ajax问题

mac2022-06-30  61

1.简单模板

  success访问成功不等于返回数据成功

$.ajax({ type:"get", data:freedata, dataType:"jsonp", url:url, success:function (data) { if (data.retcode == "00") { var item = data.freesort; numNow=data.luckytimes; $rule_number.text(numNow); var note=data.freecardname.replace(",","").replace(",","").replace(",",""); rotateFn(item,note); } else { awardAlert(data.result); }; }, error:function () { awardAlert("网络超时,请稍后尝试",{bname:"取消"}); }})

转载于:https://www.cnblogs.com/QIQIZAIXIAN/p/6928440.html

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