CODEWITHSUNDEEP
bashshell
Boby
Boby

Reputation: 145

how to Find a substring in a bash shell script variable

I have a variable like below.

variable = This script is not found

if [[ "$variable" = ~ "not found" ]];
then
echo "Not Found"
else
echo "Its there"
if

while executing im getting below err,

line 4: syntax error in conditional expression
./test.sh: line 4: syntax error near `found"'
./test.sh: line 4: `if [[ "$variable" = ~ "not found" ]]; '

could anyone point me, What im missing here?

Upvotes: 14

Views: 57501

Answers (5)

Shnookie
Shnookie

Reputation: 11

Input:

line="There\'s a substring to be found!"

if [[ "$line" =~ *"substring"* ]]; then
    echo "Not Found";
else
    echo "Found!"
fi

Output:

Found!

Upvotes: 0

Aryan
Aryan

Reputation: 69

I have tried below codes which always return same result either in true or false

if [[ "$variable" =~ "not found" ]]; then
      echo "Not Found";
else
      echo "Its there";
fi

Upvotes: -1

jfelipesp
jfelipesp

Reputation: 133

here is a correct construction of your if statement

if [[ "$variable" =~ "not found" ]]; then
      echo "Not Found";
else
      echo "Its there";
fi

Upvotes: 3

William
William

Reputation: 4925

Compare this with your version at the indicated points:

variable="This script is not found"  # <--

if [[ "$variable" =~ "not found" ]]  # <--
then
    echo "Not Found"
else
    echo "Its there"
fi  # <--

You can't put spaces around = in an assignment, and you need to quote a string literal that has spaces. You don't need a trailing ; if you're going to put then on its own line. And an if-then ends with "fi" not "if".

Upvotes: 10

Quentin Perez
Quentin Perez

Reputation: 2903

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

Good Luck ;)

Upvotes: 28

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