在我的前端JavaScript应用程序中,我发出了一个从服务器获取数据的ajax请求.一旦我获得数据,我想将该信息返回给视图.
var view_data;
$.ajax({
url:"/getDataFromServer.json",
//async: false,
type: "POST",
dataType: "json",
success:function(response_data_json) {
view_data = response_data_json.view_data;
console.log(view_data); //Shows the correct piece of @R_761_4045@ion
return view_data; // Does not work. Returns empty data
}
});
// return view_data; --> Keeping the return outside of the ajax call works but then I need to make my ajax call synchronous in order for this return clause to be executed after the ajax call fetches data.
我该怎么做?
解决方法:
而不是从成功返回数据:将数据传递给函数.
var view_data;
$.ajax({
url:"/getDataFromServer.json",
//async: false,
type: "POST",
dataType: "json",
success:function(response_data_json) {
view_data = response_data_json.view_data;
console.log(view_data); //Shows the correct piece of @R_761_4045@ion
doWork(view_data); // Pass data to a function
}
});
function doWork(data)
{
//perform work here
}
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 [email protected] 举报,一经查实,本站将立刻删除。