我有一个带有自定义视图的DialogFragment,其中包含两个文本字段,用户可以在其中输入用户名和密码。单击肯定按钮时,我想验证用户在解除对话框之前确实输入了一些内容。
public class AuthenticationDialog extends DialogFragment {
public Dialog onCreateDialog(Bundle savedInstanceState) {
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
LayoutInflater inflater = getActivity().getLayoutInflater();
builder.setView(inflater.inflate(R.layout.authentication_dialog, null))
.setPositiveButton(getResources().getString(R.string.login), new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
// TODO
}
})
.setNegativeButton(getResources().getString(R.string.reset), new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
// TODO
}
});
return builder.create();
}
}
那么如何防止对话被解雇呢?我应该覆盖一些方法吗?
答案 0 :(得分:63)
覆盖OnStart()中的默认按钮处理程序来执行此操作。
@Override
public Dialog onCreateDialog(Bundle savedInstanceState)
{
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
builder.setMessage("Test for preventing dialog close");
builder.setPositiveButton("Test",
new DialogInterface.OnClickListener()
{
@Override
public void onClick(DialogInterface dialog, int which)
{
//Do nothing here because we override this button later to change the close behaviour.
//However, we still need this because on older versions of Android unless we
//pass a handler the button doesn't get instantiated
}
});
return builder.create();
}
@Override
public void onStart()
{
super.onStart(); //super.onStart() is where dialog.show() is actually called on the underlying dialog, so we have to do it after this point
AlertDialog d = (AlertDialog)getDialog();
if(d != null)
{
Button positiveButton = (Button) d.getButton(Dialog.BUTTON_POSITIVE);
positiveButton.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View v)
{
Boolean wantToCloseDialog = false;
//Do stuff, possibly set wantToCloseDialog to true then...
if(wantToCloseDialog)
dismiss();
//else dialog stays open. Make sure you have an obvious way to close the dialog especially if you set cancellable to false.
}
});
}
}
有关其他对话框类型的更多说明和示例,请参阅我的答案https://stackoverflow.com/a/15619098/579234。
答案 1 :(得分:6)
感谢Luksprog,我找到了解决方案。
<强> AuthenticationDialog.java 强>:
public class AuthenticationDialog extends DialogFragment implements OnClickListener {
public interface AuthenticationDialogListener {
void onAuthenticationLoginClicked(String username, String password);
void onAuthenticationResetClicked(String username);
}
private AuthenticationDialogListener mListener;
private EditText mUsername;
private EditText mPassword;
private Button mReset;
private Button mLogin;
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
View view = inflater.inflate(R.layout.authentication_dialog, container);
this.getDialog().setTitle(R.string.login_title);
mUsername = (EditText) view.findViewById(R.id.username_field);
mPassword = (EditText) view.findViewById(R.id.password_field);
mReset = (Button) view.findViewById(R.id.reset_button);
mLogin = (Button) view.findViewById(R.id.login_button);
mReset.setOnClickListener(this);
mLogin.setOnClickListener(this);
return view;
}
public void onAttach(Activity activity) {
super.onAttach(activity);
// Verify that the host activity implements the callback interface
try {
// Instantiate the NoticeDialogListener so we can send events to the host
mListener = (AuthenticationDialogListener) activity;
} catch (ClassCastException e) {
// The activity doesn't implement the interface, throw exception
throw new ClassCastException(activity.toString()
+ " must implement AuthenticationDialogListener");
}
}
public void onClick(View v) {
if (v.equals(mLogin)) {
if (mUsername.getText().toString().length() < 1 || !mUsername.getText().toString().contains("@")) {
Toast.makeText(getActivity(), R.string.invalid_email, Toast.LENGTH_SHORT).show();
return;
} else if (mPassword.getText().toString().length() < 1) {
Toast.makeText(getActivity(), R.string.invalid_password, Toast.LENGTH_SHORT).show();
return;
} else {
mListener.onAuthenticationLoginClicked(mUsername.getText().toString(), mPassword.getText().toString());
this.dismiss();
}
} else if (v.equals(mReset)) {
mListener.onAuthenticationResetClicked(mUsername.getText().toString());
}
}
}
<强> authentication_dialog.xml 强>:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:orientation="vertical" >
<EditText
android:id="@+id/username_field"
android:inputType="textEmailAddress"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginTop="10dp"
android:layout_marginLeft="4dp"
android:layout_marginRight="4dp"
android:layout_marginBottom="4dp"
android:hint="@string/username"
/>
<EditText
android:id="@+id/password_field"
android:inputType="textPassword"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginTop="4dp"
android:layout_marginLeft="4dp"
android:layout_marginRight="4dp"
android:layout_marginBottom="12dp"
android:fontFamily="sans-serif"
android:hint="@string/password"
/>
<View
android:layout_width="fill_parent"
android:layout_height="1dip"
android:background="?android:attr/dividerVertical"
/>
<LinearLayout
style="?android:attr/buttonBarStyle"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:orientation="horizontal"
android:paddingTop="0dp"
android:measureWithLargestChild="true" >
<Button
android:id="@+id/reset_button"
style="?android:attr/buttonBarButtonStyle"
android:layout_height="wrap_content"
android:layout_width="0dp"
android:layout_weight="1.0"
android:text="@string/reset"
/>
<Button
android:id="@+id/login_button"
style="?android:attr/buttonBarButtonStyle"
android:layout_height="wrap_content"
android:layout_width="0dp"
android:layout_weight="1.0"
android:text="@string/login"
/>
</LinearLayout>
</LinearLayout>
答案 2 :(得分:5)
这是&#34;甜点&#34;解决了Karakuri和Sogger的答案。 Karakuri在正确的轨道上,但是如果已经显示,那么你只能获得按钮(否则,如评论中所述,它是空的)。这就是Sogger的答案有效的原因,但我更喜欢使用相同的方法设置onCreateDialog
,而不是onStart
。解决方案是将按钮的提取包装到对话框的OnShowListener
中。
public Dialog onCreateDialog(Bundle savedInstanceState) {
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
// your dialog setup, just leave the OnClick-listeners empty here and use the ones below
final AlertDialog dialog = builder.create();
dialog.setOnShowListener(new DialogInterface.OnShowListener() {
@Override
public void onShow(final DialogInterface dialog) {
Button positiveButton = ((AlertDialog) dialog).getButton(DialogInterface.BUTTON_POSITIVE);
positiveButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(final View v) {
// TODO - call 'dismiss()' only if you need it
}
});
Button negativeButton = ((AlertDialog) dialog).getButton(DialogInterface.BUTTON_NEGATIVE);
// same for negative (and/or neutral) button if required
}
});
return dialog;
}
答案 3 :(得分:0)
您可以再次弹出对话框。或者,您可以禁用正按钮,直到两个字段都有输入。如果您在onCreateVew()
中创建布局,这很容易。如果您正在使用AlertDialog.Builder
类,则可以像这样获取按钮的句柄:
AlertDialog.Builder builder = new AlertDialog.Builder(context);
/* ... */
Dialog dialog = builder.create();
Button positiveButton = ((AlertDialog) dialog).getButton(DialogInterface.BUTTON_POSITIVE);
/* now you can affect the button */