becu Posted July 18, 2007 Share Posted July 18, 2007 Hi guy, I'm very new to ajax. Please help me fix error below. This code supposes to take the url from the link, put it into showHint function and output and image. <script type="text/javascript"> var xmlHttp function showHint(str) { if (str.length==0) { document.getElementById("txtHint").innerHTML=""; return; } xmlHttp=GetXmlHttpObject() if (xmlHttp==null) { alert ("Your browser does not support AJAX!"); return; } var url="img.php"; url=url+"?img="+str; url=url+"&sid="+Math.random(); xmlHttp.onreadystatechange=stateChanged; xmlHttp.open("GET",url,true); xmlHttp.send(null); } function stateChanged() { if (xmlHttp.readyState==4) { document.getElementById("txtHint").innerHTML=xmlHttp.responseText; } } function GetXmlHttpObject() { var xmlHttp=null; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { // Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } } return xmlHttp; } </script> <a href="#" onclick="showHint(http://farm2.static.flickr.com/1123/847641124_05ad612bf7.jpg)">here</a> <span id="txtHint"></span> And this is the code of img.php <?php echo "<img src=\"{$_GET['img']}\">"; ?> the code won't work. when I paste the source code of the page into frontpage, it said I got an error near onclick. When I replace the link of the image by a random number or any other character, it works just fine. why? thanks. Quote Link to comment Share on other sites More sharing options...
becu Posted July 18, 2007 Author Share Posted July 18, 2007 i got it worked! need single quote around the url. thanks DrDoc Quote Link to comment Share on other sites More sharing options...
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