Regular Expression With No More Than 3 Alphabets Concurrently Should Not Accept 1a1a1a1a1

should not accept: 1a1a1a1a1, aaaa11111 ,2222aaa33a only allow 3 characters anywhere no more than three should not allowed I tried like below but failed var patt = new RegExp('([A-

Solution 1:

Your regex needs to account for non-letters in between the letters. I would do that with this regex:

^(?!.*([a-z].*){4,}).*$

Debuggex Demo

This way you can easily add more requirements by adding another (?!...) or (?=...)lookaround.

Solution 2:

^(?!.*(.)(?:.*\1){3,}).*$

Try this.See demo.The looahead makes sure any character is not there more than 3 times.

https://regex101.com/r/wU7sQ0/10

var re = /^(?!.*(.)(?:.*\1){3,}).*$/gm;
var str = '1a1a1a1a1\naaaa11111\n2222aaa33a\nsadsada';
var m;

while ((m = re.exec(str)) != null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
// View your result using the m-variable.// eg m[0] etc.
}

Solution 3:

functiontestNumberOfChar(str){
    var match= str.match(/[a-zA-Z]/g);
    if( match && match.length > 3 ){
       alert("No more than three alphabets are allowed"); 
       returnfalse;
    }
    returntrue;
   }

Solution 4:

You can do this:

if (/^(?:[^a-z]*[a-z]){4}/i.test(DLnumber)) {
    alert("No more than three alphabets are allowed");
    returnfalse;
}

• the i modifier makes the pattern case-insensitive, so [a-z] is the same than [a-zA-Z] and [^a-z] the same than [^a-zA-Z] for all the pattern.
• ^ is the anchor for the start of the string (not needed but improve performances in particular when the pattern must fail)
• (?:...) is an non-capturing group, (it does nothing, its only task is to put other tokens together, then you can apply a quantifier to all the group.)
• (?:[^a-z]*[a-z]){4} means a group of zero or more non-letters followed by a letter and repeated 4 times

Solution 5:

I thin this may be what you look for:

if (str.replace(/[^a-z]/gi, "").length>3) {
    alert("No more than three alphabets are allowed");
    returnfalse;
}

It counts the Alphabetical chars in the string...