我已经试着在我的应用程序中添加一些首选项几个星期了
我真的只需要复选框功能。
我试图通过选中复选框来控制单个单选按钮的可见性
这是我的首选项.java
import android.app.Activity;
import android.content.SharedPreferences;
import android.os.Bundle;
import android.preference.CheckBoxPreference;
import android.preference.Preference;
import android.preference.PreferenceActivity;
import android.preference.PreferenceManager;
import android.preference.Preference.OnPreferenceClickListener;
import android.preference.Preference.OnPreferenceChangeListener;
import android.widget.Button;
import android.widget.CheckBox;
import android.widget.CompoundButton;
import android.widget.RadioButton;
import android.widget.Toast;
import android.widget.CompoundButton.OnCheckedChangeListener;
import android.view.View;
public class Preferences extends PreferenceActivity {
private RadioButton btn01;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
addPreferencesFromResource(R.xml.prefs);
btn01 = (RadioButton)findViewById(R.id.RadioButton01);
Preference customPref = (Preference) findPreference("customPref");
customPref.setOnPreferenceClickListener(new OnPreferenceClickListener(){
public boolean onPreferenceClick(Preference preference) {
Toast.makeText(getBaseContext(),
"The Custom Preference Has Been Clicked",
Toast.LENGTH_LONG).show();
SharedPreferences customSharedPreference = getSharedPreferences(
"myCutomSharedPrefs", Activity.MODE_PRIVATE);
SharedPreferences.Editor editor = customSharedPreference
.edit();
editor.putString("myCustomPref",
"The preference has been clicked");
editor.commit();
return true;}
public void CheckBox() {
final CheckBox ThisCheckBox = (CheckBox) findViewById (R.id.checkboxPref);{
ThisCheckBox.setOnCheckedChangeListener(new OnCheckedChangeListener(){
public boolean OnCheckedChange (CompoundButton compoundButton) {
if (ThisCheckBox.isChecked()){
{
btn01.setVisibility(0);
}{
btn01.setVisibility(2);
}}
}
;
});
};}});}}
无论如何,我很确定这是完全错误的,我在这条线上得到了一个错误
此复选框.setOnCheckedChangeListener(新的onCheckedChangeListener(){
知道为什么吗?我认为这与我调用上面这一行的方式有关,但是无论我使用final、boolean还是void,它仍然会产生错误
这是错误信息
类型new compoundbutton.oncheckedchangeletener(){}必须实现继承的抽象方法compoundbutton.oncheckedchangeletener.oncheckedchanged(compoundbutton,boolean)
所以我想我要问的是
我做的选择对吗
如何正确执行if-else命令??
最佳答案
我认为您试图重写的方法的签名不正确。替换为:
public boolean OnCheckedChange (CompoundButton compoundButton) {
用这个:
@Override
public void onCheckedChanged(CompoundButton compoundButton, boolean isChecked) {
编辑
好的,这里有一个尝试来修复整个类。备份当前代码,然后选择下面的所有代码并复制它。选择类中的所有代码并将其替换为:
import android.app.Activity;
import android.content.SharedPreferences;
import android.os.Bundle;
import android.preference.Preference;
import android.preference.PreferenceActivity;
import android.preference.Preference.OnPreferenceClickListener;
import android.widget.CheckBox;
import android.widget.CompoundButton;
import android.widget.RadioButton;
import android.widget.Toast;
import android.widget.CompoundButton.OnCheckedChangeListener;
public class Preferences extends PreferenceActivity {
private RadioButton btn01;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
addPreferencesFromResource(R.xml.prefs);
btn01 = (RadioButton)findViewById(R.id.RadioButton01);
Preference customPref = (Preference) findPreference("customPref");
customPref.setOnPreferenceClickListener(new OnPreferenceClickListener(){
public boolean onPreferenceClick(Preference preference) {
Toast.makeText(getBaseContext(),"The Custom Preference Has Been Clicked",Toast.LENGTH_LONG).show();
SharedPreferences customSharedPreference = getSharedPreferences("myCutomSharedPrefs", Activity.MODE_PRIVATE);
SharedPreferences.Editor editor = customSharedPreference.edit();
editor.putString("myCustomPref","The preference has been clicked");
editor.commit();
return true;
}
public void CheckBox() {
final CheckBox ThisCheckBox = (CheckBox) findViewById (R.id.checkboxPref);
ThisCheckBox.setOnCheckedChangeListener(new OnCheckedChangeListener(){
@Override
public void onCheckedChanged(CompoundButton compoundButton,boolean test) {
if (ThisCheckBox.isChecked()){
btn01.setVisibility(0);
} else {
btn01.setVisibility(2);
}
}
});
};
});
}
}
上面的代码在eclipse中编译时对我来说没有问题