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I have following string:
{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234} and I need to get value of "scheme version", which is 1234 in this example.
I have tried
grep -Eo "\"scheme_version\":(\w*)" however it returns
"scheme_version":1234 How can I make it? I know I can add sed call, but I would prefer to do it with single grep.
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Capturing groups are a way to treat multiple characters as a single unit. They are created by placing the characters to be grouped inside a set of parentheses. For example, the regular expression (dog) creates a single group containing the letters "d" "o" and "g" .
Non-capturing groups are important constructs within Java Regular Expressions. They create a sub-pattern that functions as a single unit but does not save the matched character sequence. In this tutorial, we'll explore how to use non-capturing groups in Java Regular Expressions.
Mixing named and numbered capturing groups is not recommended because flavors are inconsistent in how the groups are numbered. If a group doesn't need to have a name, make it non-capturing using the (?:group) syntax. In . NET you can make all unnamed groups non-capturing by setting RegexOptions.
A part of a pattern can be enclosed in parentheses (...) . This is called a “capturing group”. That has two effects: It allows to get a part of the match as a separate item in the result array.
You'll need to use a look behind assertion so that it isn't included in the match:
grep -Po '(?<=scheme_version":)[0-9]+'
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