function processReqChange() { // only if req shows "loaded" if (req.readyState == 4) { // only if "OK" ...
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回复(3) 2005-07-12 16:15 来自版块 - Web 开发技术交流
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jhsea3donull(2005-07-12 16:15)
jhsea3do已自行解决,用eval /* 设置提交方法 */ WebAJAX.prototype.setReqFunc = function(oFunc) { var _oFunc = "funcWebAJAX = " + "function(){}&...(2005-07-12 16:15)
jhsea3do目前唯一的方法是 /* 执行请求 */ WebAJAX.prototype.doQuery1 = function() { try { var ajaxReq = this.core; // this.core.onreadystatechange = thi...(2005-07-12 16:15)

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