Question

我需要在应用程序的不同屏幕中检索用户位置。我编写了代码来获取屏幕活动中的位置，但是，现在我想在另一个屏幕活动中获取该位置。有什么办法可以避免再次编写相同的代码吗？

        // variables
        // These variables will be repeated
        LocationProvider locationProvider;
        double latitude;
        double longitude;
        private String userPostcode;
        // ===============

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container,
                                      Bundle savedInstanceState) {

        // This code will be repeated
        locationProvider = new LocationProvider(getActivity());
        latitude = locationProvider.getLatitude();
        longitude = locationProvider.getLongitude();
        if (isNetworkAvailable(getContext())) {
            getPostcode();
        } else {
            Toast.makeText(getActivity(), "Internet or GPS is not available. To get the best location one or both of these must be on",
                    Toast.LENGTH_LONG).show();
        }
 // =============
        return inflater.inflate(R.layout.fragment_location, container, false);
    }




      // These two methods will be repeated
        private void getPostcode(){
                    Geocoder geoCoder = new Geocoder(getActivity().getApplicationContext(), Locale.getDefault());
                    List<Address> address = null;

                    if (geoCoder != null) {
                        try {
                            address = geoCoder.getFromLocation(latitude, longitude, 1);
                        } catch (IOException e1) {
                            e1.printStackTrace();
                        }
                        if (address.size() > 0) {
                            userPostcode = address.get(0).getPostalCode();
                        }
                    }
                }


            private boolean isNetworkAvailable(final Context context) {
                    final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
                    return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
                }
    // ====================

如果可以避免重复，请有人给我看/告诉我如何才能做到。我是android新手，可能错过了一些愚蠢的东西。

Answer 1

在单独的类中添加getPostcode（）。您可以按照以下方式修改它。

yum install php-cli
Loaded plugins: fastestmirror, security
Loading mirror speeds from cached hostfile
 * base: centos.syn.co.il
 * epel: mirror.nonstop.co.il
 * extras: centos.syn.co.il
 * rpmforge: apt.sw.be
 * updates: centos.syn.co.il
Setting up Install Process
Resolving Dependencies
--> Running transaction check
---> Package php-cli.x86_64 0:5.3.3-48.el6_8 will be installed
--> Processing Dependency: php-common(x86-64) = 5.3.3-48.el6_8 for package: php-cli-5.3.3-48.el6_8.x86_64
--> Running transaction check
---> Package php-common.x86_64 0:5.3.3-48.el6_8 will be installed
--> Finished Dependency Resolution

Dependencies Resolved

=============================================================================================================================
 Package                       Arch                      Version                            Repository                  Size
=============================================================================================================================
Installing:
 php-cli                       x86_64                    5.3.3-48.el6_8                     updates                    2.2 M
Installing for dependencies:
 php-common                    x86_64                    5.3.3-48.el6_8                     updates                    530 k

Transaction Summary
=============================================================================================================================
Install       2 Package(s)

Total download size: 2.7 M
Installed size: 9.1 M
Is this ok [y/N]:

片段/活动中的

执行此操作

public class Utils {


    public static String getPostcode(Context context, double lat, double lng){
        Geocoder geoCoder = new Geocoder(context, Locale.getDefault());
        List<Address> address = null;

        if (geoCoder != null) {
            try {
                address = geoCoder.getFromLocation(lat, lng, 1);
            } catch (IOException e1) {
                e1.printStackTrace();
            }
            if (address.size() > 0) {
                return address.get(0).getPostalCode();
            }
        }
        return null;
    }


    public static boolean isNetworkAvailable(final Context context) {
        final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
        return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
    }
}

如果在片段中，

将if(Utils.isNetworkAvailable(this)) { String postCode = Utils.getPostcode(this, yourLat, yourLng); }更改为this。

Answer 2

只是简单地返回一些值的方法，你需要经常重复使用它们并且不依赖于任何对象或状态，你可以使它们保持静态。

创建某种实用类。例如

public class Utils {

    public static boolean isNetworkAvailable(final Context context) {
        final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
        return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
    }

    public static String getPostCode(Activity activity) {
        ...
    }
}

然后你可以在任何你喜欢的地方重复使用它。

boolean isNetworkAvailable = Utils.isNetworkAvailable(context);

Answer 3

创建新的类文件并将这些函数放在该类文件中

private String getPostcode(Activity activity, Double latitude, Double longitude) {
        String userPostcode = null;
        Geocoder geoCoder = new Geocoder(activity.getApplicationContext(), Locale.getDefault());
        List<Address> address = null;

        if (geoCoder != null) {
            try {
                address = geoCoder.getFromLocation(latitude, longitude, 1);
            } catch (IOException e1) {
                e1.printStackTrace();
            }
            if (address.size() > 0) {
                userPostcode = address.get(0).getPostalCode();
            }
        }
        return userPostcode;
    }

    private static boolean isNetworkAvailable(final Context context) {
        final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
        return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
    }

要访问这些功能，请创建该类的对象并调用这些功能

Answer 4

您可以创建一个实用程序类来执行所需的所有位置工作。将活动/上下文对象作为构造函数参数传递给实用程序类。然后使用类方法来获得你想要的东西。查看实用程序类的以下代码：

public class LocationHelper {
    private double mLatitude;
    private double mLongitude;
    private Context mContext;
    private LocationProvider mLocationProvider;

    // Dont know if you need Activity or Context object.
    // Pass whatever as per requirement
    public LocationHelper(Context context) {
        this.mContext = context;
        initLocation();
    }

    // Set the latitude and longitude here.
    // You can get the latitude and longitude using
    // getter methods.
    private void initLocation() {
        mLocationProvider = new LocationProvider(mContext);
        mLatitude = mLocationProvider.getLatitude();
        mLongitude = mLocationProvider.getLongitude();
    }

    public double getLatitude() {
        return mLatitude;
    }

    public double getLongitude() {
        return mLongitude;
    }

    // Get the post code as a string. Empty if failed to get any.
    public String getPostcode(){
        String userPostcode = "";
        Geocoder geoCoder = new Geocoder(mContext, Locale.getDefault());
        List<Address> address = null;

        //It is a redundant check since it will always be true
        if (geoCoder != null) {
            try {
                address = geoCoder.getFromLocation(mLatitude, mLongitude, 1);
            } catch (IOException e1) {
                e1.printStackTrace();
            }
            // Also make sure to put a null check to avoid NullPointerException
            if (address != null && address.size() > 0) {
                userPostcode = address.get(0).getPostalCode();
            }
        }
        return userPostcode;
    }

    //To check if device is offline/online.
    public boolean isNetworkAvailable() {
        final ConnectivityManager connectivityManager = ((ConnectivityManager) mContext.getSystemService(Context.CONNECTIVITY_SERVICE));
        return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
    }
}

Answer 5

是的，折射器进入一个单独的类。或者找到一种方法将数据从一个活动转移到另一个活动。（putExtra（），Singleton，DataBase，SharedPreferences等）

如何避免跨类重复Android中的代码？

5 个答案: