Regular expression in java for matching string not enclosed in single quote

I’m parsing my scripts for wrong strings and have an array list with strings that shouldn’t occur in my lines of code.

But some strings should pass when they are not exactly the string.

Examples:

list contains “FOO” and “BAR”

textlines:

This is foo
and this is BAR
but not 'foo'
and not 'BAR'
but also not FOO_BAR
but we want FOO%TEXT
and also bar.text

result

This is foo
and this is BAR
but we want FOO%TEXT
and also bar.text

I’ve tried a view examples I found online and on stackoverflow but these didn’t work for me, they don’t filter the quoted ones.

String pattern = ".*\"+strTables[i]+"\b.*[^(\w|')]";
Pattern r = Pattern.compile(pattern, Pattern.CASE_INSENSITIVE);
Matcher m = r.matcher(line);
if (m.find()) {
    System.out.println (strTables[i] + ": " + line);
    break;
}

Answer

You need b (word boundary) before and after the word, to match a whole word.

Then, a look-behind like (?<!') at the beginning of the pattern to denote “not preceded by a single quote” and a look-ahead like (?!') at the end of the pattern to denote “not followed by a single quote”.

Putting it together makes

String pattern = "(?<!')\b" + strTables[i] + "\b(?!')";