Hello,
Long time reader, first time poster. Any help is greatly appreciated.
I have crafted an AJAX query using JavaScript. The script works correctly, and the interface does what I want, but Firefox is giving me an error message related to the PHP file being hit. It's strange, because it seems to suggest there's a syntax error in the PHP, but that doesn't make any sense. This is the error:
Error: syntax error Source File: http://www.mysite.com/includes/ajax.php?action=checkpsudo&value=fd Line: 1, Column: 1 Source Code: yes
And the Javascript is below. Can anybody help me out? Thanks.
var ajaxobject = createajaxobjectObject();
function createajaxobjectObject() {
if (window.XMLHttpRequest) { // Mozilla, Safari,...
ajaxobject = new XMLHttpRequest();
if (ajaxobject.overrideMimeType) {
// set type accordingly to anticipated content type
ajaxobject.overrideMimeType('text/xml');
}
} else if (window.ActiveXObject) { // IE
try {
ajaxobject = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
ajaxobject = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {}
}
}
if (!ajaxobject) {
alrt('Cannot create XMLHTTP instance');
return false;
}
return ajaxobject;
}
function checkpsudo(value) {
if (value == "") {
document.getElementById('feedback').innerHTML = "Please select a psudonym";
document.getElementById('feedback').className = "fail";
document.getElementById('done').disabled=true;
} else {
ajaxobject.onreadystatechange = function() { check(); };
ajaxobject.open('GET', '/includes/ajax.php?action=checkpsudo&value='+value, true);
ajaxobject.send(null);
}
}
function check() {
if (ajaxobject.readyState == 4) {
//IF WE GOT OUR CHAT XML BACK CORRECTLY
if (ajaxobject.status == 200) {
var response = ajaxobject.responseText;
var value = document.getElementById('psudoentry').value;
if(response=='no') {
document.getElementById('feedback').innerHTML = "'" + value + "' is already being used";
document.getElementById('feedback').className = "fail";
document.getElementById('done').disabled=true;
} else {
document.getElementById('feedback').innerHTML = "'" + value + "' is available";
document.getElementById('feedback').className = "success";
document.getElementById('done').disabled=false;
}
} else {
alert('There was a problem with the request.');
}
}
}