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I have a string of the format [0-9]+\.[0-9]+\.[0-9]. I need to extract the first, second, and third numbers separately. As I understand it, capture groups should be capable of this. I should be able to use sed "s/\([0-9]*\)/\1/g to get the first number, sed "s/\([0-9]*\)/\2/g to get the second number, and sed "s/\([0-9]*\)/\3/g to get the third number. In each case, though, I am getting the whole string. Why is this happening?

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Capture groups capture the whole group...not individual elements in the group. You need something like 's/\([0-9]\)\([0-9]\)\([0-9]\).*/\1\2\3/' to capture individual numbers. – Munir Feb 16 at 18:10

We can't give you a full answer without an example of your input but I can tell you that your understanding of capture groups is wrong. You don't use the sequentially, they only refer to the regex on the left hand side of the same substitution operator. If you capture, for example, /(foo)(bar)(baz)/, then foo will be \1, bar will be \2 and baz will be \3. You can't do s/(foo)/\1/; s/(bar)/\2/, because, in the second s/// call, there is only one captured group, so \2 will not be defined.

So, to capture your three groups of digits, you would need to do:

sed 's/\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)/\1 : \2 : \3/'

Or, the more readable:

sed -E 's/([0-9]*)\.([0-9]*)\.([0-9]*)/\1 : \2 : \3/'
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example --

echo "123.456.78" |sed 's/\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)/\1/'

$ echo "123.456.78" |sed 's/\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)/\2/'

$ echo "123.456.78" |sed 's/\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)/\3/'

Or, all together:

$ echo "123.456.78" |sed 's/\([0-9]*\)\.\([0-9]*\)\.\([0-9]*\)/\1 : \2 : \3/'
123 : 456 : 78
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