要检查的正则表达式-不超过2个连续数字或字符，并且不超过1个相同数字或字符

Question

I want to reject user input if it contains 2 sequential numbers or characters ,for example 1234,jkl, zyxw and if it contains more than 1 same numbers or characters like aaer,0000,aabb,pp22. Thank you for insights. I have regex for the second one but dont know how to combine the two expressions:

"([a-zA-Z0-9])\\1{1,}"

Answer 1

Doing this in regex is neither sound nor practical. However, you can easily check if your input contains a sequential ( abc .. or cba ) pattern using code like that:

 function isSequencial(input) { var numpattern = '0123456789012345789'; // match ascending/descending sequence of numbers. var charpattern = 'ABCDEFGHIJKLMNOPQRSTUVWXYZYXWVUTSRQPONMLKJIHGFEDCBA'; // match ascending/descending sequence of letters. for (var i = 0; i < input.length-1; i++) { var shard = input.substring(i,i+2); if(numpattern.indexOf(shard) != -1) { console.log('sequential number pattern detected: ' + shard); return true; } if (charpattern.indexOf(shard.toUpperCase()) != -1) { console.log('sequential letter pattern detected: ' +shard); return true; } } return false; } console.log("isSequencial(a234):" + isSequencial("a234")); console.log("isSequencial(azyx):" + isSequencial("azyx")); console.log("isSequencial(xbc):" + isSequencial("xbc")); console.log("isSequencial(2435):" + isSequencial("2435")); 

This code can be optimized but is easy to understand and maintain since it does not try to do multiple things at once. You should be able to combine this with your existing approach.

Answer 2

The simplest solution for your first requirement would be to parse it, as with a regex it will be not that easy to set up, if at all possible.

Here I used charCodeAt (and check for both sequence/equal and duplicates characters)

 var input1 = "1543abc3"; var input2 = "cba23DEf"; var input3 = "ba2354cd"; console.log('test 1'); testit(input1.split('')); console.log('test 2'); testit(input2.split('')); console.log('test 3'); testit(input3.split('')); function testit (arr) { var prev = arr[0].charCodeAt(0) + 1, prev2 = -1; for (var i = 1; i < arr.length; i++) { var arritem = arr[i].charCodeAt(0); if ( (arritem == prev && arritem == (prev2+1)) || // abc (arritem == (prev-2) && arritem == (prev2-3)) // cba ) { console.log(' - sequence, more than 2: ', arr[i-2], arr[i-1], arr[i] ); //return false; } if (arr.indexOf(arr[i-1],i) > -1) { console.log(' - duplicate, more than 1: ', arr[i-1] ); //return false; } prev2 = prev; prev = arr[i].charCodeAt(0) + 1; } //return true; }