我正在寻找一段简单的代码,在我的Google Apps Script Ui中添加一个弹出窗口,当我点击提交按钮时会弹出。弹出框将显示一条消息,并有一个关闭弹出窗口的按钮。
我已经到处查找了 - 似乎所有内容都非常复杂,而且做得比我需要的要多。
这是我目前拥有的提交按钮的代码。
function doGet() {
var app = UiApp.createApplication();
app.setTitle("My Logbook");
var hPanel_01 = app.createHorizontalPanel();
var vPanel_01 = app.createVerticalPanel();
var vPanel_02 = app.createVerticalPanel();
var vPanel_03 = app.createVerticalPanel();
var submitButton = app.createButton("Submit");
//Create click handler
var clickHandler = app.createServerHandler("submitData");
submitButton.addClickHandler(clickHandler);
clickHandler.addCallbackElement(hPanel_01);
////Test PopUp Panel
var app = UiApp.getActiveApplication();
var app = UiApp.createApplication;
var dialog = app.createDialogBox();
var closeHandler = app.createClientHandler().forTargets(dialog).setVisible(false);
submitButton.addClickHandler(closeHandler);
var button= app.createButton('Close').addClickHandler(closeHandler);
dialog.add(button);
app.add(dialog);
//////
return app;
}
alert("msg");
呢?为什么要让谷歌变得如此复杂呢? - Omar