CODEWITHSUNDEEP
javascriptphpajax
user3453366
user3453366

Reputation: 51

AJAX POST is not receiving correct value

I want to post the value of a selected radio button but the $_POST variable in PHP is getting assigned 'undefined', what do I need to change so the $_POST gets the correct value? I was thinking it might be because I havent used URIencode but when I do use this my JavaScript doesn't even get the element.

JAVASCRIPT

function rappoll(){
    var xmlHTTPreq;
    var url = "poll.php";
    var params="name="+name;
    var elements = document.getElementsByName("elements");
    var checkedButton = getName(elements);

if (checkedButton) {
    var name = checkedButton.value;
}

if (window.XMLHttpRequest) {
    try {
        xmlHTTPreq = new XMLHttpRequest();
    } catch(e) {
        xmlHTTPreq = false;
    }
} else {
    try{
        xmlHTTPreq = new ActiveXObject("Microsoft.XMLHTTP");
    }catch(e){
        xmlHTTPreq = false;
    }
}

xmlHTTPreq.open("POST", url, true);
xmlHTTPreq.onreadystatechange = function()
{
    if(xmlHTTPreq.readyState == 4 && xmlHTTPreq.status == 200){
        document.getElementById('poll').innerHTML = xmlHTTPreq.responseText;
    }
}
xmlHTTPreq.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlHTTPreq.send(params);

}

function getName(group){
for (var i = 0; i < group.length; i++) {
    var button = group[i];
    if (button.checked) {
        return button;
    }
}
return undefined;

}

PHP

$name = $_POST['name'];

Upvotes: 0

Views: 61

Answers (1)

Halcyon
Halcyon

Reputation: 57709

You declare and assign params here:

var params="name="+name;

Since name is not defined params is now "name=undefined". Note that name is declared because of variable hoisting, so you wont see undefined variable 'name'.

Once you assign name you must reassign params too, or change this line:

xmlHTTPreq.send(params);

to

xmlHTTPreq.send("name=" + name);

Upvotes: 3

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