在计时器事件中,我想再次使该组件可见。有什么不同的方法来完成这个任务吗?我附上了狙击手亲爱的。
HTML 标记:在此文件中我创建警报栏
<div Class="row" wicket:id="alert_app" >
<div class="alert alert-info" role="alert" style="top: 2%; left: 50%; position: absolute;">
<a href="#" wicket:id="alert" style="color:red" >Alert - Match Found</a>
</div>
</div>
Java类:我已经实例化了WebMarkupContainer并在初始阶段使其不可见,在5秒警报栏出现后(这是我的计划),因为我使用了计时器并且陷入了计时器事件。
WebMarkupContainer informationBox = new WebMarkupContainer("alert_app");
add(informationBox);
final AjaxLink saveProfile = new AjaxLink("alert") {
private static final long serialVersionUID = 1L;
public void onClick(AjaxRequestTarget target) {
this.setResponsePage(ABC.class);
}
};
informationBox.add(saveProfile);
informationBox.setVisible(false);
Timer timer = new Timer();
timer.schedule(new TimerTask() {
@Override
public void run() {
informationBox.setVisibilityAllowed(true);
informationBox.setVisible(true); // I got error in this line
}
}, 5000);
错误:
12:37:04,735 INFO
[org.apache.wicket.response.filter.AjaxServerAndClientTimeFilter] (http-
localhost-127.0.0.1-8080-4) 1ms server time taken for request
wicket/bookmarkable/xyz.abc.vbn.class?
4&username=user+name response size: 9386
12:37:09,736 ERROR [stderr] (Timer-3) Exception in thread "Timer-3"
org.apache.wicket.WicketRuntimeException: No RequestCycle is currently set!
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.Component.getRequest(Component.java:1791)
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.markup.html.WebPage.dirty(WebPage.java:334)
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.Page.dirty(Page.java:248)
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.Page.componentStateChanging(Page.java:937)
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.Component.addStateChange(Component.java:3512)
12:37:09,736 ERROR [stderr] (Timer-3) at
org.apache.wicket.Component.setVisible(Component.java:3195)
12:37:09,736 ERROR [stderr] (Timer-3) at
java.util.TimerThread.mainLoop(Timer.java:555)
12:37:09,736 ERROR [stderr] (Timer-3) at
java.util.TimerThread.run(Timer.java:505)
最佳答案
您不能只使用java.util.Timer
。它启动一个不是 HTTP 工作线程的新线程,因此 WicketFilter 没有机会设置 ThreadLocal(应用程序、 session 和 RequestCysle)。
您必须使用org.apache.wicket.ajax.AbstractAjaxTimerBehavior
。
它将在给定的持续时间后触发一个新的 Ajax 请求,一切都会按预期工作!
关于java - 如何在类的 Timer 方法中可见 HTML 标记组件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45188579/