phpQuestioner Posted August 11, 2007 Share Posted August 11, 2007 I am new to AJAX; I am trying to learn, but I learning this slow then regular PHP. I am trying to get this script to work that I found on this site to send a query string with two php variables in it. This thing is; it's not working - so anyway, can some one tell what I need to correct to get it working? <script type="text/javascript"> function createRequestObject() { var req; if(window.XMLHttpRequest){ // Firefox, Safari, Opera... req = new XMLHttpRequest(); } else if(window.ActiveXObject) { // Internet Explorer 5+ req = new ActiveXObject("Microsoft.XMLHTTP"); } else { // There is an error creating the object, // just as an old browser is being used. alert('Problem creating the XMLHttpRequest object'); } return req; } // Make the XMLHttpRequest object var http = createRequestObject(); function sendRequest(cat,updown) { // Open PHP script for requests http.open('get', 'http://domain.com/sub1/items.php?cat='+cat&updown='+updown); http.onreadystatechange = handleResponse; http.send(null); } function handleResponse() { if(http.readyState == 4 && http.status == 200){ // Text returned FROM the PHP script var response = http.responseText; if(response) { // UPDATE ajaxTest content document.getElementById("ajaxTest").innerHTML = response; } } } </script> </head> <body> <span id="ajaxTest"></span> <br><br><br> <a href="#" onClick="sendRequest('make','asc');">Test</a> Quote Link to comment Share on other sites More sharing options...
gurroa Posted August 11, 2007 Share Posted August 11, 2007 You have syntax error here http.open('get', 'http://domain.com/sub1/items.php?cat='+cat&updown='+updown); Should be http.open('get', 'http://domain.com/sub1/items.php?cat='+cat+'&updown='+updown); Quote Link to comment Share on other sites More sharing options...
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