将可编辑内容转换为字符串时出现问题（用户名/密码测试）
我正在尝试创建一个简单的用户名/密码登录屏幕。我已经完成布局，现在，我正在尝试设置它，以便当用户名 (EditText) == "crete" 时，它应该执行某些操作。这是我的代码...：
public class Login extends Activity {
public static EditText username, password;
public Button loginbutton;
boolean accessgranted;
public String dbu, dbp, user1;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
username = (EditText) this.findViewById(R.id.username);
password = (EditText) this.findViewById(R.id.password);
loginbutton = (Button) this.findViewById(R.id.loginbutton);
user1 = "crete";
loginbutton.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
try{
dbu = (username.getText()).toString();
}
finally{
if (dbu == user1){
username.setText("SUCCESS");
}
}
}
});
}
}
遗憾的是，这不起作用。它正确地将其转换为字符串（我认为），因为当我测试此代码时：
loginbutton.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
try{
dbu = (username.getText()).toString();
}
finally{
username.setText("done" + dbu);
}
}
}
});
它正确输入您在 EditText 中输入的内容，加上单词“done”。
创建 if-then 语句似乎有问题？
I'm trying to create a simple username/password login screen. I have the layout done, and right now, I'm trying to set it so when the username (EditText) == "crete", then it should do something. Here is my code...:
public class Login extends Activity {
public static EditText username, password;
public Button loginbutton;
boolean accessgranted;
public String dbu, dbp, user1;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
username = (EditText) this.findViewById(R.id.username);
password = (EditText) this.findViewById(R.id.password);
loginbutton = (Button) this.findViewById(R.id.loginbutton);
user1 = "crete";
loginbutton.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
try{
dbu = (username.getText()).toString();
}
finally{
if (dbu == user1){
username.setText("SUCCESS");
}
}
}
});
}
}
this, sadly, doesn't work. It correctly converts it to a string (i think) because when I tested this code out :
loginbutton.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
try{
dbu = (username.getText()).toString();
}
finally{
username.setText("done" + dbu);
}
}
}
});
It correctly enters what you entered into the EditText, plus the word "done".
There seems to be a problem with creating if-then statements??
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您可以使用
.equals("String")方法测试String相等性。使用
==您可以测试对象的引用是否相等。You test for
Stringequality with the method.equals("String").With
==you are testing if the references to the objects are equal.尝试使用
equalsIgnoreCase(String)而不是==比较器。像这样：
dbu.equalsIgnoreCase(user1)Try using
equalsIgnoreCase(String)instead of the==comparator.Like this:
dbu.equalsIgnoreCase(user1)dub和user1是两个独立的 String 对象。您可以这样比较它们：dbu == user1。这将始终返回 false。相反，请将其替换为dbu.equals(user1)。dubanduser1are two separate String objects. You're comparing them like this:dbu == user1. This will always return false. Instead, replace it withdbu.equals(user1).