I have the following strings in bash
str1="any string"
str2="any"
I want to check if str2 is a substring of str1
I can do it in this way:
c=`echo $str1 | grep $str2`
if [ $c != "" ]; then
...
fi
Is there a more efficient way of doing this?
You can use wild-card expansion *.
str1="any string"
str2="any"
if [[ "$str1" == *"$str2"* ]]
then
echo "str2 found in str1"
fi
Note that * expansion will not work with single [ ].
[[]] over [] was making quoting unnecessary...
str1="any string"
str2="any"
Old school (Bourne shell style):
case "$str1" in *$str2*)
echo found it
esac
New school (as speakr shows), however be warned that the string to the right will be viewed as a regular expression:
if [[ $str1 =~ $str2 ]] ; then
echo found it
fi
But this will work too, even if you're not exactly expecting it:
str2='.*[trs].*'
if [[ $str1 =~ $str2 ]] ; then
echo found it
fi
Using grep is slow, since it spawns a separate process.
case expression. But that's not old-school, it's the current state-of-the-art since it's as POSIXly portable as it gets. BTW, no need to quote $str1; there's no word splitting done after a case.
case way of looking for substrings works with any Bourne shell, so is state-of-the-art and to be unconditionally preferred. Anyone using bashisms like [[ ]] is in need for a portability lesson. :-)You can use bash regexp matching without using grep:
if [[ $str1 =~ $str2 ]]; then
...
fi
Note that you don't need any surrounding slashes or quotes for the regexp pattern. If you want to use glob pattern matching just use == instead of =~ as operator.
Some examples can be found here.
if echo $str1 | grep -q $str2 #any command
then
.....
fi
