4

使用 JavaScript 将 JSON A 转换为 JSON B 的最简单方法是什么？

JSON A：

{
    "d":
    [
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"0","value":"one"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"1","value":"two"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"2","value":"three"}
    ]
}

JSON B：

{
    data:
    [
        {"key":"1", "value":"one"},
        {"key":"2", "value":"two"},
        {"key":"3", "value":"three"}
    ]
}

====================

2012 年 8 月 1 日更新（使用 Ext JS 并且您有 ASP.NET 代理时回答：

我没有在关于我为 JavaScript 框架使用什么的问题中提供这一点，但事实证明，您可以通过在根属性中指定值“d”来隐式消除“d”键

var statusDropdownStore = new Ext.data.Store({
    proxy: new Ext.ux.AspWebAjaxProxy({
        url: '/track/Controls/Shared/GeneralService.asmx/GetDropdownOptions',
        actionMethods: {
            create: 'POST',
            destroy: 'DELETE',
            read: 'POST',
            update: 'POST'
        },
        extraParams: {
            user_login: authUser,
            table_name: '[status]'
        },
        reader: {
            type: 'json',
            model: 'DropdownOption',
            root: 'd'
        },
        headers: {
            'Content-Type': 'application/json; charset=utf-8'
        }
    })
});

代理：

Ext.define('Ext.ux.AspWebAjaxProxy', {
    extend: 'Ext.data.proxy.Ajax',
    require: 'Ext.data',

    buildRequest: function (operation) {
        var params = Ext.applyIf(operation.params || {}, this.extraParams || {}),
                                request;
        params = Ext.applyIf(params, this.getParams(params, operation));
        if (operation.id && !params.id) {
            params.id = operation.id;
        }

        params = Ext.JSON.encode(params);

        request = Ext.create('Ext.data.Request', {
            params: params,
            action: operation.action,
            records: operation.records,
            operation: operation,
            url: operation.url
        });
        request.url = this.buildUrl(request);
        operation.request = request;
        return request;
    }
});

组合框（下拉）配置：

                    {
                        xtype: 'combo',
                        fieldLabel: 'Status',
                        emptyText: 'Select a status...',
                        store: statusDropdownStore,
                        valueField: 'key',
                        displayField: 'value',
                        mode: 'remote',  // or 'local'
                        renderTo: document.body
                    },
4

4 回答 4

2

这是一个示例

var old = {
    "d":
    [
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"0","value":"one"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"1","value":"two"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"2","value":"three"}
    ]
};

old.data = old.d;
delete old.d;
for(var i=0,l=old.data.length;i<l;i++){
    delete old.data[i].__type;
}
于 2012-08-01T15:12:58.497 回答
1

您可以尝试https://stackoverflow.com/a/1219633/832457中概述的解决方案-delete用于删除密钥。

尝试遍历数组并使用delete，然后重命名数组（通过创建新属性并删除旧属性

于 2012-08-01T15:06:45.857 回答
1

我认为这会做到： 

var TheJsonA = JSON.parse(JsonA);
TheJsonA = TheJsonA.d;

var TheJsonB = {};
TheJsonB.data = [];
var TheObject = {};

if (TheJsonA.length > 0) { 

  for (var i = 0, LoopTimes = TheJsonA.length; i < LoopTimes; i++) {
      TheObject = {};
      TheObject.key = TheJsonA[i].key;
      TheObject.value = TheJsonA[i].value;
      TheJsonB.data.push(TheObject);
  }
}

TheJsonA = null; // if you need to discard the initial object

我也这个 JsonB 不应该是一个包含对象数组的对象；我认为它应该只是一个像这样的对象数组：

[
     {"key":"1", "value":"one"},
     {"key":"2", "value":"two"},
     {"key":"3", "value":"three"}
]
于 2012-08-01T15:25:45.957 回答
1

试试这个:) http://jsfiddle.net/daewon/LnpXb/

var jsonA = {
    d: [
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"0","value":"one"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"1","value":"two"},
        {"__type":"Web.Controls.Shared.GeneralService+DropdownKeyValuePair","key":"2","value":"three"}
   ]
};

var jsonB = {
    data: []
};

var d = jsonA.d;
for (var i=0; i<​d.length​; i++){
    var obj = {
        key : d[i].key,
        value : d[i].value
    };

    jsonB.data.push(obj);    
}
console.log(JSON.stringify(jsonB));
=> {"data":[{"key":"0","value":"one"},{"key":"1","value":"two"},{"key":"2","value":"three"}]}

​</p>

于 2012-08-01T16:00:51.160 回答