Java Android:如何在类文件中使用Onclicklistener
MainActivity.javaJava Android:如何在类文件中使用Onclicklistener,java,android,class,onclicklistener,Java,Android,Class,Onclicklistener,MainActivity.java import android.app.Activity; import android.os.Bundle; import android.view.View; import android.widget.Toast; public class MainActivity extends Activity { protected void onCreate(Bundle paramBundle) { super.onCreate(paramBundl
import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Toast;
public class MainActivity extends Activity {
protected void onCreate(Bundle paramBundle) {
super.onCreate(paramBundle);
setContentView(R.layout.lr);
new PINT();
}
}
Pint.java
public class Pint {
puclic void Pint()
{
}
public void PINT(View view) {
switch (view.getId()) {
case (R.id.rightleft):
X(0);
break;
case (R.id.rightleft):
X(1);
break;
}
}
public void X(int x) {
if (x == 0) Toast.makeText(getApplicationContext(), "Left",
Toast.LENGTH_SHORT).show();
if (x == 1) Toast.makeText(getApplicationContext(), "Right",
Toast.LENGTH_SHORT).show();
}
}
}
}
运行上面的代码失败了,我希望能够从类文件中使用onclicklister,我该如何做呢
xml中只有两个按钮这是我的建议:
import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Toast;
public class MainActivity extends Activity implements OnClickListener {
protected void onCreate(Bundle paramBundle) {
super.onCreate(paramBundle);
setContentView(R.layout.lr);
}
@Override
public void onClick(View v) {
switch (view.getId()) {
case (R.id.rightleft):
X(0);
break;
case (R.id.rightleft):
X(1);
break;
}
}
public void X(int x) {
(x == 0)
Toast.makeText(getApplicationContext(), "Left",
Toast.LENGTH_SHORT).show();
(x == 1)
Toast.makeText(getApplicationContext(), "Right",
Toast.LENGTH_SHORT).show();
}
}
类文件必须扩展其中一个视图类,OnClickListener才能工作 除此之外,这是什么( 我想如果政治家们:
public void X(int x) {
if(x == 0) {Toast.makeText(getApplicationContext(), "Left",
Toast.LENGTH_SHORT).show();}
else
if(x == 1) {Toast.makeText(getApplicationContext(), "Right",
Toast.LENGTH_SHORT).show();}
}
为什么不在您的活动中定义OnClickListener?为什么要在单独的类中定义它?主要是为了可读性,实际上这段代码不可编译。@Shlubu如果您阅读了我所说的,您就会知道,我说“运行上面的代码失败”,您所做的对可读性没有好处。您只会以这种方式生成更多的代码。
public void X(int x) {
if(x == 0) {Toast.makeText(getApplicationContext(), "Left",
Toast.LENGTH_SHORT).show();}
else
if(x == 1) {Toast.makeText(getApplicationContext(), "Right",
Toast.LENGTH_SHORT).show();}
}