Ip Address Validation

[bruccutler]

So - what is the right way to do this? I think I will set it aside, since I'm sure someone out there has solved this problem with Qt.
- BRC

[Jimmy2775]

I think you could do it with the combination of a masked line edit, and a QRegExpValidator that checks for a regex like this: (?:1\d?\d?|2(?:[0-4]\d?|[6789]|5[0-5]?)?|[3-9]\d?|0)(?:\.(?:1\d?\d?|2(?:[0-4]\d?|[6789]|5[0-5]?)?|[3-9]\d?|0)){3}

That said, I haven't tried this myself, so who knows...

[wysota]

Oh my God, what an expression.... How about:

class IP4Validator : public QValidator {
public:
  IP4Validator(QObject *parent=0) : QValidator(parent){}
  void fixup(QString &input) const {}
  State validate(QString &input, int &pos) const {
    if(input.isEmpty()) return Acceptable;
      QStringList slist = input.split(".");
      int s = slist.size();
      if(s>4) return Invalid;
      bool emptyGroup = false;
      for(int i=0;i<s;i++){
        bool ok;
        if(slist[i].isEmpty()){
          emptyGroup = true;
          continue;
        }
        int val = slist[i].toInt(&ok);
        if(!ok || val<0 || val>255) return Invalid;
      }
      if(s<4 || emptyGroup) return Intermediate;
      return Acceptable;
  }
};

class IP4Validator : public QValidator {
public:
  IP4Validator(QObject *parent=0) : QValidator(parent){}
  void fixup(QString &input) const {}
  State validate(QString &input, int &pos) const {
    if(input.isEmpty()) return Acceptable;
      QStringList slist = input.split(".");
      int s = slist.size();
      if(s>4) return Invalid;
      bool emptyGroup = false;
      for(int i=0;i<s;i++){
        bool ok;
        if(slist[i].isEmpty()){
          emptyGroup = true;
          continue;
        }
        int val = slist[i].toInt(&ok);
        if(!ok || val<0 || val>255) return Invalid;
      }
      if(s<4 || emptyGroup) return Intermediate;
      return Acceptable;
  }
};

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[bruccutler]

This worked perfectly! Thanks so much! It took me a while to get back to it, but this really did the job!

[przemoc]

I think you could do it with the combination of a masked line edit, and a QRegExpValidator that checks for a regex like this: (?:1\d?\d?|2(?:[0-4]\d?|[6789]|5[0-5]?)?|[3-9]\d?|0)(?:\.(?:1\d?\d?|2(?:[0-4]\d?|[6789]|5[0-5]?)?|[3-9]\d?|0)){3}

That said, I haven't tried this myself, so who knows...

I know that problem is already solved, but... it's definetely too long regexp. I love the power of regular expressions, so here is my shorter version (w/o non-capturing parentheses, for increased readability), where each octet range from 0 to 255:

^0*(2(5[0-5]|[0-4]\d)|1?\d{1,2})(\.0*(2(5[0-5]|[0-4]\d)|1?\d{1,2})){3}$

QRegExpValidator of course doesn't need ^ and $ assertions.