在jqgrid列中显示数据

Question

我正在使用Struts 2.3.7，struts2-jquery-grid-plugin-3.5.0和hibernate 3.6。我使用一对多映射来连接两个实体。 这些实体

Issue.java

    public class Issue implements Serializable
    {
    private Integer issue_id;
    private String  issue_description;
    private Date issue_raised_date;
    private Set<Issue_Tracker> issueTracker = new HashSet<Issue_Tracker>(0);

getter and setters

Issue_Tracker.java

public class Issue_Tracker implements Serializable
{  
     private Integer issue_id;
     private String assignedTo;

如何在 assignedTo 列的jqgrid中显示此数据。我的jqgrid如下：

<sjg:grid
    id="gridtable"
    caption="Issue-Summary"
    dataType="json"
    href="%{remoteurl}"
    pager="true"
    gridModel="gridModel"
    rowList="10,15,20"
    rowNum="15"
    rownumbers="true"
    reloadTopics="reloadGrid"
>
    <sjg:gridColumn name="issue_id"  id="issueId"  index="id" title="Issue-ID" formatter="integer"  sortable="false"/>
    <sjg:gridColumn name="issue_description" index="id" title="Issue-Details"  sortable="false"/>
    <sjg:gridColumn name="issue_raised_date" index="date" title="Issue-Date"  formatter="date"  sortable="false"/>

    <sjg:gridColumn name="issueTracker"  index="assigned" title="Assigned To"  sortable="false"/>
</sjg:grid>

但输出如下，

如何在 assignedTo 列中显示实际数据而不是此对象。

Answer 1

假设assignedTo不是多个

试试这个

更改

 private Set<Issue_Tracker> issueTracker = new HashSet<Issue_Tracker>(0);

要

  public  IssueTracker issueTracker;

在你的jsp页面中

更改

 <sjg:gridColumn name="issueTracker"  index="assigned" title="Assigned To"  sortable="false"/>

要

 <sjg:gridColumn name="issueTracker.assignedTo"  index="issueTracker.assigned" title="Assigned To"  sortable="false"/>