How to check if a string contains a substring in Bash

Question

I have a string in Bash:

string=\"My string\"

How can I test if it contains another string?

if [ $string ?? \'foo\' ]; then         


        

Answer 1

One is:

[ $(expr $mystring : ".*${search}.*") -ne 0 ] && echo 'yes' ||  echo 'no'

Answer 2

Extension of the question answered here How do you tell if a string contains another string in POSIX sh?:

This solution works with special characters:

# contains(string, substring)
#
# Returns 0 if the specified string contains the specified substring,
# otherwise returns 1.
contains() {
    string="$1"
    substring="$2"

    if echo "$string" | $(type -p ggrep grep | head -1) -F -- "$substring" >/dev/null; then
        return 0    # $substring is in $string
    else
        return 1    # $substring is not in $string
    fi
}

contains "abcd" "e" || echo "abcd does not contain e"
contains "abcd" "ab" && echo "abcd contains ab"
contains "abcd" "bc" && echo "abcd contains bc"
contains "abcd" "cd" && echo "abcd contains cd"
contains "abcd" "abcd" && echo "abcd contains abcd"
contains "" "" && echo "empty string contains empty string"
contains "a" "" && echo "a contains empty string"
contains "" "a" || echo "empty string does not contain a"
contains "abcd efgh" "cd ef" && echo "abcd efgh contains cd ef"
contains "abcd efgh" " " && echo "abcd efgh contains a space"

contains "abcd [efg] hij" "[efg]" && echo "abcd [efg] hij contains [efg]"
contains "abcd [efg] hij" "[effg]" || echo "abcd [efg] hij does not contain [effg]"

contains "abcd *efg* hij" "*efg*" && echo "abcd *efg* hij contains *efg*"
contains "abcd *efg* hij" "d *efg* h" && echo "abcd *efg* hij contains d *efg* h"
contains "abcd *efg* hij" "*effg*" || echo "abcd *efg* hij does not contain *effg*"

Answer 3

You can use Marcus's answer (* wildcards) outside a case statement, too, if you use double brackets:

string='My long string'
if [[ $string == *"My long"* ]]; then
  echo "It's there!"
fi

Note that spaces in the needle string need to be placed between double quotes, and the * wildcards should be outside. Also note that a simple comparison operator is used (i.e. ==), not the regex operator =~.

Answer 4

As Paul mentioned in his performance comparison:

if echo "abcdefg" | grep -q "bcdef"; then
    echo "String contains is true."
else
    echo "String contains is not true."
fi

This is POSIX compliant like the 'case "$string" in' the answer provided by Marcus, but it is slightly easier to read than the case statement answer. Also note that this will be much much slower than using a case statement. As Paul pointed out, don't use it in a loop.

Answer 5

Exact word match:

string='My long string'
exactSearch='long'

if grep -E -q "\b${exactSearch}\b" <<<${string} >/dev/null 2>&1
  then
    echo "It's there"
  fi

Answer 6

I use this function (one dependency not included but obvious). It passes the tests shown below. If the function returns a value > 0 then the string was found. You could just as easily return 1 or 0 instead.

function str_instr {
   # Return position of ```str``` within ```string```.
   # >>> str_instr "str" "string"
   # str: String to search for.
   # string: String to search.
   typeset str string x
   # Behavior here is not the same in bash vs ksh unless we escape special characters.
   str="$(str_escape_special_characters "${1}")"
   string="${2}"
   x="${string%%$str*}"
   if [[ "${x}" != "${string}" ]]; then
      echo "${#x} + 1" | bc -l
   else
      echo 0
   fi
}

function test_str_instr {
   str_instr "(" "'foo@host (dev,web)'" | assert_eq 11
   str_instr ")" "'foo@host (dev,web)'" | assert_eq 19
   str_instr "[" "'foo@host [dev,web]'" | assert_eq 11
   str_instr "]" "'foo@host [dev,web]'" | assert_eq 19
   str_instr "a" "abc" | assert_eq 1
   str_instr "z" "abc" | assert_eq 0
   str_instr "Eggs" "Green Eggs And Ham" | assert_eq 7
   str_instr "a" "" | assert_eq 0
   str_instr "" "" | assert_eq 0
   str_instr " " "Green Eggs" | assert_eq 6
   str_instr " " " Green "  | assert_eq 1
}