如何使用切换按钮启用或禁用在webview中加载图像?我已经使用了这段代码,但图像永远都是启用的。
toggle=(ToggleButton)findViewById(R.id.tglbtn1);
toggle.setOnClickListener(new OnClickListener() {
public void onClick(View v) {
// TODO Auto-generated method stub
if(toggle.isChecked())
{
Toast.makeText(getApplicationContext(), "The state is changed to on", Toast.LENGTH_LONG).show();
webView.getSettings().setLoadsImagesAutomatically(true);// Enable Image
// Loading
}
else
{
Toast.makeText(getApplicationContext(), "The state is changed to off", Toast.LENGTH_LONG).show();
webView.getSettings().setLoadsImagesAutomatically(false);// Enable Image
// Loading
}
}
});
答案 0 :(得分:0)
在android中实现切换按钮的更好方法,
布局xml,
<ToggleButton
android:id="@+id/leave_toggle"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:background="@drawable/custom_toggle"
android:minHeight="1dp"
android:minWidth="1dp"
android:textOff=""
android:textOn="" />
Drawable xml, @绘制/ custom_toggle
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:drawable="@drawable/notification_on" android:state_checked="true"/>
<item android:drawable="@drawable/notification_off" android:state_checked="false"/>
</selector>
notification_on和notification_off是图像
内部活动,
private ToggleButton getToggleButton(int id) {
return (ToggleButton) findViewById(id);
}
将oncheckedchagelistener设置为切换按钮,当用户启用或禁用带缺陷的按钮时,它会通知
getToggleButton(R.id.leave_toggle).setOnCheckedChangeListener(
new OnCheckedChangeListener() {
@Override
public void onCheckedChanged(CompoundButton toggleButton,
boolean isChecked) {
if(isChecked)
{
// enabled
}else
{
//disabled
}
}
});