I'm trying to get grep/sed out the following output: "name":"test_backup_1" from the below response
{"backups":[{"name":"test_backup_1","status":"CORRUPTED","creationTime":"2019-11-08T15:03:49.460","id":"test_backup_1"}]}
I have been trying variations of the following grep -Eo 'name:"\w+\"' but no joy.
I'm not sure if it would be easier to achieve this using grep or sed?
The way I am running this is curling a response from the server and saving it to a local variable, then echo out the variable and pipe grep/sed
example of what I am running
echo ${view_backup} | grep -Eo '"name":"\w+\"'
Referencing #sundeep answer
grep -Eo '"name":"[^"]+"'
resulted in the expected output
Make sure to transform the file to one line before grep
and pipe from your curl
echo `curl --silent https://someurl | tr -d '\n' | grep -oP "(?<=name\":\")[^\"]+"`
will return
test_backup_1
If you want more variables you can chain the -oP grep like in this example where I get some data on a danish license plate (bt419329)
curl --silent https://www.tjekbil.dk/api/v2/nummerplade/bt41932 | grep -oP -m 1 "(?<=\"RegNr\":\")[^\"]+|(?<=\"MaerkeTypeNavn\":\")[^\"]+|(?<=\"MaksimumHastighed\":)[^,]+"| tr '\n' ' '
returns
BT41932 SKODA 218
Related
I'm trying to fetch the first line in a log file which contain a date.
Here is an example of the log file :
SOME
LOG
2021-1-1 21:50:19.0|LOG|DESC1
2021-1-4 21:50:19.0|LOG|DESC2
2021-1-5 21:50:19.0|LOG|DESC3
2021-1-5 21:50:19.0|LOG|DESC4
In this context I need to get the following line:
2021-1-1 21:50:19.0|LOG|DESC1
An other log file example :
SOME
LOG
21-1-3 21:50:19.0|LOG|DESC1
21-1-3 21:50:19.0|LOG|DESC2
21-1-4 21:50:19.0|LOG|DESC3
21-1-5 21:50:19.0|LOG|DESC4
I need to fetch :
21-1-3 21:50:19.0|LOG|DESC1
At the moment I tried the following command :
cat /path/to/file | grep "$(date +"%Y-%m-%d")" | tail -1
cat /path/to/file | grep "$(date +"%-Y-%-m-%-d")" | tail -1
cat /path/to/file | grep -E "[0-9]+-[0-9]+-[0-9]" | tail -1
In case you are ok with awk, could you please try following. This will find the matched regex first line and exit from program, which will be faster since its NOT reading whole Input_file.
awk '
/^[0-9]{2}([0-9]{2})?-[0-9]{1,2}-[0-9]{1,2} [0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]+/{
print
exit
}' Input_file
Using sed, being not too concerned about exactly how many digits are present:
sed -En '/^[0-9]+-[0-9]+-[0-9]+ [0-9]+:[0-9]+:[0-9]+[.][0-9]+[|]/ {p; q}' file
$ grep -m1 '^[0-9]' file1
2021-1-1 21:50:19.0|LOG|DESC1
$ grep -m1 '^[0-9]' file2
21-1-3 21:50:19.0|LOG|DESC1
If that's not all you need then edit your question to provide more truly representative sample input/output.
A simple grep with -m 1 (to exit after finding first match):
grep -m1 -E '^([0-9]+-){2}[0-9]+ ([0-9]{2}:){2}[0-9]+\.[0-9]+' file1
2021-1-1 21:50:19.0|LOG|DESC1
grep -m1 -E '^([0-9]+-){2}[0-9]+ ([0-9]{2}:){2}[0-9]+\.[0-9]+' file2
21-1-3 21:50:19.0|LOG|DESC1
This sed works with either GNU or POSIX sed:
sed -nE '/^[[:digit:]]{2,4}-[[:digit:]]{1,2}-[[:digit:]]{1,2}/{p;q;}' file
But awk, with the same BRE, is probably better:
awk '/^[[:digit:]]{2,4}-[[:digit:]]{1,2}-[[:digit:]]{1,2}/{print; exit}' file
echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" | grep -oE "([^\/]+$)"
This prints just the filename, without the directory structure, but I cannot manage to print just mainbinary from that string. Suggestions?
And a sed alternative to PS.'s great grep -oP
echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" |sed -r 's#^.*/([^-]+).*#\1#'
mainbinary
echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" |grep -oP '.*/\K[^-]+'
mainbinary
This will scan till last / and ignore everything to its left and keep moving until - (excluding)
With any awk in any shell on any UNIX machine:
$ echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" | awk -F'[/-]' '{print $3}'
mainbinary
How do I grep strings in between nested brackets using bash? Is it possible without the use of loops? For example, if I have a string like:
[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]
I wish to grep only the two target strings inside the [[]]:
TargetString1
TargetString2
I tried the following command which cannot get TargetString2
grep -o -P '(?<=\[\[).*(?=\]\])'|cut -d ':' -f1
With GNU's grep P option:
grep -oP "(?<=\[\[)[\w\s]+"
The regex will match a sequence of word characters (\w+) when followed by two brackets ([[). This works for your sample string, but will not work for more complicated constructs like:
[[[[TargetString1]]TargetString2:SomethingIDontWantAfterColon[[TargetString3]]]]
where only TargetString1 and TargetString3 are matched.
To extract from nested [[]] brackets, you can use sed
#!/bin/bash
str="[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]"
echo $str | grep -o -P '(?<=\[\[).*(?=\]\])'|cut -d ':' -f1
echo $str | sed 's/.*\[\([^]]*\)\].*/\1/g' #which works only if string exsit between []
Output:
TargetString1
TargetString2
You can use grep regex grep -Eo '\[\[\w+' | sed 's/\[\[//g' for doing this
[root#localhost ~]# echo "[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]" | grep -Eo '\[\[\w+' | sed 's/\[\[//g'
TargetString1
TargetString2
[root#localhost ~]#
I have some code I'm using with Windows and SED to give me the first set of eight characters in a file name that keeps giving me the second set only that I cannot figure out what I'm doing wrong.
My Code:
echo JiggySauce_20161208_21325005_Meat.txt | sed -r "s/.*_([0-9]*)_.*/\1/g"
Addition Example (so regex per underbar delimiters won't always work):
echo JiggySauce_Mustard_Mayo_20161208_21325005_Meat.txt | sed -r "s/.*_([0-9]*)_.*/\1/g"
I keep getting this wrong result (at least not what I need):
21325005
My expected result:
20161208
I could even live with (preferrably not but could work with that I suppose):
20161208_21325005
Please help me with this if you have an answer as I'm at a standstill looking dumb and stumped over here like UHHH....
With GNU sed:
echo JiggySauce_20161208_21325005_Meat.txt | sed -r 's/^[^_]*_([^_]*).*/\1/'
Output:
20161208
Post Initial Answer Update:
I suggest: sed -r 's/[^0-9]*([0-9]{8}).*/\1/'
Cyrus
Output:
20161208
See: The Stack Overflow Regular Expressions FAQ
Using grep:
echo JiggySauce_20161208_21325005_Meat.txt | grep -Eo '[0-9]+' | head -1
or
echo JiggySauce_20161208_21325005_Meat.txt | tr '_' '\n' | grep -m1 -Eo '[0-9]+'
I'm trying to detect Russian characters with grep, but what I have at the moment does not appear to be doing anything:
echo "Ёё" | grep -Eo "/[А-Яа-яЁё]/u"
No output is returned. Is there anything I have to do to tell grep to return the output?
there is no output because grep is looking for pattern /yourletters/u
try this:
echo "Ёё" | grep -Eo "[А-Яа-яЁё]*"
test here:
kent$ echo "Ёё" | grep -Eo "[А-Яа-яЁё]*"
Ёё