提问者:小点点

自定义JSON来填充Datatable


我正在从Ajax JSON创建一个DataTable。

resultTable = $('#changeTable').DataTable({
            "order": [[0, "desc"]],
            "pageLength": 50,
            "scrollX": true,
            "lengthMenu":[[50,100,250, -1], [50, 100, 250, "All"]],
            "dom":'<"toolbar">ltipr', //write ltfipr to show a search bar
            "ajax":{
                url:"api/changes",
                "dataType":"json",
                timeout:15000
            }
    });

将创建DataTables,但它显示一个错误:

DataTables警告:表ID=CHANGETABLE-为行0,列0请求未知参数“0”。有关此错误的详细信息,请参阅http://datatables.net/TN/4

我的JSON看起来如下所示

{"data":
    [
       {"id":1,
        "createdDate":"Apr 18, 2018 4:10:58 PM",
        "source":"manual upload",
        "emailId":"manual upload",
        "attachmentId":"manual upload",
        ...,},
       {next objet}]}

在我的Java控制器中创建了这样的JSON对象:

@RequestMapping(value = "/api/changes", method = RequestMethod.GET, produces = "application/json")
    @ResponseBody
    public String getChanges(){
        Optional<List<PriceChange>> priceChangeList = pcService.findAllPriceChanges();
        JsonObject result = new JsonObject();
        if (priceChangeList.isPresent()) {
            result.add("data", new Gson().toJsonTree(priceChangeList.get()));
            return  result.toString();
        }
        return null;

    }

我不知道如何将此信息与属性一起使用,以使其适用于DataTable。有什么想法吗?


共1个答案

匿名用户

您只需要为表定义。如果你有

<table id="changeTable"></table>

将此内容添加到DataTables选项中:

resultTable = $('#changeTable').DataTable({
  ...,
  columns: [
     { data: 'id', title: 'id' },
     { data: 'createdDate', title: 'createdDate' },
     { data: 'source', title: 'source' },
     { data: 'emailId', title: 'emailId' },
     { data: 'attachmentId', title: 'attachmentId' }
   ]
})

如果您指定了部分,您可以跳过