我有一个用户对话框提示,用户输入将在代码的下一步中使用。我试图使用OnDismiss使其余的代码接受用户输入。以下是代码段:
AlertDialog.Builder alert = new AlertDialog.Builder(this);
alert.setTitle("Title");
alert.setMessage("Message");
alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 3;
}
});
alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 1;
}
});
alert.show();
}
public void OnDismiss(DialogInterface dialogInterface)
{
Toast.makeText(this, "Input: " + input, Toast.LENGTH_LONG)
.show();
}
但是,当我按下对话框按钮时,对话框将被取消,但不显示吐司。请在这里建议我做错了什么。
答案 0 :(得分:1)
首先将OnDismissListener
设置为AlertDialog
,然后在按下对话框按钮时调用dismiss()
,代码如下:
AlertDialog.Builder alert = new AlertDialog.Builder(this);
alert.setTitle("Title");
alert.setMessage("Message");
alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 3;
dialog.dismiss();
}
});
alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 1;
dialog.dismiss();
}
});
alert.setOnDismissListener(new OnDismissListener(){
public void OnDismiss(DialogInterface dialogInterface)
{
Toast.makeText(this, "Input: " + input, Toast.LENGTH_LONG)
.show();
}
});
alert.show();
答案 1 :(得分:0)
你必须使用dialog.dismiss();
alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 3;
dialog.dismiss();
}
});
alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
input = 1;
dialog.dismiss();
}
});