Regex match a string with spaces (use quotes?) in an if statement

Question

How would I do a regex match as shown below but with quotes around the (\"^This\") as in the real world \"This\" will be a string that can have spaces in it.

#!/         


        

Answer 1

You can use \ before spaces.

#!/bin/bash

text="This is just a test string"
if [[ "$text" =~ ^This\ is\ just ]]; then
  echo "matched"
else
  echo "not matched"
fi

Answer 2

can you make your problem description clearer?

text="This is just a test string"
case "$text" in
    "This is"*) echo "match";;
esac

the above assume you want to match "This is" at exactly start of line.

Answer 3

I did not manage to inline the expression like this:

if [[ "$text" =~ "^ *This " ]]; then

but if you put the expression in a variable you could use normal regex syntax like this:

pat="^ *This "
if [[ $text =~ $pat ]]; then

Note that the quoting on $text and $pat is unnessesary.

Edit: A convinient oneliner during the development:

pat="^ *This is "; [[ "   This is just a test string" =~ $pat ]]; echo $?

Answer 4

Have you tried:

^[\s]*This