Chrome下xmlHttpRequest的open方法报DOMException
RT!我贴出我的代码,在IE下各种访问毫无压力.Chrome 版本 22.0.1229.94 m.
<html><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312" /><title>Login</title></head><script>function createNewRequest(){var HttpRequest = null ;try {//if (window.XMLHttpRequest) { // Mozilla, Safari, ...HttpRequest = new XMLHttpRequest();//} else if (window.ActiveXObject) { // IE//HttpRequest = new ActiveXObject("Microsoft.XMLHTTP");//}if (HttpRequest.overrideMimeType) {// set type accordingly to anticipated content typeif(type.toUpperCase() == "XML")HttpRequest.overrideMimeType('text/xml');elseHttpRequest.overrideMimeType('text/html');}} catch (e) {//showWarnningMessage(errorText, e.message);}return HttpRequest ;}function ajaxCall(urlString,data){var pageRequest = createNewRequest();pageRequest.open('GET', urlString);//pageRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');pageRequest.setRequestHeader("Content-Type","application/x-www-form-urlencoded;charset=UTF-8");var encodedData = encodeURI(data);//pageRequest.send("");pageRequest.send(encodedData);pageRequest.onreadystatechange=function(){//alert("readyState = " + pageRequest.readyState + "; HTTP Status = " + pageRequest.status);//var returnText = pageRequest.responseText ;if (pageRequest.readyState == 4 && pageRequest.status == 200){document.getElementById("f1").src="http://172.16.5.158:9080/portal/jsp/viewScoreboard.do?id=2015.209";}}}function doSubmit(){ajaxCall("http://172.16.5.158:9080/portal/j_security_check?j_username=admin&j_password=admin","abcd");}</script><body><input type="button" value="Report" onclick="doSubmit()"><iframe style="display:inline" src="" id="f1" name="f1" width=100% height=100%></iframe></body> </html>