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java - 尝试在android中传递参数

转载 作者:行者123 更新时间:2023-12-01 10:45:46 25 4
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所以我有两个文件( MainActivity.javaHomeFragment.java ),我试图传递 public static voiddisplayPromptForEnablingGPS来自HomeFragmentMainActivity 。这是代码

HomeFragment.java (我试图传递的代码)

public static void displayPromptForEnablingGPS(
final Activity activity)
{
final AlertDialog.Builder builder =
new AlertDialog.Builder(activity);
final String action = Settings.ACTION_LOCATION_SOURCE_SETTINGS;
final String message = "Enable either GPS or any other location"
+ " service to find current location. Click OK to go to"
+ " location services settings to let you do so.";


builder.setMessage(message)
.setPositiveButton("OK",
new DialogInterface.OnClickListener() {
public void onClick(DialogInterface d, int id) {
activity.startActivity(new Intent(action));
d.dismiss();
}
})
.setNegativeButton("Cancel",
new DialogInterface.OnClickListener() {
public void onClick(DialogInterface d, int id) {
d.cancel();
}
});
builder.create().show();
}

MainActivity.java (我想如何调用它)

public void showMainView() {
HomeFragment.displayPromptForEnablingGPS();
}

但是我收到一条错误消息

"HomeFragment.displayPromptForEnablingGPS();" is invalid

最佳答案

您要调用的方法需要一个参数,但您没有传递它。

public static void displayPromptForEnablingGPS(
final Activity activity)

MainActivity 中的方法应如下所示:

public void showMainView() {
HomeFragment.displayPromptForEnablingGPS(this);
}

关于java - 尝试在android中传递参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34181490/

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