Question

function addphoto()
{
    var ajaxRequest = initAjax();
    if (ajaxRequest == false)
    {
        return false;
    }

    // Return Ajax result when the state changes later
    ajaxRequest.onreadystatechange = function()
    {
        if(ajaxRequest.readyState == 4)
        {
            alert(ajaxRequest.responseText);
            return ajaxRequest.responseText;
        }
    }

    // Capture form elements
    var values = {
        "category" : encodeURIComponent(document.addphoto.category.options[document.addphoto.category.selectedIndex].value),
        "photo_title" : encodeURIComponent(document.addphoto.photo_title.value),
        "photo_descrip" : encodeURIComponent(document.addphoto.photo_descrip.value)
    }

    var queryString = '?', i = 0;
    for (var key in values)
    {
        if (i != 0)
        {
            queryString += '&'
        }
        queryString += key + '=' + values[key];
        i++;
    }

    // Execute Ajax
    ajaxRequest.open("POST", "ajaxcheckform.php" + queryString, true);
    ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    ajaxRequest.setRequestHeader("Content-length", queryString.length);
    ajaxRequest.setRequestHeader("Connection", "close");
    ajaxRequest.send(null);
}

function ajaxCheckform(formname)
{
    var response = addphoto(); // <--This is undefined and not sure why
    var responseObj = JSON.parse(response);

    if (responseObj.success == 1)
    {
        // Successful form!
        alert(responseObj.success_text);
    }
    else
    {
        // Error!
        alert(responseObj.error);
    }
}

I’m sure I must be making some basic error somewhere, but I’m having trouble locating it. In this script, ajaxCheckform() is a function that executes one of several similar functions. Above, I included addphoto(), which is one of several functions I’ll need that look like this.

On a side note, I’d love to know I can call upon a function dynamically. The addphoto() function will be only one such function being called up at that moment and I’m trying to find a way to pass formname as the function I need. I’ve searched Stackoverflow and Google. I’ve found nothing that works.

Note, I’m aware of jQuery, but I’m not there yet. I need this function to work first.

Answer 1

It is not addphoto() thats undefined but response is undefined. ajaxRequest is asynchronous and the addphoto() function will return before the request completes.

try this

function addphoto() {...

    // Return Ajax result when the state changes later
    ajaxRequest.onreadystatechange = function()
    {
        if(ajaxRequest.readyState == 4)
        {
            alert(ajaxRequest.responseText);            
            var responseObj = JSON.parse(ajaxRequest.responseText);

            if (responseObj.success == 1) {
                 // Successful form!
                 alert(responseObj.success_text);
            }
           else {
               // Error!
               alert(responseObj.error);
           }        
        }    
    }
....
}

function ajaxCheckform(formname) {
    addphoto();    
}