创建一个1337密码检查器（javascript）

Question

I'm having an issue with some javascript that I am working on. The purpose is to check the input for the list of known bad passwords. That part was easy. But then it should also be able to check for the "1337" versions of these passwords. Basically it needs to be able to check for numbers in the password and convert that to english, to then run through the known bad passwords function. I tried to set up a function to do that from Making a leet speak translator but it is only partially working and I am confused as to why, as it is not iterating properly. Letters that appear more then once only get replaced the first time. There is also the issue with l or i being the same number. I am hoping someone would be able to take a look at it?

<script>    
var LettersEnglish = 
    new Array('o', 'i', 'l', 'e', 'a', 's', 't');
var LettersLeet = 
    new Array('0', '1', '1', '3', '4', '5', '7');

function changeLetters() { // change all letters
    var text = document.getElementById("password").value;
    for (var i = 0; i < text.length; i++) {
        text = text.replace(LettersLeet[i], LettersEnglish[i]);
        }englishPasswords(text);
}

function englishPasswords(convertedstring) {
    var myarr = ["password", "letmein", "dragon", "shadow", "joker"];
    var string = (myarr.indexOf(convertedstring) > -1);
    if (string === true){
        document.getElementById("output").innerHTML = convertedstring + " is a bad password";
    }else{
        document.getElementById("output").innerHTML = convertedstring + " is an acceptable password";
    }
}
</script> 

Answer 1

You appear to be iterating over text.length instead of LettersEnglish.length or LettersLeet.length . This means that if you have a password shorter than the length of these arrays, not all the characters in them will be substituted.

Answer 2

@MTCoaster is correct in that you're iterating over the wrong thing. You would need something like this instead:

for (var i = 0; i < LettersLeet.length; i++) {
  text = text.replace(new RegExp(LettersLeet[i], 'g'), LettersEnglish[i]);
}

Note that the replace() call now uses a regular expression as the first argument, and that the regex is created with the "global" flag. This allows replace() to replace all occurrences of the match. The way you call it, with a string as the first argument, only the first occurrence would be replaced.

Regarding the i/l mapping, you're not mapping them both to 1, but rather the other way around, you need 1 to be treated both as i and l when checking for bad passwords. 1 can't possibly map to both at the same time. A simplistic approach could be:

var LettersEnglish1 = 
    new Array('o', 'i', 'e', 'a', 's', 't');
var LettersEnglish2 = 
    new Array('o', 'l', 'e', 'a', 's', 't');
var LettersLeet = 
    new Array('0', '1', '3', '4', '5', '7');

and then, inside changeLetters() :

var text1 = document.getElementById("password").value,
    text2 = text;
for (var i = 0; i < LettersLeet.length; i++) {
  text1 = text1.replace(new RegExp(LettersLeet[i], 'g'), LettersEnglish1[i]);
  text2 = text2.replace(new RegExp(LettersLeet[i], 'g'), LettersEnglish2[i]);
}
//Now check both text1 and text2 for validity

A complete solution should probably check for all possible permutations of i and l though. For example, if the password is 1n1t1a1 , there are 2^4 (16) different replaced passwords to check for.

Finally, a style advice:

var LettersEnglish1 = 
    new Array('o', 'i', 'e', 'a', 's', 't');
var LettersEnglish2 = 
    new Array('o', 'i', 'l', 'e', 'a', 's', 't');
var LettersLeet = 
    new Array('0', '1', '3', '4', '5', '7');

is normally written as

var LettersEnglish1 = ['o', 'i', 'e', 'a', 's', 't'],
    LettersEnglish2 = ['o', 'l', 'e', 'a', 's', 't'],
    LettersLeet = ['0', '1', '3', '4', '5', '7'];

Some may prefer multiple var statements like you had it but most use the array literal syntax instead of new Array() .

Answer 3

Because we had a list of known bad passwords and were later informed there would only be once instance of 1's in the password here is the answer I went with. For the check of all letters except L, I used @PeterHerdenborg's answer.

var LettersEnglish1 = ['o', 'e', 'a', 's', 't', 'i'],
    LettersLeet = ['0', '3', '4', '5', '7', '1'];

function changeLetters() {
    var text = document.getElementById("password").value;
    for (var i = 0; i < LettersLeet.length; i++) {
        text = text.replace(new RegExp(LettersLeet[i], 'g'), LettersEnglish1[i]);
        //console.log(text);
        englishPasswords(text);
    }
}

function englishPasswords(text) {
    var myarr = ["password", "letmein", "dragon", "shadow", "joker"];
    var string = (myarr.indexOf(text) > -1);
    if (string === true){
        document.getElementById("output").innerHTML = text + " is a bad password";
    }else{
        checkForLs(text);
    }
}

function checkForLs(text){
    text = text.replace('i','l')
    if(text == 'letmein'){
        document.getElementById("output").innerHTML = text + " is a bad password";
    }else{
        document.getElementById("output").innerHTML = text + "thats an acceptable password";
    }
}