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关于js调用ajax和弹出confirm提示框的有关问题

热度:44   发布时间:2016-05-05 04:50:19.0
关于js调用ajax和弹出confirm提示框的问题
用ajax获取出一个值后,弹出带有该值的confirm提示框,
问题是,提示框中没有获取到值,点confirm提示框的确定之后,才出来这个值

代码如下
                       var sturenshu = null;
                        try {
                            // Firefox, Opera 8.0+, Safari
                            var xml = new XMLHttpRequest();
                        }
                        catch (e) {
                            // Internet Explorer
                            try {
                                xml = new ActiveXObject("Msxml2.XMLHTTP");
                            }
                            catch (e) {
                                try {
                                    xml = new ActiveXObject("Microsoft.XMLHTTP");
                                }
                                catch (e) {
                                    alert("您的浏览器不支持AJAX!");
                                    return false;
                                }
                            }
                        }
                        xml.open("GET", "ajax.ashx?id=" + frm.elements[i].value + "&t=" + Math.random(), true);
                        xml.onreadystatechange = function () {
                            if (xml.readyState == 4 && xml.status == 200) {
                                sturenshu = xml.responseText;
                            }
                        }
                        xml.send();
                    }
                }
            }
            var returnVal = window.confirm("" + sturenshu + "个人,您确定吗?", "标题");
            if (!returnVal) {
                return false;
            }
请问我应该怎样在confirm提示框中获取ajax的值呢?各位大虾请指教,多谢多谢~!
------解决思路----------------------
  xml.open("GET", "ajax.ashx?id=" + frm.elements[i].value + "&t=" + Math.random(), false);


------解决思路----------------------
把你的

var returnVal = window.confirm("" + sturenshu + "个人,您确定吗?", "标题");
            if (!returnVal) {
                return false;
            }

放入到

if (xml.readyState == 4 && xml.status == 200) {
里面去

或者你可以async=false,强制禁止ajax异步
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