How to find substring inside a string (or how to grep a variable)?

Question

I'm using BASH, and I don't know how to find a substring. It keeps failing, I've got a string (should this be an array?)

Below, LIST is a string list of database names, SOURCE is the reply, one of those databases. The following still doesn't work:

echo "******************************************************************"
echo "*                  DB2 Offline Backup Script                     *"
echo "******************************************************************"
echo "What's the name of of the  database you would like to backup?"
echo "It will be named one in this list:"
echo ""
LIST=`db2 list database directory | grep "Database alias" | awk '{print $4}'`
echo $LIST
echo ""
echo "******************************************************************"
echo -n ">>> "
read -e SOURCE

if expr match "$LIST" "$SOURCE"; then
    echo "match"
    exit -1
else
    echo "no match"
fi
exit -1

I've also tried this but doesn't work:

if [ `expr match "$LIST" '$SOURCE'` ]; then

Solution 1

LIST="some string with a substring you want to match"
SOURCE="substring"
if echo "$LIST" | grep -q "$SOURCE"; then
  echo "matched";
else
  echo "no match";
fi

Solution 2

You can also compare with wildcards:

if [[ "$LIST" == *"$SOURCE"* ]]

Solution 3

This works in Bash without forking external commands:

function has_substring() {
   [[ "$1" != "${2/$1/}" ]]
}

Example usage:

name="hello/world"
if has_substring "$name" "/"
then
   echo "Indeed, $name contains a slash!"
fi

Solution 4

If you're using bash you can just say

if grep -q "$SOURCE" <<< "$LIST" ; then
    ...
fi

Solution 5

expr match "$LIST" '$SOURCE'

don't work because of this function search $SOURCE from begin of the string and return the position just after pattern $SOURCE if found else 0. So you must write another code:

expr match "$LIST" '.*'"$SOURCE" or expr "$LIST" : '.*'"$SOURCE"

The expression $SOURCE must be double quoted so as a parser may set substitution. Single quoted not substitute and the code above will search textual string $SOURCE from the beginning of the $LIST. If you need the beginning of the string subtract the length $SOURCE e.g ${#SOURCE}. You may write also

expr "$LIST" : ".*\($SOURCE\)"

This function just extract $SOURCE from $LIST and return it. You'll get empty string else. But they problem with double double quote. I don't know how it resolve without using additional variable. It's light solution. So you may write in C. There is ready function strstr. Don't use expr index, So is very attractive. But index search not substring and only first char.

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Comments

[SiegeX]

It's hard to answer any sort of question without knowing what LIST and SOURCE looks like

[SourceSeeker]

In Bash, there's almost always no reason to use expr which is an external utility.

[edumike]

yes that one solved it: stackoverflow.com/questions/229551/string-contains-in-bash

[jww]

Possible duplicate of String contains a substring in Bash.

[maganap]

You should always quote the strings, like: if [[ "$list" == *"$source"* ]]

[wrlee]

The advantage to this solution is that it doesn't rely on any external commands nor file descriptors (via redirection or pipes).

[codesniffer]

2 issues to consider with this approach: (1) grep uses regex so some chars will match unexpected or cause errors; (2) Using grep (or anything that's not built in to bash) spawns a separate process which will run slower (only matters in larger scale cases).

[codesniffer]

Note that this is using a regex to search for the substring (and also does an out-of-place replace), so special chars in the search (substring) can cause unexpected results. But if regex is fine, then why not just do a regex search (ie. no need for the replace)? Syntax example: if [[ "$string" =~ $substring ]]

[Leo]

Note that if [[ *"$SOURCE"* == "$LIST$" ]] is false.

[Manuel Jordan]

I've confirmed that the order is important/mandatory - so *"$SOURCE"* must be in the right - otherwise always returns false - why that behavior?