How to check if a string contains a substring in B

Question

I have a string in Bash:

string="My string"

How can I test if it contains another string?

if [ $string ?? 'foo' ]; then
  echo "It's there!"
fi

Where ?? is my unknown operator. Do I use echo and grep?

if echo "$string" | grep 'foo'; then
  echo "It's there!"
fi

That looks a bit clumsy.

Answer 1

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
}

Answer 2

I am not sure about using an if statement, but you can get a similar effect with a case statement:

case "$string" in 
  *foo*)
    # Do stuff
    ;;
esac

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.