道场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)'/> 

当我开始输入到FilteringSelect来调用的URL，并返回

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

，但没有被填充进过滤选择 - 服务器需要返回的json的格式是什么？

Answer 1

尝试返回一组对象。快速测试是将您的JsonRest目标修改为一个简单的php文件。如果您的服务器进程名“.php”文件，例如，创建一个名为像的welcome.php有以下内容的文件：

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

然后改变你的JsonRest目标是的welcome.php