
JavaScript
不太热心的市民王大勇
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JS获取url后面的参数
var Request = new Object();Request = GetRequest();var xxxx = Request [“xxxx”];function GetRequest() { //函数var url = location.search; //获取url中"?“符后的字串var theRequest = new Object();if (url.indexOf(”?") != -1) {var str = url.substr(1);strs = str.spli原创 2021-10-18 11:26:07 · 315 阅读 · 0 评论 -
js请求url交互,获取返回值执行window.open(),打开新窗口
<script> function fengyin(){ var httpRequest = new XMLHttpRequest(); httpRequest.open('GET','http://www.baidu.com',true); httpRequest.send(); var json = null; httpRequest.onreadystatechange = function(){ //read原创 2021-07-08 09:52:04 · 2546 阅读 · 1 评论