我希望第二个活动中textview的数据在第三个活动中传递,但我不想在第三个活动中显示它。在最后一个活动中单击按钮后,我想直接发送第二个活动中的数据以传入我的sqldb。是否可以实现?这是我的第一个活动。
public class MainActivity extends AppCompatActivity {
private EditText email, password;
private Button btn_login;
private ProgressBar loading;
private static String URL_LOGIN ="http://192.168.1.83/Attendance/login.php";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
loading= findViewById(R.id.loading);
email = findViewById(R.id.editTextUserEmail);
password = findViewById(R.id.editTextPassword);
btn_login = findViewById(R.id.buttonRegister);
btn_login.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
String mEmail = email.getText().toString().trim();
String mPassword = password.getText().toString().trim();
if (!mEmail.isEmpty() || !mPassword.isEmpty()) {
Login(mEmail, mPassword);
}else{
email.setError("Please Enter Email");
password.setError("Please Enter Password");
}
}
});
}
private void Login(final String email, final String password) {
loading.setVisibility(View.VISIBLE);
btn_login.setVisibility(View.GONE);
StringRequest stringRequest = new StringRequest(Request.Method.POST, URL_LOGIN,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONObject jsonObject = new JSONObject(response);
Log.d("JSON", jsonObject.toString());
String success = jsonObject.getString("success");
JSONArray jsonArray =
jsonObject.getJSONArray("login");
if (success.equals("1"))
{
for (int i = 0; i<jsonArray.length(); i++){
JSONObject object =
jsonArray.getJSONObject(i);
String name =
object.getString("name").trim();
String email =
object.getString("email").trim();
// Toast.makeText(MainActivity.this,
"Success Login \nYour name: "+name+"\nYour Email: "+email,
// Toast.LENGTH_LONG).show();
Intent intent = new
Intent(MainActivity.this, HomeActivity.class);
intent.putExtra("name",name);
intent.putExtra("email", email);
startActivity(intent);
loading.setVisibility(View.GONE);
}
}
} catch (JSONException e) {
e.printStackTrace();
loading.setVisibility(View.GONE);
btn_login.setVisibility(View.VISIBLE);
Toast.makeText(MainActivity.this, "Error"
+e.toString(),
Toast.LENGTH_LONG).show();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(MainActivity.this, "Error"
+error.toString(),
Toast.LENGTH_LONG).show();
}
})
{
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("email", email);
params.put("password", password);
return params;
}
};
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
这是我的第二项活动
public class HomeActivity extends AppCompatActivity {
private TextView name,email;
private TextView welcome;
private Button Send;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_home);
name = findViewById(R.id.name);
email = findViewById(R.id.email);
Send = findViewById(R.id.btn_send);
welcome = findViewById(R.id.welcome);
final Intent intent = getIntent();
String extraName = intent.getStringExtra("name");
String extraEmail = intent.getStringExtra("email");
name.setText(extraName);
email.setText(extraEmail);
Send.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
Intent intent1 = new Intent(HomeActivity.this,StatusActivity.class);
intent1.putExtra("wel", welcome.getText().toString());
intent1.putExtra("name1", name.getText().toString());
intent1.putExtra("email1", email.getText().toString());
startActivity(intent1);
}
});
}
这是我的最后一个活动,但这里我不想显示文本视图,我只想从这里发布它。
public class StatusActivity extends AppCompatActivity {
private TextView welcome, name, email;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_status);
welcome = findViewById(R.id.abcd);
name = findViewById(R.id.name1);
email = findViewById(R.id.email1);
final Intent intent = getIntent();
String extraName1 = intent.getStringExtra("name1");
String extraEmail1 = intent.getStringExtra("email1");
String extraWelcome = intent.getStringExtra("wel");
name.setText(extraName1);
email.setText(extraEmail1);
welcome.setText(extraWelcome);
}
1条答案
按热度按时间wi3ka0sx1#
你的问题很清楚,直到你把mysql的主题。为什么要通过数据库将数据从一个活动传递到另一个活动。为什么需要花费时间和资源的数据库操作。
您可以通过将数据从一个活动传递到另一个活动
Intent
s(基本值,String
s和Parcelable
s) 哦!你所要做的就是:
获取数据(从textview获取案例)。
创建导航到下一个活动的意图。
把数据放到目的地。
开始活动。
从意图中检索数据。
例如,如果将通过按钮单击导航到下一个活动,则在该按钮的
onClick()
: