异常在android系统对话框内创建对话框 [英] Exception creating dialog inside a dialog in android

问题描述

我有一对夫妇一个对话框，对话框内被抛出异常的：

I have a couple of dialogs inside a dialog that are throwing the exception:

02-10 15:52:45.592: ERROR/AndroidRuntime(321): java.lang.IllegalArgumentException: Activity#onCreateDialog did not create a dialog for id 2

在code是：

dialog.setButton(DialogInterface.BUTTON_POSITIVE, "Yes",
            new DialogInterface.OnClickListener() {
                public void onClick(DialogInterface dialog, int which) {

                ....

           showDialog(ID_DIALOG_SEND);

对话框定义如下：

The dialog is defined as follows:

 @Override
 protected Dialog onCreateDialog(int id) {
        switch (id) {

    case ID_DIALOG_SEND:

        ProgressDialog loadingDialog = new ProgressDialog(this);
        loadingDialog.setMessage("Sending...");
        loadingDialog.setIndeterminate(true);
        loadingDialog.setCancelable(false);
        return null;
      ....

和它并不适用于该对话框工作，要么：

And it doesn't work for this dialog either:

Builder b = new AlertDialog.Builder(this);

case ID_DIALOG_ERR:
b.setMessage("Error")
        .setCancelable(false)
        .setPositiveButton("OK",
            new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int which) {
                    }
                });
        return null;

任何提示？

感谢

推荐答案

由于您要使用的ShowDialog（ID_DIALOG_SEND）; 在一个匿名内部类，它会给你一个问题，因为它认为ShowDialog的是内部类的方法。您需要引用外Activity类，其中包含为它工作。所以你的情况我会做：

Because you are trying to use showDialog(ID_DIALOG_SEND); in an Anonymous Inner class, it will give you a problem, as it thinks ShowDialog is a method of the inner class. You need to reference the outer Activity class, in which it is contained for it to work. So in your case i would do :

dialog.setButton(DialogInterface.BUTTON_POSITIVE, "Yes", 
 new DialogInterface.OnClickListener() {                 
    public void onClick(DialogInterface dialog, int which) {

           ....             

     thisActivityClassName.showDialog(ID_DIALOG_SEND);

或 thisActivityClassName.this.showDialog（ID_DIALOG_SEND）;

否则，如果失败，则可能需要创建一个处理程序与外部类进行通信。

Otherwise if this fails, you might need to create a Handler to communicate with the outer class.

另外，我完全同意，并重申dave.c的答复。

Also, i fully agree and reiterate dave.c's reply.

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