Js脚本不往下执行了
var DataOfLeft = (function() {
var retcontent ="";
$.ajax({
url: 'bin/getDataConfig.php',
type: "POST",
async: false,
timeout: null,
contentType: "application/json; charset=utf-8",
data: { type:"dataConfig"
},
dataType: "json",
success: function(data)
{
if(data.result == true)
{
retcontent = eval(data.content);
}
}
})
// alert(retcontent);
return retconnect;
})();
alert(DataOfLeft);
alert(typeof(DataOfLeft));
在上面的代码中,为什么函数ajax同步执行完后,alert(DataOfLeft)和alert(typeof(DataOfLeft))都不执行了,弹不出对话框?