防止按钮表单提交Javascript

Question

This form keeps on submitting even if it executes the return value, what is the problem with my code?

function formhash (form, password)
{
    var pass1 = document.getElementById("password").value;
    var pass2 = document.getElementById("cpassword").value;
    var ok = true;
    if (password != cpassword) {
        //alert("Passwords Do not match");
        document.getElementById("password").style.borderColor = "#E34234";
        document.getElementById("cpassword").style.borderColor = "#E34234";
        ok = false;
        return;
    }
    else
    {
        $.post('insert_home.php'
            {PRIMAID:PRIMAID,EDITCAP:EDITCAP,EDITIMG:EDITIMG,EB_TITLE:EB_TITLE}).done(function(data){
            alert ("Book Successfully Updated");
            location.reload();
        });


        var p = document.createElement("input");    
        form.appendChild(p);
        p.name="p";
        p.type="hidden";
        p.value=hex_sha512(password.value);
        password.value="";
        form.submit();
    }
}

Answer 1

You are calling form.submit(); , remove it and it won't submit.

Answer 2

You are using the wrong variable names "password" and cpassword". You created pass1 and pass2 so you need to use those.

Change to this:

//You were using the WRONG variable names
if (pass1 != pass2) {
    //alert("Passwords Do not match");
    document.getElementById("password").style.borderColor = "#E34234";
    document.getElementById("cpassword").style.borderColor = "#E34234";
    ok = false;
    return false;
}

Answer 3

I don't see where formhash() is used.

The code calls the submit, not the function.

Answer 4

If it submits the form, it' because it's falling in the else condition.

Add lots of "console.log("debug_X")" where "X" is a number different each time.

You have to add inside a console.log the value of pass1 and pass2.

Maybe you are passign a value instead of an object or maybe the reversed situation can happen also.

My instinct tells me to check both password object, and their values. Also it tells me that you didn't check the content of pass1 and pass2 before posting a comment to Don Rhummy.

Check carefully and please search more before posting your next answer.

You can have the value by doing "console.log(pass1);" and by opening firebug and by checking the javascript console.

Answer 5

从窗体中删除操作属性。