我遇到了在頁面上顯示按鈕的問題。有兩個按鈕叫做「上傳」和「保存」。開始時「上傳」按鈕是可見的,而保存按鈕具有.setVisible(false)。Wicket:在Ajax響應中顯示按鈕
…
<tr>
<td width="35%" align="right">
<input type="submit" wicket:id="createUploadButton" value="Upload" class="ui-button ui-button-text-only ui-widget ui-state-default ui-corner-all"/>
</td>
<td width="30%" align="right">
</td>
<td width="35%" align="left">
<input type="submit" wicket:id="createCancelButton" value="Cancel" class="ui-button ui-button-text-only ui-widget ui-state-default ui-corner-all"/>
</td>
</tr>
在上傳AjaxRequest按鈕期間,需要顯示「保存」按鈕並隱藏上傳按鈕,但出現錯誤。代碼片段如下所示:
AjaxButton createSaveButton=new IndicatingAjaxButton("createSaveButton"){
private static final long serialVersionUID = 1L;
@Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
// TODO Auto-generated method stub
}
};
createSaveButton.setVisible(uploaded);
createSaveButton.setOutputMarkupId(true);
form.add(createSaveButton);
AjaxButton createUploadButton=new IndicatingAjaxButton("createUploadButton"){
private static final long serialVersionUID = 1L;
@Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
…
createUploadButton.setVisible(false);
createSaveButton.setVisible(true);
target.addComponent(createUploadButton);
target.addComponent(createSaveButton);
}
createUploadButton.setOutputMarkupId(true);
form.add(createUploadButton);
有人知道問題在哪裏嗎?
謝謝! Sonja
你能正確格式化你的代碼嗎? – ireddick 2010-08-05 12:57:09