How to determine if a string is a substring of another in bash?

Question

I want to see if a string is inside a portion of another string.
e.g.:

'ab' in 'abc' -> true
'ab' in 'bcd' -> false

How can I do this in a conditional of a bash script?

Answer 1

[[ "bcd" =~ "ab" ]]
[[ "abc" =~ "ab" ]]

the brackets are for the test, and as it is double brackets, it can so some extra tests like =~.

So you could use this form something like

var1="ab"
var2="bcd"
if [[ "$var2" =~ "$var1" ]]; then
    echo "pass"
else
    echo "fail"
fi

Edit: corrected "=~", had flipped.

Answer 2

You can use the form ${VAR/subs} where VAR contains the bigger string and subs is the substring your are trying to find:

my_string=abc
substring=ab
if [ "${my_string/$substring}" = "$my_string" ] ; then
  echo "${substring} is not in ${my_string}"
else
  echo "${substring} was found in ${my_string}"
fi

This works because ${VAR/subs} is equal to $VAR but with the first occurrence of the string subs removed, in particular if $VAR does not contains the word subs it won't be modified.

Answer 3

Using bash filename patterns (aka "glob" patterns)

substr=ab
[[ abc == *"$substr"* ]] && echo yes || echo no    # yes
[[ bcd == *"$substr"* ]] && echo yes || echo no    # no

Answer 4

The following two approaches will work on any POSIX-compatible environment, not just in bash:

substr=ab
for s in abc bcd; do
    if case ${s} in *"${substr}"*) true;; *) false;; esac; then
        printf %s\\n "'${s}' contains '${substr}'"
    else
        printf %s\\n "'${s}' does not contain '${substr}'"
    fi
done

substr=ab
for s in abc bcd; do
    if printf %s\\n "${s}" | grep -qF "${substr}"; then
        printf %s\\n "'${s}' contains '${substr}'"
    else
        printf %s\\n "'${s}' does not contain '${substr}'"
    fi
done

Both of the above output:

'abc' contains 'ab'
'bcd' does not contain 'ab'

The former has the advantage of not spawning a separate grep process.

Note that I use printf %s\\n "${foo}" instead of echo "${foo}" because echo might mangle ${foo} if it contains backslashes.

Answer 5

shell case statement

This is the most portable solution, will work even on old Bourne shells and Korn shell

#!/bin/bash
case "abcd" in
    *$1*) echo "It's a substring" ;;
    *) echo "Not a substring" ;;
esac

Sample run:

$ ./case_substr.sh "ab"                                                                                           
It's a substring
$ ./case_substr.sh "whatever"                                                                                     
Not a substring

Note that you don't have to specifically use echo you can use exit 1 and exit 0 to signify success or failure.

What we could do as well, is create a function (which can be used in large scripts if necessary) with specific return values ( 0 on match, 1 on no match):

$ ./substring_function.sh                                  
ab is substring

$ cat substring_function.sh                                
#!/bin/sh

is_substring(){
    case "$2" in
        *$1*) return 0;;
        *) return 1;;
    esac
}

main(){
   if is_substring "ab" "abcdefg"
   then
       echo "ab is substring"
   fi
}

main $@

grep

$ grep -q 'ab' <<< "abcd" && echo "it's a substring" || echo "not a substring"                                    
it's a substring

This particular approach is useful with if-else statements in bash. Also mostly portable

AWK

$ awk '$0~/ab/{print "it is a substring"}' <<< "abcd"                                                             
it is a substring

Python

$ python -c 'import sys;sys.stdout.write("it is a substring") if "ab" in sys.stdin.read() else exit(1)' <<< "abcd"
it is a substring

Ruby

$ ruby -e ' puts "is substring" if  ARGV[1].include? ARGV[0]'  "ab" "abcdef"                                             
is substring

Answer 6

Mind the [[ and ":

[[ $a == z* ]]   # True if $a starts with an "z" (pattern matching).
[[ $a == "z*" ]] # True if $a is equal to z* (literal matching).

[ $a == z* ]     # File globbing and word splitting take place.
[ "$a" == "z*" ] # True if $a is equal to z* (literal matching).

So as @glenn_jackman said, but mind that if you wrap the whole second term in double quotes, it will switch the test to literal matching.

Source: http://tldp.org/LDP/abs/html/comparison-ops.html

Answer 7

Similar to edwin's answer, but with improved portability for posix & ksh, and a touch less noisy than Richard's:

substring=ab

string=abc
if [ "$string" != "${string%$substring*}" ]; then
    echo "$substring IS in $string"
else
    echo "$substring is NOT in $string"
fi

string=bcd
if [ "$string" != "${string%$substring*}" ]; then
    echo "$string contains $substring"
else
    echo "$string does NOT contain $substring"
fi

Output:

abc contains ab
bcd does NOT contain ab