Checking form input for an AJAX request

Question

I've made a few functions to check form input for an AJAX request. I am still getting used to JavaScript and some of these AJAX requests. I am looking for any suggestions to better work with JavaScript functions, and passing variables through a function level scoping.

//Form Handler
$("#form-submit").submit(function(e) {
    e.preventDefault(); 
    $.ajax({    
        beforeSend: function(){ 
            return (checkAll());
            return (passwordCheck());
        },
        complete: function(){
        },
           type: "POST",
           url: '',
           data: $("#form-submit").serialize(),
           success: function(data)
           {
           //window.location = '/dashboard.php';
           // Set Session Php Etc 
           alert('php ran');
           return true;
           }
         });
}); 

//Pushing all Input Values to an Array  
function checkAll() {
    var arr = [];
    $('#form-submit :input').each(function() {
        arr.push($(this).val());
    });
    return (checkArray(arr));   
}

//Checks Array for empty strings 
function checkArray(arr){
    for(var i=0; i < arr.length; i++) {
        console.log(arr[i]);
        if (arr[i].trim() == '') {
            alert('Please Enter All Fields');   
            return false;
        } else {
            return true;
        }   
    }
}

//Matches Password
function passwordCheck() {
    var pass1 = $('#password').val();
    var pass2 = $('#password-check').val();
    if(pass1 != pass2) {
        $('#password').addClass('highlight');
        $('#password-check').addClass('highlight');
        alert("Passwords don't match");
        return false;
    } else {
        $('#password').removeClass('highlight');
        $('#password-check').removeClass('highlight');  
        return true;
    }
}

How would you go about checking an ajax based off of two Boolean based function? is there a method of concatenating function returns? Or is my methodology off? — miasyntax
– miasyntax, Commented Sep 6, 2014 at 8:45
The methodology is wrong. Use return (passwordCheck() && checkAll());? — hjpotter92
– hjpotter92, Commented Sep 6, 2014 at 8:46

janos · Accepted Answer · 2014-09-06 09:20:41Z

Bugs

You may think this code works, but it probably doesn't.

Example 1:

beforeSend: function () {
    return (checkAll());
    return (passwordCheck());
},

The passwordCheck function will never be called. Most probably you wanted this:

return checkAll() && passwordCheck();

Example 2:

for (var i = 0; i < arr.length; i++) {
    console.log(arr[i]);
    if (arr[i].trim() == '') {
        alert('Please Enter All Fields');
        return false;
    } else {
        return true;
    }
}

This for statement doesn't loop: it returns after checking the first value in arr, it won't check the others. Most probably you wanted this:

for (var i = 0; i < arr.length; i++) {
    console.log(arr[i]);
    if (arr[i].trim() == '') {
        alert('Please Enter All Fields');
        return false;
    }
}
return true;

Also, a sligthly simpler way to check for blank value:

if (!arr[i].trim()) {

Efficiency

In passwordCheck you lookup $('#password') and $('#password') twice. It would be better to lookup only once and cache in a local variable, for example:

function passwordCheck() {
    var $password = $('#password');
    var $password_check = $('#password-check');
    var pass1 = $password.val();
    var pass2 = $password_check.val();
    if (pass1 != pass2) {
        $password.addClass('highlight');
        $password_check.addClass('highlight');
        alert("Passwords don't match");
        return false;
    } else {
        $password.removeClass('highlight');
        $password_check.removeClass('highlight');
        return true;
    }
}

After testing it, I found exactly what you were referring to it not calling the check password function. I reworked it so that it would return undefined, and ran it as return (passwordCheck() && checkAll()); But this in an of itself was a dirty fix. I didn't think of moving the return outside of the for loops. Thank You, I am still pretty new to programming this was very helpful. — miasyntax
– miasyntax, Commented Sep 6, 2014 at 19:08

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