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我試圖讓wicket在單擊保存按鈕後顯示一個信息對話框,它調用一個無法訪問AjaxRequestTarget目標的onsubmit。這裏是代碼片段檢票口onsubmit信息對話框
if (trainingmode() && !recordDecision.equalsIgnoreCase("Primary")) {
if (trainingEvalService.compareDecisions(recordDecision, recordSet.getRecordSetId())) {
System.out.println("Validity matchesMaserati: " + trainingEvalService.getTrainingEval().getActual_validity_decision_comment());
// Dialog associated with save button
dialog = new MessageDialog("dialog", "Notice", "Decision Matches " + trainingEvalService.getTrainingEval().getActual_validity_decision_comment() , DialogButtons.OK_CANCEL, DialogIcon.WARN) {
public void onClose(AjaxRequestTarget target, DialogButton button) {
}
};
dialog.open(target) // breaks here without reference to AjaxTarget
} else {
}
}
如何獲得對當前AjaxRequestTarget的引用?
那麼問題是什麼? –
我假設你正在尋找RequestCycle.get()。find(AjaxRequestTarget.class) – svenmeier
@svenmeir是你接近我試圖做到這一點下面,但得到了一個錯誤引起:java.lang.NullPointerException dialog = new MessageDialog( 「dialog」,「Notice」,「Decision Matches」+ trainingEvalService.getTrainingEval()。getActual_validity_decision_comment(),DialogButtons.OK_CANCEL,DialogIcon.WARN){public void onClose(AjaxRequestTarget target,DialogButton button){ } } dialog.open(RequestCycle.get()。find(AjaxRequestTarget.class)); – Toosmooth