How to extract substring from string

Question

I have a string, that does not always look the same, and from this string I want to extract some information if it exists. The string might look like one of the following:

myCommand -s 12 moreParameters
myCommand -s12 moreParamaters
myCommand -s moreParameters

I want to get the number, i.e. 12 in this case, if it exists. How can I accomplish that?

Many thanks!

EDIT: There is a fourth possible value for the string:

myCommand moreParameters

How can I modify the regex to cover this case as well?

Fredrik Pihl · Accepted Answer · 2011-09-14 20:21:45Z

2

$ a="myCommand -s 12 moreParameters"
$ b="myCommand -s12 moreParamaters"
$ echo $(expr "$a" : '[^0-9]*\([0-9]*\)')
12
$ echo $(expr "$b" : '[^0-9]*\([0-9]*\)')
12

answered Sep 14, 2011 at 20:21

Fredrik Pihl

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Liz · Accepted Answer · 2011-09-15 10:20:45Z

1

Try this:

n=$(echo "$string"|sed 's/^.*-s *\([0-9]*\).*$/\1/')

This will match a -s eventually followed by spaces and digits; and replace the whole string by the digits.

myCommand -s 12 moreParameters => 12
myCommand -s12 moreParamaters => 12
myCommand -s moreParameters => empty string

EDIT: There is a fourth possible value for the string:

myCommand moreParameters

How can I modify the regex to cover this case as well?

edited Sep 15, 2011 at 10:20

Liz

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answered Sep 14, 2011 at 20:15

Arnaud Le Blanc

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3 Comments

Jill Over a year ago

Thank you so much for a quick response. Much appreciated!

glenn jackman Over a year ago

in bash, you can use "here-strings": n=$(sed 's/^.*-s *$[0-9]*$.*$/\1/' <<< "$string")

tripleee Over a year ago

You can handle the additional case e.g. by only printing matching lines. sed -n 's/^.*-s *$[0-9]*$.*$/\1/p'

bash-o-logist · Accepted Answer · 2011-09-15 11:22:09Z

1

You can do all these without the need for external tools

$ shopt -s extglob
$ string="myCommand -s 12 moreParameters"
$ string="${string##*-s+( )}"
$ echo "${string%% *}"
12

answered Sep 15, 2011 at 11:22

bash-o-logist

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