显示PopupPanel

Question

三个问题：

1. 为什么我的popupPanel无法正确显示？
2. 如果多次点击，我怎么才能让它出现一次？
3. 如何向popupPanel添加“关闭”按钮？

---

function showPassword(e){
  var app = UiApp.getActiveApplication();
  var vrtPanel = app.getElementById("vrtPanel");
  //Create Spreadsheet Source

  var spSheet = SpreadsheetApp.openById('0Aur3owCpuUY-dF92dGp3c2RORGNkY011dGFnMjBXbXc');
  var spTeacherList = spSheet.getSheetByName('TeacherList');

  //Create the form elements
  var hdlTeacherName = app.createServerHandler('getTeacherName').addCallbackElement(vrtPanel);
  var lbxTeacherName = app.createListBox().setId('lbxTeacherName').setName('lbxTeacherName').addChangeHandler(hdlTeacherName);
  var lstTeacherNames = spTeacherList.getRange(1,1,spTeacherList.getLastRow(),1).getValues();
  lstTeacherNames.sort();

  for (var l = 0; l < lstTeacherNames.length; l++) {
    lbxTeacherName.addItem(lstTeacherNames[l],l);
  }

  var lblTeacherName = app.createLabel('Teacher Name:');
  var txtTeacherName = app.createTextBox().setName('txtTeacherName').setId('txtTeacherName').setVisible(false);

  var lblExt = app.createLabel('Ext:');
  var txtExt = app.createTextBox().setName('txtExt').setId('txtExt');

  var lblEmail = app.createLabel('Email:');
  var txtEmail = app.createTextBox().setName('txtEmail').setId('txtEmail');

  var lblSchool = app.createLabel('School:');
  var txtSchool = app.createTextBox().setName('txtSchool').setId('txtSchool');

  var btnCreate = app.createButton('Create Event');

  //Create validation handler
  var valSubmit = app.createServerClickHandler('valSubmit');
  valSubmit.addCallbackElement(vrtPanel);

  //Add this handler to the button
  btnCreate.addClickHandler(valSubmit);

  //Add all the elemnts to the panel
  var formTable = app.createFlexTable().setCellPadding(3);
  vrtPanel.add(formTable);
  formTable
  .setWidget(0,0,lbxTeacherName)
  .setWidget(0,1,txtExt)
  .setWidget(0,2,txtTeacherName)
  .setWidget(1,0,txtEmail)
  .setWidget(2,0,btnCreate);

  //Add all the panel to the popup
  var popPassword = app.createDecoratedPopupPanel(false, true).setId("popPassword");
  popPassword.add(vrtPanel);

  app.add(vrtPanel);
  app.add(popPassword);
  return app;
}

Answer 1

使用popupPanel的方式存在一些混乱，您必须将其添加到主面板并将要显示的表添加到弹出窗口中。 您还应该决定要显示它的位置。 以下是您的代码的相关部分：

  //Add all the elemnts to the panel
  var formTable = app.createFlexTable().setCellPadding(3);
  formTable
  .setWidget(0,0,lbxTeacherName)
  .setWidget(0,1,txtExt)
  .setWidget(0,2,txtTeacherName)
  .setWidget(1,0,txtEmail)
  .setWidget(2,0,btnCreate);

  var popPassword = app.createDecoratedPopupPanel(false, true).setPopupPosition(150,300).setId("popPassword");

  vrtPanel.add(popPassword.add(formTable));

  popPassword.show();
  return app;
} 

回答第二点：il不会出现多次，只有一次。

第3点：如果你想用一个按钮关闭它，你将不得不添加一个按钮和一个服务器处理程序来关闭它（通过使用ID获取它就像任何其他小部件一样）但它会更容易和用户友好利用自动隐藏功能，只需更改定义参数，如下所示：

  var popPassword = app.createDecoratedPopupPanel(true).setPopupPosition(150,300).setId("popPassword");

---

上例中的

编辑也可以使用

app.getElementById('popPassword').hide() 

在你的'valSubmit'函数中（由按钮调用），即使没有自动隐藏，弹出窗口也会在处理函数执行时隐藏。