ajax - wicket 6.0.0-beta2 使用 AjaxButton 提交表单时更新 DataTable 的内容

Question

我想根据表单的内容更改 DataTable 的内容(将其视为搜索栏功能)。我曾经在 wicket 1.5.x 中这样做，但我似乎无法让它在 wicket 6.0.0-beta2 中工作。在AjaxButton的onSubmit方法中好像没有输入。其他一切正常，每个组件都正确呈现，并且在页面加载时 dataTable 填充了正确的数据，但是当我单击按钮时，没有任何 react 。

任何帮助将不胜感激。这是我的代码的样子:

数据表:

public SubscriberPage(PageParameters parameters) { 
super(parameters); 
add(new SearchForm("searchForm")); 

List<IColumn<Subscriber, String>> columns = new ArrayList<IColumn<Subscriber, String>>(); 
columns.add(new PropertyColumn<Subscriber, String>(new Model<String>("Telephone Number"), 
                                                   "tn", 
                                                   "tn")); 
[...] 
columns.add(new PropertyColumn<Subscriber, String>(new Model<String>("Initialized MB"), 
                                                   "initializedMB")); 

table = new AjaxFallbackDefaultDataTable<Subscriber, String>("table", 
                                                             columns, 
                                                             subscriberDataProvider, 
                                                             40); 
table.setOutputMarkupId(true); 
add(table); 
} 

这是带有 AjaxButton 的表单:

private class SearchForm extends Form<String> { 
private static final long serialVersionUID = 1L; 

private String tnModel; 
private Label tnLabel = new Label("tnLabel", "Telephone Number :"); 
private TextField<String> tn; 

public SearchForm(String id) { 
  super(id); 
  tn = new TextField<String>("tnTextField", new PropertyModel<String>(this, "tnModel")); 
  tn.setOutputMarkupId(true); 
  add(tnLabel); 
  add(tn); 

  AjaxButton lSearchButton = new AjaxButton("searchButton") { 
    private static final long serialVersionUID = 1L; 

    @Override 
    protected void onSubmit(AjaxRequestTarget target, Form<?> form) { 
      SubscriberFilter filter = new SubscriberFilter(); 
      target.add(table); 
      if (!(tn.getValue() == null) && !tn.getValue().isEmpty()) { 
        filter.setTn(tn.getValue()); 
      } 
      // giving the new filter to the dataProvider 
      subscriberDataProvider.setFilterState(filter); 
    } 

    @Override 
    protected void onError(AjaxRequestTarget target, Form<?> form) { 
      // TODO Implement onError(..) 
      throw new UnsupportedOperationException("Not yet implemented."); 
    } 

  }; 
  lSearchButton.setOutputMarkupId(true); 
  this.setDefaultButton(lSearchButton); 
  add(lSearchButton); 
} 
} 

Answer 1

要刷新的组件需要添加到容器中。提交时，需要将容器添加到目标。这样您的组件将被刷新。就像是:

WebMarkupContainer outputContainer = new WebMarkupContainer("searchResult");
outputContainer.setOutputMarkupId(true);
outputContainer.add(table);
add(outputContainer);

@Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
    //change table ..... stuff ..... ...

    //refresh container
    target.add(outputContainer);
}


<div wicket:id="searchResult"></div>