How can I display all jobs that ended OK only?
When I try the command below, it shows both OK and NOTOK since both have "OK"
ctmpsm -listall application | grep OK 8 Answers
You need a more specific expression. Try grep " OK$" or grep "[0-9]* OK". You want to choose a pattern that matches what you want, but won't match what you don't want. That pattern will depend upon what your whole file contents might look like.
You can also do: grep -w "OK" which will only match a whole word "OK", such as "1 OK" but won't match "1OK" or "OKFINE".
$ cat test.txt | grep -w "OK" 1 OK 2 OK 4 OK This may work for you
grep -E '(^|\s)OK($|\s)' You can use the -v switch to exclude strings.
This will grab any line ending in OK and then remove any line that includes NOT:
(If OK doesn't appear at the end, just remove the $.)
cat test.txt | grep 'OK$' | grep -v 'NOT' This worked for me:
grep "\bsearch_word\b" text_file > output.txt ## \b indicates boundaries. This is much faster. or,
grep -w "search_word" text_file > output.txt Try this:
Alex Misuno@hp4530s ~ $ cat test.txt 1 OK 2 OK 3 NOTOK 4 OK 5 NOTOK Alex Misuno@hp4530s ~ $ cat test.txt | grep ".* OK$" 1 OK 2 OK 4 OK Try the below command, because it works perfectly:
grep -ow "yourstring" crosscheck:- Remove the instance of word from file, then re-execute this command and it should display empty result.
2try this:
grep -P '^(tomcat!?)' tst1.txt It will search for specific word in txt file. Here we are trying to search word tomcat
Recently I came across an issue in grep. I was trying to match the pattern x.y.z and grep returned x.y-z.Using some regular expression we may can overcome this, but with grep whole word matching did not help. Since the script I was writing is a generic one, I cannot restrict search for a specific way as in like x.y.z or x.y-z ..
Quick way I figured is to run a grep and then a condition check var="x.y.z" var1=grep -o x.y.z file.txt if [ $var1 == $var ] echo "Pattern match exact" else echo "Pattern does not match exact" fi
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