Ashish Singh Ashish Singh - 1 year ago 54
Bash Question

Same pattern in grep and sed select different values

I want to create a pattern for sed, which will find out 'type=""
For this I tried to use the pattern


echo 'aa type="none" stretchChildren="first"' | sed s/'type=".*\?"'/hello/

Above is the sed command which prints

aa hello

Which means it selects 'type="none" stretchChildren="first"' for 'type=".*\?"'

Now below is the grep command using same pattern on same string

echo 'aa type="none" stretchChildren="first"' | grep -oP 'type=".*?"'

It gives output


Don't know what I am missing in sed pattern

Can some one help me out here
Output of sed should be

aa hello stretchChildren="first"

Answer Source

sed doesn't have non-greedy pattern matching, so using *? or *\? won't work.

If you want to have the same output as grep then use a grouping without the " - [^"]+ instead of ".*?":

sed -r 's/type="[^"]+"/hello/'

[, ] is a group of characters, ^ is a negation, so [^"] means any character that is not a ".

For OSX use -E instead of -r. (-E also works on latest GNU sed, but it is not documented in --help nor in man sed so I don't recommend it)

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