我向返回数据库查询结果的 servlet 发出 ajax 请求。我可以用 firebug 看到响应,我想将这些结果放在我已经创建的列表(或其他..)中。
我试图阅读这篇文章,但它没有帮助我..
代码:
这是ajax请求:
Ext.Ajax.request({
url: '/VIProject/Container',
success: function (action){alert('Lista caricata!'); console.debug(action); },
failure: function (){alert('Errore nel caricamento...');},
headers: {
'my-header': 'foo'
},
params: { action: "GETCONTAINERLIST" }
});
来自 servlet(萤火虫)的响应:
{"message":"OK","container":[{"idOrdine":"1","numLotto":"123"},{"idOrdine":"2","numLotto":"321"},{"idOrdine":"3","numLotto":"876"}],"success":true}
列表:
var listView = Ext.create('Ext.grid.Panel', {
width:425,
height:250,
id: 'lista',
collapsible:true,
title:'Simple ListView <i>(0 items selected)</i>',
store: //???
multiSelect: true,
viewConfig: {
emptyText: 'No images to display'
},
columns: [{
text: 'idOrdine',
flex: 15,
sortable: true,
dataIndex: 'idOrdine'
},{
text: 'Last Modified',
flex: 20,
sortable: true,
dataIndex: 'numLotto'
}]
});
我能怎么做?