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 =~ $pattern ]]
thanks
Your regular expression is not anchored which means that [^;]* will still match against all characters up to a possible ; (and thus match as a whole). If you anchor the regex against the end of the line ([^;]*$) 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.
how about straightforwardly:
pattern="^[^;]*\bclass\b[^;]*$"
\b word boundary was added, for matching xxx class xxx only, not matching superclass xxx
Use ^[^;]+($|\s*//). This means any number of non-semicolon characters (at least one) from the start of the string until either the end of the line or any number of spaces followed by two slashes.
I think you need:
pattern="^[^;]*class [^;]*$"`
This ensures the line don't have a ; before or after your [^;]* match.
class test // test class;?