qml and regexp; non-repeating digits and no leading zero

[djvujke]

Thanks for you answer! this is the regular expression i was looking for. it passes my tests on the page you provided

now i'm trying to implement it
ui->setupUi(this);
QRegExp regex("^([1-9])(?!\\1)([0-9])(?!\\1|\\2)([0-9])(?!\\1|\\2|\\3)([0-9])$");
QValidator *validator = new QRegExpValidator(regex, this);
ui->lineEdit->setValidator(validator);
ui->lineEdit->setFocus();

but lineEdit only accepts first digit...do you maybe know why?

i followed the example that i've found in the documentation
// regexp: optional '-' followed by between 1 and 3 digits
QRegExp rx("-?\\d{1,3}");
QValidator *validator = new QRegExpValidator(rx, this);

QLineEdit *edit = new QLineEdit(this);
edit->setValidator(validator);

[wysota]

QRegExp vs QRegularExpression. Read about differences between the two.

[jefftee]

but lineEdit only accepts first digit...do you maybe know why?

QRegExp doesn't understand or fully implement Perl compatible regular expressions, so when it gets to the next negative look-ahead, it doesn't understand it.

You can stop trying to make the regex I gave you work with QRegExp as it simply won't work. If you are using Qt 5.1 or higher, use QRegularExpressionValidator instead of QRegExpValidator.

If you are still using Qt 4.x, then sorry, I don't have a regex for you that will accomplish what you want using QRegExp.

Edit: I do see where QRegExp supposedly supports pcre negative lookaheads (?!...) but it simply doesn't appear to work. It works fine with QRegularExpression though.

[djvujke]

Thanks

This is working code
QRegularExpression regex("^([1-9])(?!\\1)([0-9])(?!\\1|\\2)([0-9])(?!\\1|\\2|\\3)([0-9])$");
QValidator *validator = new QRegularExpressionValidator(regex, this);
ui->lineEdit->setValidator(validator);


case closed