java 在Android中检查用户名和密码
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4780197/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Checking username and password in Android
提问by theJava
I have a username and password field and now i need to check and redirect him to the next page in Android.
我有一个用户名和密码字段，现在我需要检查并将他重定向到 Android 中的下一页。
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
final EditText loginText = (EditText) findViewById(R.id.widget44);
final EditText loginPassword = (EditText) findViewById(R.id.widget47);
final Button button = (Button) findViewById(R.id.widget48);
button.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
Intent myIntent = null;
if(loginText.getText().equals("admin") &&
loginPassword.getText().equals("admin")) {
System.out.println("Entering");
myIntent = new Intent(view.getContext(), Page1.class);
} else {
}
startActivity(myIntent);
}
});
}
Temp now i am checking by hardcoding the values, but this too does not work for me. Why? I usually check in Javalike this, why does it not accept me the same way in Android
Temp 现在我正在通过对值进行硬编码来检查，但这对我也不起作用。为什么？我通常像这样检查Java，为什么它在Android中不以同样的方式接受我
采纳答案by alex.zherdev
I think it's because EditText#getText()returns an Editableobject. Try
我认为这是因为EditText#getText()返回一个Editable对象。尝试
if(loginText.getText().toString().equals("admin") &&
loginPassword.getText().toString().equals("admin")) {
...
}
回答by Beasly
Don't you need to put toString()to your textelements?
Like this:
您不需要放入toString()文本元素吗？
像这样：
if(loginText.getText().toString().equals("admin") &&
loginPassword.getText().toString().equals("admin")) {
...
Edit: neutrino was faster (+1) :)
编辑：中微子更快（+1）:)
回答by chikka.anddev
Friend i am finding the only small prob. in ur code is showed below.correct it and i think it will work for u:
朋友，我正在寻找唯一的小问题。在你的代码中显示如下。更正它，我认为它对你有用：
if(loginText.getText().**toString()**.equals("admin") &&
loginPassword.getText()**.toString()**.equals("admin")) {
System.out.println("Entering");
myIntent = new Intent(view.getContext(), Page1.class);
} else {
...
}
see the correction in bold letters
请参阅以粗体显示的更正
