Finding a particular string from a file

Question

I have a file that contains one particular string many times. How can I print all occurrences of the string on the screen along with letters following that word till the next space is encountered.

Suppose a line contained:

example="123asdfadf" foo=bar

I want to print example="123asdfadf".

I had tried using less filename | grep -i "example=*" but it was printing the complete lines in which example appeared.

please provide a minimal reproducible example together with what you tried so far — fedorqui
– fedorqui, Commented Aug 26, 2016 at 13:13
@Jean-FrançoisFabre -- not working, it's returning nothing. Though, Can you please explain this, I can try with it. Edit : Sorry it worked too. I had to use "-i" to include all cases, my bad. — mb47
– mb47, Commented Aug 26, 2016 at 13:21
@fedorqui thanks for pointing out. Edited the question with what I tried. — mb47
– mb47, Commented Aug 26, 2016 at 13:28

James Brown · Accepted Answer · 2016-08-26 13:15:36Z

3

$ grep -o "example[^ ]*" foo example="abc" example="123asdfadf"

answered Aug 26, 2016 at 13:15

James Brown

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2 Comments

mb47 Over a year ago

Thank you, this worked. Used -i along with -o to include all cases. Thanks a lot.

agc Over a year ago

@mb47, please check the green arrow to mark this answered.

hek2mgl · Accepted Answer · 2016-08-26 17:12:34Z

Since -o is only supported by GNU grep, a portable solution would be to use sed:

sed -n 's/.*\(example=[^[:space:]]*\).*/\1/p' file

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