[Konversation-devel] [Bug 218332] New: No way to output '%1' in autoreplace

[Nicolás Alvarez]

https://bugs.kde.org/show_bug.cgi?id=218332

           Summary: No way to output '%1' in autoreplace
           Product: konversation
           Version: Git
          Platform: Compiled Sources
        OS/Version: Linux
            Status: UNCONFIRMED
          Severity: normal
          Priority: NOR
         Component: general
        AssignedTo: konversation-devel at kde.org
        ReportedBy: nicolas.alvarez at gmail.com


In a regex autoreplace, it's impossible to output a literal percent sign
followed by a number. If I have a regex "foo-(\S*)" and replacement "bar%23%1",
"foo-x" will output "bar3x", because the %2 is interpreted as a capturing
subpattern, even though the regex only has one subpattern.

The current code is:
 // replace %0 - %9 in regex groups
 for (int capture=0;capture<captures.count();capture++)
 {
     replacement.replace(QString("%%1").arg(capture),captures[capture]);
 }
 replacement.remove(QRegExp("%[0-9]"));

Removing the last line would allow the previous example to work. However, it
still doesn't solve the problem if I *do* have a second subpattern (or if I
want to output %1). What we really need is a way to escape the percent sign.

Instead of using replace(), I propose looping over the characters in the
string. Pseudocode:
foreach (char in string) {
    if (char == %) {
        look at the following character;
        if (char == %) output += '%'; //this lets you escape a % by doubling it
        if (char is digit) output += capture[digit];
        else output += '%' + char;
    } else {
        output += char;
    }
}
I may implement this myself later.

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