使用“selectRecords(Record record)”方法在 SmartGWT 中设置 ListGrid 选择

发布于 2024-10-26 02:33:35 字数 1166 浏览 1 评论 0原文

我正在尝试使用记录在 SmartGWT 中设置 ListGrid 表对象的选定记录，但我找不到任何方法。我想选择记录，而不是索引。我想使用 selectRecord(Record record) 方法。

举个例子；

public void onModuleLoad() 
{
    VLayout main = new VLayout();
    final ListGrid grid = new ListGrid();
    grid.setHeight(500);
    grid.setWidth(400);
    grid.setFields(new ListGridField("name", "Name"));
    grid.setData(createRecords());

    final IButton button = new IButton("Select some");  
    button.addClickHandler(new ClickHandler() {  
        public void onClick(ClickEvent event) 
        {  
          grid.selectRecord(createRecord("orange")); 
        }  
    });  

    main.addMember(grid);
    main.addMember(button);
    RootPanel.get().add(main);
}

private ListGridRecord[] createRecords() 
{
    return new ListGridRecord[]{
      createRecord("monkey"),
      createRecord("banana"),
      createRecord("orange"),
      createRecord("sun")
    };
}

private ListGridRecord createRecord(String name) 
{
    ListGridRecord record = new ListGridRecord();
    record.setAttribute("name", name);
    return record;
}

在本例中，我想选择橙色，但此代码选择任何内容。有可能吗？如果可以的话怎么办？

提前致谢。

原文

I'm trying to set the selected records of a ListGrid table object in SmartGWT using records, but I can't find any way of doing it. I want to select with record, not index. I want to use selectRecord(Record record) method.

As an example;

public void onModuleLoad() 
{
    VLayout main = new VLayout();
    final ListGrid grid = new ListGrid();
    grid.setHeight(500);
    grid.setWidth(400);
    grid.setFields(new ListGridField("name", "Name"));
    grid.setData(createRecords());

    final IButton button = new IButton("Select some");  
    button.addClickHandler(new ClickHandler() {  
        public void onClick(ClickEvent event) 
        {  
          grid.selectRecord(createRecord("orange")); 
        }  
    });  

    main.addMember(grid);
    main.addMember(button);
    RootPanel.get().add(main);
}

private ListGridRecord[] createRecords() 
{
    return new ListGridRecord[]{
      createRecord("monkey"),
      createRecord("banana"),
      createRecord("orange"),
      createRecord("sun")
    };
}

private ListGridRecord createRecord(String name) 
{
    ListGridRecord record = new ListGridRecord();
    record.setAttribute("name", name);
    return record;
}

In this case I want to select orange, But this code select anything.
Is it posible? If possible how?

Thanks in advance.

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远昼 2024-11-02 02:33:35

找到了这个解决方案；

selectRecord(grid.getRecordList().find("name", "orange"));

Found this solution;

selectRecord(grid.getRecordList().find("name", "orange"));

回复收藏 0 原文

手长情犹 2024-11-02 02:33:35

你的代码有问题：
当您编写

grid.selectRecord(record);

时，它会搜索网格具有的同一记录实例。如果记录的两个实例相等，则仅 &然后它选择记录。否则什么也不会发生，就像你现在所面临的那样。这里你需要做的是：

ListGridRecord[] records = countryGrid.getRecords();
int i;
for (i = 0; i < records.length; i++)
{
    if (records[i].getAttribute("name").equalsIgnoreCase("orange"))
    {
        break;
    }
}
countryGrid.selectRecord(i);

There's a problem with your code:
When you write

grid.selectRecord(record);

it goes to search the same record instance that the grid has. If both instances of record are equal, only then & then it selects the record. Otherwise nothing happens as you're facing right now. Here what you need to do is:

ListGridRecord[] records = countryGrid.getRecords();
int i;
for (i = 0; i < records.length; i++)
{
    if (records[i].getAttribute("name").equalsIgnoreCase("orange"))
    {
        break;
    }
}
countryGrid.selectRecord(i);

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