我如何访问 json 响应对象字段

Question

我正在尝试在响应后构建 table 并尝试将响应数据附加到 table。我的回复正确返回如下所示

Object {__type: "DomainLayer.BusinessLogic.UserOnlineCommunity", CommunityID: "15bb29d4-b451-4034-be7e-8197723e0a89", Description: "name", Detail: "yes", Link: "test"}

但我正在尝试访问描述或详细信息等字段，但是当我调试这些字段并将这些字段记录到控制台中时，就像我在下面的代码中所做的那样，我得到了未定义。我如何获得每个字段的值。

function UpdateOnlineCommunity(communityId, communityName, communityDescription, communityLink) {
var serviceData = JSON.stringify({
    'communityId': communityId,
    'communityName': communityName,
    'communityDescription': communityDescription,
    'communityLink': communityLink
});
$.ajax({
    type: "POST",
    url: "PresentationService.asmx/UpdateUserCommunity",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    data: serviceData,
    success: function (response) {
        var data = response.d;
        console.log(data);
        $.each(data, function () {
            var tblCommunity = $('#tblCommunity');
            console.log(this['Description']);
            var rowDescription = $('</tr></tr>');
            rowDescription.append('<td>' + this['Description'] + '</td>');

            var rowDetail = $('</tr></tr>');
            rowDetail.append('<td>' + this['Detail'] + '</td>');

            var rowLink = $('</tr></tr>');
            rowLink.append('<td>' + this['Link'] + '</td>');

            var rowUpdateBtn = $('</tr></tr>');
            rowUpdateBtn.append('<button>Update</button>');

            tblCommunity.append(rowDescription);
            tblCommunity.append(rowDetail);
            tblCommunity.append(rowLink);
        });
    },
    error: function (response) {
        alert(response);
    }
});

Answer 1

您的响应是单个对象，因此使用 each 会导致问题。例如，您可以使用 data.Description 简单地访问对象的属性。试试这个：

success: function (response) {
    var data = response.d;
    var tblCommunity = $('#tblCommunity');
    var rowDescription = $('<tr />').append('<td>' + data.Description + '</td>');
    var rowDetail = $('<tr />').append('<td>' + data.Detail + '</td>');
    var rowLink = $('<tr />').append('<td>' + data.Link + '</td>');
    var rowUpdateBtn = $('<tr />').append('<button>Update</button>');
    tblCommunity.append(rowDescription, rowDetail, rowLink);
},

Example fiddle

另请注意，您用于创建元素的语法 - </tr></tr> 不正确。

我如何访问 json 响应对象字段

How do i access json response object fields

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