ExtJS的菜單從數據庫綁定

Question

請提供一下，如何動態地從Json的綁定菜單一些示例代碼或想法導致 我從數據庫中獲得的結果爲JSON，因此如何從JSON綁定菜單（家長和孩子的）

在此先感謝

Answer 1

其實很容易。當您從服務器返回數據時，您需要做的只是在您的JSON中包含一個定義記錄結構的元數據字段。

看到這個文檔：http://dev.sencha.com/deploy/ext-3.3.1/docs/?class=Ext.data.JsonReader

該示例從文檔如下：

{ 

    metaData: { 
     "idProperty": "id", 
     "root": "rows", 
     "totalProperty": "results" 
     "successProperty": "success", 
     "fields": [ 
      {"name": "name"}, 
      {"name": "job", "mapping": "occupation"} 
     ], 
     // used by store to set its sortInfo 
     "sortInfo":{ 
      "field": "name", 
      "direction": "ASC" 
     }, 
     // paging data (if applicable) 
     "start": 0, 
     "limit": 2, 
     // custom property 
     "foo": "bar" 
    }, 
    // Reader's configured successProperty 
    "success": true, 
    // Reader's configured totalProperty 
    "results": 2000, 
    // Reader's configured root 
    // (this data simulates 2 results per page) 
    "rows": [ // *Note: this must be an Array 
     { "id": 1, "name": "Bill", "occupation": "Gardener" }, 
     { "id": 2, "name": "Ben", "occupation": "Horticulturalist" } 
    ] 
}