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$echo "foo 65 bar" | sed -n -e 's/.*\([0-9]\+\).*/\1/p'
5

Why is the output not 65? Shouldn't sed greedily match the [0-9]\+ part? How do I tell sed to match all of 65?

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1 Answer 1

up vote 9 down vote accepted

The .* is greedy first -- it's matching foo 6. The only reason it stops there is because matching any further would stop the whole pattern from matching, so it leaves the 5 for the ([0-9]+). If you made it ([0-9]*) instead the .* would match the whole line and you'd get nothing in your group. One way around it is to tell the first part not to match numbers:

$ echo "foo 65 bar" | sed -n -e 's/[^0-9]*\([0-9]\+\).*/\1/p'
65
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