getting a substring from a regular expression

Question

Here is the line of text:

SRC='999'

where 999 can be any three digits.

I need a grep command that will return me the 999. How do I do this?

Answer 1

Platform grep or general regular expression?

Regex

SRC\=\'(\d{3})\'

Answer 2

why grep? How about..

substr("SRC='999'",6,3)

Answer 3

Are the lines to match always in the format SRC='nnn' ? Then you could use

grep SRC | cut -d"'" -f2

Answer 4

You can't do it with plain grep. As the man page on my box states: "grep - print lines matching a pattern" grep only prints lines, not part of lines.

I would recommend awk since it can do both the pattern matching and sub-line extracting:

awk -F\' ' /SRC/ {print $2}'

Answer 5

Here is how to do it using sed

grep SRC=\'.*\' | sed 's/SRC=.\(.*\)./\1/'

Answer 6

depends on your platform / language.

in Ruby:

string = "SRC = '999'"

string.match(/([0-9]{3})/).to_s.to_i

will return an integer

Answer 7

just sed will do

$ echo SRC='999' | sed '/SRC/s/SRC=//'
999

Answer 8

You can use the -o option on grep to return only the part of the string that matches the regex:

echo "SRC='999'" | grep -o -E '[0-9]{3}'