using sed to parse log to eliminate lines by regex - sed

I would like to be able to clean up a log file using sed and regex
a) any line which matches /\d{13,13},\d*,[\w\s]+,.+\n/ I would like to delete
b) remove any empty lines that are remaining.
How can I do this ?

sed -re '/\d{13,13},\d*,[\w\s]+,.+$/d' -e '/^$/d' logfile

Related

sed remove line if neither pattern provided don't match

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).

sed replace 1 line in file with all lines in file

Lets say I have a line #SYM
I need to replace it with all lines from file1.txt
Is it possible to do that with sed?
I have tried sed 's/#SYM/file1.txt/' updater
But that doesn't work, because I need to load file1.txt as string, and I do not know how to do that.
EDIT: I believe that there could be a way to do it in a shell script somehow.
EDIT2: I also just tried this:
#!/bin/bash
value=$(<tools/symlink)
sed -i 's/#SYM/$value/' META-INF/com/google/android/updater-script
Use r command:
sed -e '/#SYM/ {r tools/symlink' -e 'd}' META-INF/com/google/android/updater-script
/#SYM/ {r tools/symlink if a line contains #SYM, append the contents of tools/symlink
d} then delete the matching line
the two commands are separated using -e option because everything after r is considered as part of filename
Add the -i option once you are satisifed that it is working

sed: remove matching line and everything after

I'm trying to remove all lines after the first blank line in a file with a git filter using sed.
This seems to remove everything after the blank line
sed -i '/^$/q' test.rpt
How do I also include the blank line itself to be deleted?
If this is GNU sed, just use Q instead of q.
sed -i '/^$/Q' test.rpt
For BSD sed, use -n switch to suppress automatic printing, and print lines manually. E.g:
sed -n -i '/^$/q;p' test.rpt
PS: You might want to change the regex to ^[[:blank:]]*$ to regard lines of all blank characters as blank lines as well.
Try this:-
sed -i '/^$/,$ d' inputfile

Replacing the test with sed

I'm trying to replace the text using the sed, but it's showing some error. Not getting where I'm getting wrong.
sed -i 's/process.env.REDIRECT_URI/http:\/\/test-domain.apps.io/\callback/g' input.txt
Have this :
process.env.REDIRECT_URI
Replace this with :
http://test-domain.apps.io
Try:
sed -i 's/process.env.REDIRECT_URI/http:\/\/test-domain.apps.io/g' input.txt
Notes:
The original command has a spurious string /\callback. All that was needed to make the code work was to remove it.
. is a wildcard. If you want to be sure that you are matching periods, they should be escaped:
sed -i 's/process\.env\.REDIRECT_URI/http:\/\/test-domain.apps.io/g' input.txt
Sometimes, its clearer if one doesn't have to escape /. One can use a separator of one's choice. For example, use #:
sed -i 's#process\.env\.REDIRECT_URI#http://test-domain.apps.io#g' input.txt
If you did want /callback in the output, use:
sed -i 's/process\.env\.REDIRECT_URI/http:\/\/test-domain.apps.io\/callback/g' input.txt
or:
sed -i 's#process\.env\.REDIRECT_URI#http://test-domain.apps.io/callback#g' input.txt

Sed Remove 3 last digits from string

27211;18:05:03479;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
Second string after ; is time. gg:mm:sssss:. I just want to be gg:mm:ss:
Like so:
27211;18:05:03;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
I tried with cut but it deletes everything after n'th occurance of character, and for now I am stuck, please help.
give this one liner a try:
awk -F';' -v OFS=";" 'sub(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file