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Plus the syntax is simpler (so it may be more quickly understood by many readers/maintainers). For example with jq, the following command would extract the current summary : jq '.currently.Since you don't know much about regex, I suggest you instead learn to use a JSON parser, which would be more appropriate for this task. With that way, it's clear your goal is to select lines that contain one thing but not another. You use grep's -o flag, which makes it output only the matched part. In other words match foo11, foo12, foo22 and so on, enter: grep 'foo 0-9 0-9' filename. grep -r -l string1 > tmp while read p do grep -l string2 p done < tmp rm tmp. In this example match two numeric digits. If you want to display all lines that contain a sequence of four digits that is itself not part of any longer sequence of digits, one way is: grep -P '(?the above command prints nothing and the reason I believe is it is not able to match anything. echo 'This is 02G05 a test string 2' sed -n '/\d+G\d+/p'. For that I tried the following regex with sed. use \s\S (match spaces and non-spaces) instead. What system are you on More generally, what you're asking for is more a job for awk or sed. ![]() Only some grep implementations (like pcregrep or GNU grep) have provision to print parts of the matching lines as an extension. ![]() In those languages, you can use a character class such as \s\S to match any character.' Instead of the. grep is the command to print the lines that match a pattern. Now from the above string I want to extract 02G05. From that link: 'JavaScript and VBScript do not have an option to make the dot match line break characters. that contain a sequence of four digits that is itself not part of any longer sequence of digits, or My example string is as follows: This is 02G05 a test string 2.There are two ways to interpret this question I'll address both cases. The -E flag in grep activates regex (which is what egrep does by default), and the -o flag prints only the matching string.
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