道場FilterSelect不JsonRest

Question

填充嗨，我有以下

<script type='text/javascript'>if (dijit.byId('assignedUserId') != undefined) { 
    dijit.byId('assignedUserId').destroy();}require([ 'dojo/store/JsonRest',  
'dijit/form/FilteringSelect', 'dojo/domReady!'], function(JsonRest, FilteringSelect){ 
    var jsonRest =  new JsonRest({  target: 'Welcome.do?call=JS&actionRefId=142' }); 
    var filteringSelect = new  FilteringSelect({  id: 'assignedUserId',  name: 
'assignedUserId',  value: '25',    store: jsonRest,  searchAttr: 'name', 
labelAttr: 'label' },  'assignedUserIdSelect').startup();}); 
</script> 
<input id='assignedUserIdSelect' name='value(assignedUserId)'/> 

當我開始輸入到FilteringSelect來調用的URL，並返回

{"identifier": "id", "label": "label", "items": [{ "name": "", "id": "0" , "label": "" },{ 
"name": "Lea M Test", "id": "26" , "label": "Lea M Test" }]} 

，但沒有被填充進過濾選擇 - 服務器需要返回的json的格式是什麼？

Answer 1

嘗試返回一組對象。快速測試是將您的JsonRest目標修改爲一個簡單的php文件。如果您的服務器進程名「.php」文件，例如，創建一個名爲像的welcome.php有以下內容的文件：

<?php 
    $data = '[{ "name": "Abi Normal", "id": "0" , "label": "Abi Normal" },{"name": "Lea M Test", "id": "26" , "label": "Lea M Test" }]'; 
    echo $data; 
?> 

然後改變你的JsonRest目標是的welcome.php