Question

<script language="JavaScript">
    function validate(x) {



        var cur_p = document.getElementById('current').value;
        var new_p = document.getElementById('new').value;
        var con_p = document.getElementById('confirm').value;

        document.getElementById('msg_p').innerHTML = '';
        document.getElementById('msg_cur').innerHTML = '';


        if(x != cur_p)
        {   document.getElementById('msg_cur').innerHTML = ' Your password was incorrect';
            return false;
        }   

        if(new_p != con_p)
        {   document.getElementById('msg_p').innerHTML = 'Passwords do not match';
            return false;
        }

        return (true);

    }
</script>

html

<form action='change-password.php' method='post' onsubmit="return validate('<?=$current?>')" >

I have these on my code.
I can’t show the result of these ifs simultaenously.

if(x != cur_p)
and 
if(new_p != con_p)

if I place the

if(x != cur_p){} 

at the top of

if(new_p != con_p){}

the response result of if(x != cur_p) will show and the latter will not

and vice versa.

how can i show the result of those two ifs
(assuming that the condition for those were meet)

Answer 1

First you have a typo in your code

document.getElementById('msg_p').innerHTM = '';  <-- Missing an L

Second, of course if you return, it exits the function. So the code will not execute both statements.

Change

    if(x != cur_p)
    {   document.getElementById('msg_cur').innerHTML = ' Your password was incorrect';
        return false;
    }   

    if(new_p != con_p)
    {   document.getElementById('msg_p').innerHTML = 'Passwords do not match';
        return false;
    }

    return (true);

to

    var isValid = true;
    if(x != cur_p)
    {   document.getElementById('msg_cur').innerHTML = ' Your password was incorrect';
        isValid = false;
    }   

    if(new_p != con_p)
    {   document.getElementById('msg_p').innerHTML = 'Passwords do not match';
        isValid = false;
    }

    return isValid;