Android 无法将TextInputText转换为TextInputLayout
我的活动中有此代码Android 无法将TextInputText转换为TextInputLayout,android,view,android-edittext,Android,View,Android Edittext,我的活动中有此代码 public class RegisterActivity extends AppCompatActivity { private static final String TAG = RegisterActivity.class.getSimpleName(); private TextInputLayout mDisplayName; private TextInputLayout mEmail; private TextInputLayout mPassword; p
public class RegisterActivity extends AppCompatActivity {
private static final String TAG = RegisterActivity.class.getSimpleName();
private TextInputLayout mDisplayName;
private TextInputLayout mEmail;
private TextInputLayout mPassword;
private Button mCreateBtn;
private FirebaseAuth mAuth;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_register);
mAuth = FirebaseAuth.getInstance();
mDisplayName = (TextInputLayout)findViewById(R.id.reg_display_name);
mEmail = (TextInputLayout)findViewById(R.id.reg_email);
mPassword = (TextInputLayout)findViewById(R.id.reg_password);
mCreateBtn = (Button) findViewById(R.id.reg_create_btn);
mCreateBtn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
String displayName = mDisplayName.getEditText().getText().toString();
String email = mEmail.getEditText().getText().toString();
String password = mPassword.getEditText().getText().toString();
registerUser(displayName,email,password);
}
});
}
private void registerUser(String displayName, String email, String password) {
mAuth.signInWithEmailAndPassword(email,password)
.addOnCompleteListener(this, new OnCompleteListener<AuthResult>() {
@Override
public void onComplete(@NonNull Task<AuthResult> task) {
if(task.isSuccessful()){
Intent mainIntent = new Intent(RegisterActivity.this,MainActivity.class);
startActivity(mainIntent);
finish();
}else{
Toast.makeText(RegisterActivity.this,"You got some error.",Toast.LENGTH_SHORT).show();
}
}
});
}
}
我还提供了gradle文件,以防设计版本出现问题
apply plugin: 'com.android.application'
android {
compileSdkVersion 27
defaultConfig {
applicationId "theo.tziomakas.lapitchat"
minSdkVersion 21
targetSdkVersion 27
versionCode 1
versionName "1.0"
testInstrumentationRunner
"android.support.test.runner.AndroidJUnitRunner"
}
buildTypes {
release {
minifyEnabled false
proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
}
}
}
dependencies {
implementation fileTree(dir: 'libs', include: ['*.jar'])
implementation 'com.android.support:appcompat-v7:27.1.1'
implementation 'com.android.support:design:27.1.1'
implementation 'com.android.support.constraint:constraint-
layout:1.1.0'
implementation 'com.google.firebase:firebase-auth:11.6.0'
testImplementation 'junit:junit:4.12'
androidTestImplementation 'com.android.support.test:runner:1.0.2'
androidTestImplementation
'com.android.support.test.espresso:espresso-core:3.0.2'
}
apply plugin: 'com.google.gms.google-services'
那么,有没有办法解决这个异常呢
谢谢
西奥。试试这个:
public class RegisterActivity extends AppCompatActivity {
private static final String TAG = RegisterActivity.class.getSimpleName();
private TextInputEditText mDisplayName;
private TextInputEditText mEmail;
private TextInputEditText mPassword;
private Button mCreateBtn;
private FirebaseAuth mAuth;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_register);
mAuth = FirebaseAuth.getInstance();
mDisplayName = (TextInputEditText)findViewById(R.id.reg_display_name);
mEmail = (TextInputEditText)findViewById(R.id.reg_email);
mPassword = (TextInputEditText)findViewById(R.id.reg_password);
mCreateBtn = (Button) findViewById(R.id.reg_create_btn);
mCreateBtn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
String displayName = mDisplayName.getEditText().getText().toString();
String email = mEmail.getEditText().getText().toString();
String password = mPassword.getEditText().getText().toString();
registerUser(displayName,email,password);
}
});
}
private void registerUser(String displayName, String email, String password) {
mAuth.signInWithEmailAndPassword(email,password)
.addOnCompleteListener(this, new OnCompleteListener<AuthResult>() {
@Override
public void onComplete(@NonNull Task<AuthResult> task) {
if(task.isSuccessful()){
Intent mainIntent = new Intent(RegisterActivity.this,MainActivity.class);
startActivity(mainIntent);
finish();
}else{
Toast.makeText(RegisterActivity.this,"You got some error.",Toast.LENGTH_SHORT).show();
}
}
});
}
}
公共类注册表活动扩展了AppCompatActivity{
私有静态最终字符串标记=RegisterActivity.class.getSimpleName();
私有文本输入文本mDisplayName;
私有文本输入文本mEmail;
私有文本输入文本mPassword;
专用按钮mCreateBtn;
私人消防队;
@凌驾
创建时受保护的void(Bundle savedInstanceState){
super.onCreate(savedInstanceState);
setContentView(R.layout.activity\u寄存器);
mAuth=FirebaseAuth.getInstance();
mDisplayName=(textInputItemText)findViewById(R.id.reg\u display\u name);
mEmail=(textinputtext)findviewbyd(R.id.reg_电子邮件);
mPassword=(textinputtext)findviewbyd(R.id.reg\u密码);
mCreateBtn=(按钮)findViewById(R.id.reg\u create\u btn);
mCreateBtn.setOnClickListener(新视图.OnClickListener(){
@凌驾
公共void onClick(视图){
String displayName=mDisplayName.getEditText().getText().toString();
字符串email=mEmail.getEditText().getText().toString();
字符串密码=mPassword.getEditText().getText().toString();
注册服务器(显示名称、电子邮件、密码);
}
});
}
私有无效注册表服务器(字符串显示名称、字符串电子邮件、字符串密码){
mAuth.使用电子邮件和密码登录(电子邮件,密码)
.addOnCompleteListener(这是新的OnCompleteListener(){
@凌驾
未完成的公共void(@NonNull任务){
if(task.issusccessful()){
Intent maintent=新的Intent(RegisterActivity.this,MainActivity.class);
星触觉(主旨);
完成();
}否则{
Toast.makeText(RegisterActivity.this,“您有一些错误。”,Toast.LENGTH_SHORT.show();
}
}
});
}
}
尝试转换为
private TextInputEditText mDisplayName;
private TextInputEditText mEmail;
private TextInputEditText mPassword;
mDisplayName = (TextInputEditText)findViewById(R.id.reg_display_name);
mEmail = (TextInputEditText)findViewById(R.id.reg_email);
mPassword = (TextInputEditText)findViewById(R.id.reg_password);
您正在尝试将TextInputInputText id与java文件中的TextInputLayout类绑定。确保用相同类型的小部件和类名绑定id 在java文件中,使用TextInputInputText而不是TextInputLayout声明变量
private TextInputEditText mDisplayName;
private TextInputEditText mEmail;
private TextInputEditText mPassword;
在cast中,将TextInputLayout转换为textInputItemText
mDisplayName = (TextInputEditText)findViewById(R.id.reg_display_name);
mEmail = (TextInputEditText)findViewById(R.id.reg_email);
mPassword = (TextInputEditText)findViewById(R.id.reg_password);
在更改TextInputLayouyt的所有ID后,我修复了该问题。我采用了相同的方法,更改了所有字段的所有ID,问题得到了解决 剪切ID表单TextInputItemText并粘贴到TextInputLayout
它不起作用。在字符串displayName=mDisplayName.getEditText().getText().toString()中找不到getEditText()…mDisplayName
是一个TextInputItemText
,请替换为mDisplayName.getText().toString()
您的答案与上一个答案重复。
private TextInputEditText mDisplayName;
private TextInputEditText mEmail;
private TextInputEditText mPassword;
mDisplayName = (TextInputEditText)findViewById(R.id.reg_display_name);
mEmail = (TextInputEditText)findViewById(R.id.reg_email);
mPassword = (TextInputEditText)findViewById(R.id.reg_password);
<com.google.android.material.textfield.TextInputLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
style="@style/Widget.MaterialComponents.TextInputLayout.OutlinedBox"
android:hint="Username"
android:textColorHint="@color/black"
app:boxStrokeColor="@color/black"
app:boxStrokeWidthFocused="2dp"
app:endIconTint="@color/black"
app:hintTextColor="@color/black"
app:endIconMode="clear_text"
app:startIconTint="@color/black"
android:id="@+id/signup_username_editText"
>
<com.google.android.material.textfield.TextInputEditText
android:layout_width="match_parent"
android:layout_height="match_parent"
android:inputType="text"
android:textColor="@color/black"
android:textCursorDrawable="@null" />
</com.google.android.material.textfield.TextInputLayout>