我有以下
PHP代码片段:
原文链接:https://www.f2er.com/php/240188.htmlif (!($result = MysqL_query($query,$link))) { die("Invalid sql query: " . $query); }
我有以下JQuery代码片段:
$.ajax({ url: "....search.PHP",data: ...,async: false,//to trigger error alert success: function(xml) { ... },error: function(xml) { foundError = true; },dataType: "xml" }); if(foundError) { setProgress("Could not complete the search because an error was found",ProgressBar.ERROR); }