/^192\.168\.1\.102/b' -e 'd'
The b command will skip all following commands if the regex matches and the d command will thus delete the lines for which the regex did not match.