日期格式 JSON jqgrid

Posted 2023-05-08

【中文标题】日期格式 JSON jqgrid

【英文标题】：Date format JSON jqgrid

【发布时间】：2018-05-13 21:24:58

【问题描述】：

我正在用 Java 开发一个 Web 应用程序。我正在用 JSON 数据填充 jqgrid。问题是“生日”列显示的格式不正确。在 mysql 数据库中，该字段为 DATE 类型，但在 jqgrid 中显示为：“Dec 30, 1989”。我需要以下格式：'1989-12-30'

我的 servlet 代码：

 protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException 

    String parameter = request.getParameter("parameter");
    String username = request.getParameter("username");
    int option = Integer.parseInt(request.getParameter("option"));

    DBConnection db = new DBConnection();
    StringBuilder sb = new StringBuilder();
    Connection con = (Connection) db.getCon();
    PreparedStatement ps = null;
    ResultSet rs = null;
    ArrayList<Customers> customerList = new ArrayList<Customers>();

    try 
        sb.append("call SP_MSTCUSTOMERS_FIND(?,?,?)");
        ps = con.prepareCall(sb.toString());
        ps.setString(1, parameter);
        ps.setString(2, username);
        ps.setInt(3, option);
        rs = ps.executeQuery();
        while (rs.next()) 
            Customers customer = new Customers();
            customer.setIdMstCustomers(rs.getInt("IDMSTCUSTOMERS"));
            customer.setFirstName(rs.getString("FIRST_NAME"));
            customer.setLastName(rs.getString("LAST_NAME"));
            customer.setBirthdate(rs.getDate("BIRTHDATE"));       
            customerList.add(customer);
        
     catch (SQLException e) 
        e.printStackTrace();
    

    Gson gson = new Gson();
    JsonElement element = gson.toJsonTree(customerList, new TypeToken<List<Customers>>() 
    .getType());

    JsonArray jsonArray = element.getAsJsonArray();
    response.setContentType("application/json");
    response.getWriter().print(jsonArray);

jqgrid的代码：

 $("#list").jqGrid(
    url: 'PopulateCustomersTable',
    postData: 
        parameter: function () 
            return $("#parameter").val();
        ,
        username: function () 
            return $("#txtusrID").val();
        ,
        option: function () 
            return $("input[name='srch']:checked").val(); //$('[name="srch"]').val();
        
    ,
    datatype: "json",
    mtype: 'POST',
    colNames: ['Id', 'First Name', 'Last Name', 'Birthdate'],
    colModel: [
            name: 'IdMstCustomers',
            index: 'IdMstCustomers',
            width: 50,
            fixed: true,
            align: 'center'
        , 
            name: 'FirstName',
            index: 'FirstName',
            width: 100,
            fixed: true,
            align: 'center',
            editable: true
        , 
            name: 'LastName',
            index: 'LastName',
            width: 100,
            fixed: true,
            align: 'center',
            editable: true
        , 
            name: 'Birthdate',
            index: 'Birthdate',
            width: 100,
            editable: true,                
        ],
    pager: '#pager',
    rowNum: 10,
    rowList: [10, 20, 30],
    sortname: 'FirstName',
    sortorder: 'desc',
    viewrecords: true,
    gridview: true,
    caption: 'Customers',
    jsonReader: 
        repeatitems: false
    ,
    editurl: "CustomersServlet"
);
$('#list').trigger('reloadGrid');
$('#list').jqGrid('setGridWidth', '900'); 
jQuery("#list").jqGrid('navGrid', '#pager', 
    edit: true,
    add: false,
    del: false,
    search: true
);

The jqgrid data looks like this

我的客户类

public Customers(int IdMstCustomers, String FirstName, String LastName, Date Birthdate) 
    this.setIdMstCustomers(IdMstCustomers);
    this.setFirstName(FirstName);
    this.setLastName(LastName);
    this.setBirthdate(Birthdate);


public Customers() 


private int IdMstCustomers;
private String FirstName;
private String LastName;
private Date Birthdate;

public void setIdMstCustomers(int IdMstCustomers) 
    this.IdMstCustomers = IdMstCustomers;


public void setFirstName(String FirstName) 
    this.FirstName = FirstName;


public void setLastName(String LastName) 
    this.LastName = LastName;


public void setBirthdate(Date Birthdate) 
    this.Birthdate = Birthdate;


public int getIdMstCustomers() 
    return IdMstCustomers;


public String getFirstName() 
    return FirstName;


public String getLastName() 
    return LastName;


public Date getBirthdate() 
    return Birthdate;

【问题讨论】：

请发布您的源 JSON 数据，而不是 jqGrid 中显示的数据。还有一些其他注意事项：绝对不需要运行命令 $('#list').trigger('reloadGrid'); - 使用此命令，您在构建网格时再次重新加载网格，但什么也没做

【参考方案1】：

您可以添加formatter 以根据您所需的格式格式化日期值：

    name: 'Birthdate',
    index: 'Birthdate',
    width: 100,
    editable: true,
    formatter: 'date', formatoptions:  srcformat: 'M, d, Y', newformat: 'Y-m-d'

您可以在此处阅读更多信息：http://www.trirand.com/jqgridwiki/doku.php?id=wiki:custom_formatter

【讨论】：

对我不起作用，仍然显示不正确的格式：70707070-JanJan-0101

谢谢@Waqas！为我做了一个非常小的调整：formatoptions: srcformat: 'M d, Y', newformat: 'Y-m-d' M of month 部分后面的逗号是问题

很好，那您能把它标记为正确答案吗？谢谢