My code does the following:

1. Picks an image with pickImage() (from image_picker package)
2. Then it takes the picked image and crop it with cropImage() (image_cropper package)

After being picked, the function pickImage() saves it in this path :

/data/user/0/com.example.myapp/cache/9930cfca-6eb2.jpg

And then the cropImage() function saves the returned image here:

/data/user/0/com.example.myapp/cache/image_cropper_1924443.jpg

Here is my function

  Future pickImage(source) async {
    try {
      final pickedFile = await ImagePicker().pickImage(source: source);
      if (pickedFile == null) return;

      File? croppedImage = await ImageCropper.cropImage(
          sourcePath: pickedFile.path,
          aspectRatio: CropAspectRatio(ratioX: 1.0, ratioY: 1.0)
      );
      if(croppedImage == null) return;

      this.image = File(croppedImage.path);
      setState((){});
     } on PlatformException catch (e) {
      print('Failed to catch image: $e');
    }
  }

Now what i want to do is to save the cropped image to a specific folder. Is there any way to implement this.

CodePudding user response：

One of the solutions is to copy the image after that cropImage() stores it in its default directory:

After getting our cropped image:

croppedImage = await ImageCropper.cropImage(...);

Do the following:

1. Get the application Folder on the phone disk using the path_provider package:

final Directory dir = await getApplicationDocumentsDirectory();
final String appDir = dir.path

2. Check if the file already exists, then delete it.

final File imageFile = File(appDir   '/profile_picture.jpg');
if (await imageFile.exists()) {
  imageFile.delete();
}

3. Preferably clear the cache to avoid some problems:

imageCache!.clearLiveImages();
imageCache!.clear();

4. Copy the image using the copy method which is provided by the File class:

final File copiedImage = await croppedImage.copy('$appDir/profile_picture.jpg');

5. Store the new image in a variable of type File:

File myImage = File(copiedImage.path);