match a line using bash regex

Question

I want to match a line that contains a word, but does not have semi-colon in it

This should match:

class test

this should not match

class test;

this should not match either

class test; // test class

this is what I was expecting to work, but it doesn't:

pattern="class [^;]*"

if [[ $line =~ $pattern ]]

thanks

Answer 1

Your regular expression is not anchored which means that [^;]* will still match against all characters up to a possible ; (and thus match as a whole). If you anchor the regex against the end of the line ([^;]*$) it will produce the results you are after:

$ cat t.sh
#!/bin/bash

pattern='class [^;]*$'
while read -r line; do
    printf "testing '${line}': "
    [[ $line =~ $pattern ]] && echo matches || echo "doesn't match"
done <<EOT
class test
class test;
class test; // test class
EOT

---

$ ./t.sh
testing 'class test': matches
testing 'class test;': doesn't match
testing 'class test; // test class': doesn't match

TL;DR: In other words, the bold part in

class test; foo bar quux

matches your regex even though the string contains a semicolon which is why it always matches. The anchor makes sure that the regular expression only matches if there is no semicolon until the very end of the string.

Answer 2

how about straightforwardly:

 pattern="^[^;]*\bclass\b[^;]*$"

\b word boundary was added, for matching xxx class xxx only, not matching superclass xxx