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javascript - 如何通过java获取JS Grid值

问题描述

我正在使用 js 网格（http://js-grid.com/docs/），当用户单击提交按钮时，我想获取所有 js-grid 内容，如下代码：

<link type="text/css" rel="stylesheet" href="jsgrid.min.css" />
<link type="text/css" rel="stylesheet" href="jsgrid-theme.min.css" />

<script type="text/javascript" src="jquery-3.3.1.min.js"></script>    
<script type="text/javascript" src="jsgrid.min.js"></script>

<form name="form1"  method="post" action="MyJavaBackend">
<div id="jsGrid"></div>
<Input type="submit">
</form>
<script>
    var clients = [
        { "Name": "Otto Clay", "Age": 25, "Country": 1, "Address": "Ap #897-1459 Quam Avenue", "Married": false },

    ];

    var countries = [
        { Name: "", Id: 0 },
        { Name: "United States", Id: 1 },
        { Name: "Canada", Id: 2 },
        { Name: "United Kingdom", Id: 3 }
    ];

    $("#jsGrid").jsGrid({
        width: "100%",
        height: "400px",

        inserting: true,
        editing: true,
        sorting: true,
        paging: true,

        data: clients,

        fields: [
            { name: "Name", type: "text", width: 150, validate: "required" },
            { name: "Age", type: "number", width: 50 },
            { name: "Address", type: "text", width: 200 },
            { name: "Country", type: "select", items: countries, valueField: "Id", textField: "Name" },
            { name: "Married", type: "checkbox", title: "Is Married", sorting: false },
            { type: "control" }
        ]
    });
</script>

但我的后端文件：

public ActionForward create(ActionMapping mapping, ActionForm form,
            HttpServletRequest request,
            HttpServletResponse response)
    throws Exception {

        String[] datas = request.getParameterValues("data");// **get null**
        return mapping.findForward("create");
    }

我检查了以下文章：http: //zetcode.com/articles/jsgridservlet/

但它只得到一行编辑的记录。（我想获取所有表格数据）

用户通过 request.getParameterValues 或 request.getParameter 或.....

标签： javascriptjavajspservletsjsgrid

解决方案

在字体结束javascript：

function onSubmit(){    

    var items = $("#jsGrid").jsGrid("option", "data");//$("#jsGrid") must match <div id="jsGrid"></div>
    var json = JSON.stringify(items);
    document.form1.json.value=json;
}

<form name="form1"  method="post" action="MyJavaBackend">
<div id="jsGrid"></div>
<input type="hidden" name="json"  >
<Input type="button" onClick="onSubmit();">
</form>


在后端 Java 中：

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

.....
String json_data = request.getParameter("json");
JSONArray jsonArray = new JSONArray(json_data);
for(int i=0;i<jsonArray.length();i++) {
                    JSONObject jsonObject=jsonArray.getJSONObject(i);

                    String name=jsonObject.getString("Name");
                    String age=jsonObject.getString("Age");
                }
....

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