我试图设置从数据库中获取的Edittext小部件的文本（数据），但是我得到了上述错误，Iam试图解决但没有得到实际的想法，请帮我这个，提前谢谢，
这是我的MainActivity.java
package com.mycompany.newlogin;
import android.content.Intent;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
public class MainActivity extends AppCompatActivity implements View.OnClickListener{
Button btLogout;
EditText etName,etPhone,etEmail,etUsername;
private static String url = "jdbc:mysql://31.170.160.74:3306/a5582611_mane";
private static String user = "a5582611_nanu";
private static String password = "sonne0";
UserLocalStore userLocalStore;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
etName=(EditText)findViewById(R.id.etName);
etPhone=(EditText)findViewById(R.id.etPhone);
etEmail=(EditText)findViewById(R.id.etEmail);
etUsername=(EditText)findViewById(R.id.etUsername);
btLogout=(Button)findViewById(R.id.btLogout);
userLocalStore=new UserLocalStore(this);
btLogout.setOnClickListener(this);
}
@Override
protected void onStart(){
super.onStart();
if(authenticate()==true){
//go to your question window
displayUserDetails();
}else {
startActivity(new Intent(MainActivity.this,Login.class));
}
}
private void displayUserDetails(){
User user=userLocalStore.getLoggedInUser();
etName.setText(user.name);
etEmail.setText(user.email);
etUsername.setText(user.username);
}
private boolean authenticate(){
return userLocalStore.getUserLoggedIn();
}
@Override
public void onClick(View v) {
switch (v.getId()){
case R.id.btLogout:
userLocalStore.clearUserData();
userLocalStore.setUserLoggedIn(false);
startActivity(new Intent(this, Login.class));
break;
}
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
我的activity_xml文件是这样的，
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Name"/>
<EditText
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:id="@+id/etName"
android:layout_marginBottom="10dp"
/>
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Email"/>
<EditText
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:id="@+id/Email"
android:layout_marginBottom="10dp"
/>
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Username"/>
<EditText
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:id="@+id/etUsername"
android:layout_marginBottom="10dp"
/>
<Button
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:id="@+id/btLogout"
android:text="Logout"
android:layout_gravity="center"/>
这是我的错误详情，
请帮我解决此问题
答案 0 :(得分：1)
您的etEmail编辑文本为空：
代替 - etEmail=(EditText)findViewById(R.id.etEmail);
应该是 - etEmail=(EditText)findViewById(R.id.Email);
由于您的etEmail ID是“电子邮件”而不是“etEmail”。