如何使用jquery ajax()回调后获取一些值

本文介绍了如何使用jquery ajax()回调后获取一些值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

尊敬的专家，

在mvc3-i中，我希望某些东西在后台执行，并且其回调再次返回到同一页面，得到2件事-"Exception"& 成功".为此，我正在使用Ajax()-

Dear Experts,

In mvc3-i want some stuff to be perform in background and its callback comes again to the same page getting 2 things-"Exception" & "IsSuccessfull". For that i am using Ajax()-

$.ajax({
                        url: '//localhost:1787/Table/Index/',
                        type: "POST",
                        data: JSON.stringify(tb),
                        dataType: "json",
                        contentType: "application/json; charset=utf-8",
                        async:true,
                        success: function (data) {
                            $('.result').html(data);
                            alert('success');
                        },
                        error: function () { alert("error"); }
                    });

此动作方法工作正常. TableModel有2个属性Error(类型异常)和IsSuccessfull(布尔)，这些是我实际上想要的属性.因此，在ajax()的error function()中，我可以获得Error属性.怎么走，不知道?

This action method working fine. TableModel have 2 property Error(of type exception) and IsSuccessfull(bool),these are the properties which i actually want..so that in error function() of ajax() i can get the Error property. how to go,don''t know??

public ActionResult Index(TableModel tb)
        {
            string tablename = tb.TableName;
            TableModel tableModel = new TableModel();
            try
            {
               ....doning some stuff, getting the result in "var result".
               if (!result.IsOperationSuccessful)
               {
                   tableModel.Error=result.Error;
               }
            }
            catch (ArgumentNullException ex)
            {
                tableModel.Error = ex;
            }
            catch (ArgumentException ex)
            {
                tableModel.Error = ex;
            }
            return View(tableModel);
        }

推荐答案

public ActionResult Index(TableModel tb)
        {
            string tablename = tb.TableName;
            TableModel tableModel = new TableModel();
            try
            {
               ....doning some stuff, getting the result in "var result".
               if (!result.IsOperationSuccessful)
               {
                   tableModel.Error=result.Error;
               }
            }
            catch (ArgumentNullException ex)
            {
                tableModel.Error = ex;
            }
            catch (ArgumentException ex)
            {
                tableModel.Error = ex;
            }
            return View(tableModel);
        }

这篇关于如何使用jquery ajax()回调后获取一些值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！