Android7.0 uri转file问题

适配7.0的时候，我们需要编写

 permission_file_paths.xml
<?xml version="1.0" encoding="utf-8"?>
<paths>
    <root-path
        name="root_path"
        path="." />

    <files-path
        name="file_path"
        path="." />

    <cache-path
        name="cache_path"
        path="." />

    <external-path
        name="external_path"
        path="." />

    <external-files-path
        name="external_files_path"
        path="." />

    <external-cache-path
        name="external_cache_path"
        path="." />

    <external-media-path
        name="external_media_path"
        path="." />
</paths>

manifrest.xml

<provider
            android:name="androidx.core.content.FileProvider"
            android:authorities="${applicationId}.fileprovider"
            android:exported="false"
            android:grantUriPermissions="true">
            <meta-data
                android:name="android.support.FILE_PROVIDER_PATHS"
                android:resource="@xml/permission_file_paths" />
        </provider>

用到uri的时候，我们可能要转下

   public static Uri toUri(String filePath) {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
            return FileProvider.getUriForFile(App.getInstance(), App.getInstance().getPackageName() + ".fileprovider", new File(filePath));
        }
        return Uri.fromFile(new File(filePath));
    }

有的时候，我们还需要uri转回file，这时候，我们需要适配下

 public static File toFile(Uri uri) {
        String pathStr = uri.getPath();
        if (pathStr != null) {
            //7.0路径适配问题
            if (pathStr.startsWith("/external_path")) {
                pathStr = pathStr.replace("/external_path", Environment.getExternalStorageDirectory().getPath());
            }
            File file = new File(pathStr);
            return file;
        }
        return null;
    }