Question

我创建了简单的android表单，其中我获得了用户名和密码如果我输入用户名和密码是deepak，那么它也会转到其他条件。但是我的if条件在任何情况下都是假的，我不知道为什么。

package com.example.dac.simple_intent_form;

import android.content.Intent;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

public class MainActivity extends AppCompatActivity implements View.OnClickListener {

Button button;
EditText email, password;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    button = (Button) findViewById(R.id.button);
    email = (EditText) findViewById(R.id.eemail);
    password = (EditText) findViewById(R.id.epassword);

    button.setOnClickListener(this);
}

@Override
public void onClick(View v) {
    String ema = email.getText().toString();
    String pass = password.getText().toString();

    if (ema == "deepak" && pass == "deepak")
    {
        Bundle b = new Bundle();
        b.putString("email", ema);
        b.putString("password", pass);
        Intent i = new Intent(getApplicationContext(),Activity2.class);
        i.putExtras(b);
        startActivity(i);
    }
    else
    {
        Toast.makeText(getApplicationContext(),"Plase Enter Right Username and Password",Toast.LENGTH_SHORT).show();
    }
}
}

Answer 1

使用此

您的ema.equals("deepak") && pass.equals("deepak")声明中的

if

Answer 2

尝试使用此

String ema = email.getText().toString();
    String pass = password.getText().toString();

    if (ema.toLowerCase().equals("deepak") && pass.toLowerCase().equals("deepak"))
    {


    }

Answer 3

检查这样的情况，它会正常工作

if (ema.equals("deepak") && pass.equals("deepak")){}

Answer 4

if（＆＃34; deepak＆＃34; .equals（ema）＆amp;＆amp;＆＃34; deepak＆＃34; .equals（pass））将完成这项工作。这也将确保对象ema和通过不为空并且可以避免空检查

Answer 5

使用.equals（）

if (ema.equals("deepak") && pass.equals("deepak"))
{
    Bundle b = new Bundle();
    b.putString("email", ema);
    b.putString("password", pass);
    Intent i = new Intent(getApplicationContext(),Activity2.class);
    i.putExtras(b);
    startActivity(i);
}

Android如果条件错误

5 个答案: