收听DataSource操作并通知SmartGWT中的侦听器

时间:2011-10-28 09:01:19

标签: gwt smartgwt

想象一下,DataSource字段依赖于另一个DataSource的值。如何使用该DataSource的每个ListGrid自动通知任何更改? 如果可能,无需重绘整个ListGrid,只需重新绘制受影响的记录?

将以下类视为 observer

public class ObserverDataSource extends DataSource {

    public ObserverDataSource() {
    // This field needs to update/notify every ListGrid that uses 
    // this DataSource when a change occurs in the CountryDataSource.
        DataSourceField countryField = new DataSourceTextField("country", "Country");
        addField(countryField);
        // Other fields...
    }

    public void update() {
    // invalidateCache() doesn't work on its own.
    // What will make each object (ListGrid) that uses the DataSource refresh itself?
    // Even better if it only refreshes the changed records.
    // E.g. now a full redraw of the ListGrid.
    }

}

我们的可观察 DataSource:

public class ObservableDataSource extends DataSource {

    public ObservableDataSource() { 
    DataSourceField idField = new DataSourceIntegerField("id", "Id");
    idField.setPrimaryKey(true);       
    DataSourceField countryField = new DataSourceTextField("country", "Country");
    DataSourceField codeField = new DataSourceIntegerField("code", "Country code");
    setFields(idField, countryField, codeField);
    }

    public executeFetch(...) {
    // Doesn't change anything, don't notify observers.
    // Do logic...
    processResponse(requestId, response);
    }

    public executeAdd(...) {
    // Changed the data, notify the observer (MyDataSource instance).
    // Do logic...
    myDataSourceInstance.update();
    processResponse(requestId, response);
    }

}

注意: DataSource模板基于GwtRpcDataSource,可以是found here

1 个答案:

答案 0 :(得分:0)

经过一些搜索和实验,我发现你不能轻易地通知DataSource的每个用户。调用fireCallback时,数据源会在每个单独的侦听器上调用processResponse。唉,我无法重新创建功能,但幸运的是我找到了适合这种情况的解决方法:

您需要手动使每个侦听器的缓存无效。对于ListGrid,可以使用invalidateCache,但使用invalidateDisplayValueCache进行ComboBoxItem。 E.g:

public class ObserverDataSource extends DataSource {

    private Set<MyChangeHandler> handlers = new HashSet<MyChangeHandler>();

    public void update() {
        for(MyChangeHandler handler : handlers){
            handler.onChanged(/* Calls invalidateCache on itself */);
        }
    }

}

其他可能的解决方案可能是重绘整个ListGrid,但这更有效。