我需要将意图推送到默认相机应用程序以使其拍照、保存并返回 URI。有没有办法做到这一点?
问问题
147982 次
6 回答
181
private static final int TAKE_PICTURE = 1;
private Uri imageUri;
public void takePhoto(View view) {
Intent intent = new Intent(MediaStore.ACTION_IMAGE_CAPTURE);
File photo = new File(Environment.getExternalStorageDirectory(), "Pic.jpg");
intent.putExtra(MediaStore.EXTRA_OUTPUT,
Uri.fromFile(photo));
imageUri = Uri.fromFile(photo);
startActivityForResult(intent, TAKE_PICTURE);
}
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
switch (requestCode) {
case TAKE_PICTURE:
if (resultCode == Activity.RESULT_OK) {
Uri selectedImage = imageUri;
getContentResolver().notifyChange(selectedImage, null);
ImageView imageView = (ImageView) findViewById(R.id.ImageView);
ContentResolver cr = getContentResolver();
Bitmap bitmap;
try {
bitmap = android.provider.MediaStore.Images.Media
.getBitmap(cr, selectedImage);
imageView.setImageBitmap(bitmap);
Toast.makeText(this, selectedImage.toString(),
Toast.LENGTH_LONG).show();
} catch (Exception e) {
Toast.makeText(this, "Failed to load", Toast.LENGTH_SHORT)
.show();
Log.e("Camera", e.toString());
}
}
}
}
于 2010-04-29T13:43:25.427 回答
22
尝试以下我在这里找到的
Intent intent = new Intent("android.media.action.IMAGE_CAPTURE");
startActivityForResult(intent, 0);
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
if (resultCode == Activity.RESULT_OK && requestCode == 0) {
String result = data.toURI();
// ...
}
}
于 2010-04-28T12:32:54.920 回答
7
我花了几个小时才完成这项工作。代码几乎是来自developer.android.com的复制粘贴,略有不同。
请求此权限AndroidManifest.xml
:
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
在你的 上Activity
,首先定义这个:
static final int REQUEST_IMAGE_CAPTURE = 1;
private Bitmap mImageBitmap;
private String mCurrentPhotoPath;
private ImageView mImageView;
然后Intent
在一个onClick
:
Intent cameraIntent = new Intent(MediaStore.ACTION_IMAGE_CAPTURE);
if (cameraIntent.resolveActivity(getPackageManager()) != null) {
// Create the File where the photo should go
File photoFile = null;
try {
photoFile = createImageFile();
} catch (IOException ex) {
// Error occurred while creating the File
Log.i(TAG, "IOException");
}
// Continue only if the File was successfully created
if (photoFile != null) {
cameraIntent.putExtra(MediaStore.EXTRA_OUTPUT, Uri.fromFile(photoFile));
startActivityForResult(cameraIntent, REQUEST_IMAGE_CAPTURE);
}
}
添加以下支持方法:
private File createImageFile() throws IOException {
// Create an image file name
String timeStamp = new SimpleDateFormat("yyyyMMdd_HHmmss").format(new Date());
String imageFileName = "JPEG_" + timeStamp + "_";
File storageDir = Environment.getExternalStoragePublicDirectory(
Environment.DIRECTORY_PICTURES);
File image = File.createTempFile(
imageFileName, // prefix
".jpg", // suffix
storageDir // directory
);
// Save a file: path for use with ACTION_VIEW intents
mCurrentPhotoPath = "file:" + image.getAbsolutePath();
return image;
}
然后收到结果:
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == REQUEST_IMAGE_CAPTURE && resultCode == RESULT_OK) {
try {
mImageBitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), Uri.parse(mCurrentPhotoPath));
mImageView.setImageBitmap(mImageBitmap);
} catch (IOException e) {
e.printStackTrace();
}
}
}
使它起作用的是,它与developer.android.comMediaStore.Images.Media.getBitmap(this.getContentResolver(), Uri.parse(mCurrentPhotoPath))
的代码不同。原始代码给了我一个.FileNotFoundException
于 2015-08-10T22:50:41.997 回答
2
我找到了一个非常简单的方法来做到这一点。使用按钮打开它,使用on click
侦听器启动函数openc()
,如下所示:
String fileloc;
private void openc()
{
Intent takePictureIntent = new Intent(MediaStore.ACTION_IMAGE_CAPTURE);
File f = null;
try
{
f = File.createTempFile("temppic",".jpg",getApplicationContext().getCacheDir());
if (takePictureIntent.resolveActivity(getPackageManager()) != null)
{
takePictureIntent.putExtra(MediaStore.EXTRA_OUTPUT,FileProvider.getUriForFile(profile.this, BuildConfig.APPLICATION_ID+".provider",f));
fileloc = Uri.fromFile(f)+"";
Log.d("texts", "openc: "+fileloc);
startActivityForResult(takePictureIntent, 3);
}
}
catch (IOException e)
{
e.printStackTrace();
}
}
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data)
{
super.onActivityResult(requestCode, resultCode, data);
if(requestCode == 3 && resultCode == RESULT_OK) {
Log.d("texts", "onActivityResult: "+fileloc);
// fileloc is the uri of the file so do whatever with it
}
}
uri
您可以使用位置字符串做任何您想做的事情。例如,我将其发送到图像裁剪器以裁剪图像。
于 2018-05-31T16:16:04.063 回答
0
尝试以下我发现这是一个链接
如果您的应用程序以 M 及以上为目标并声明为使用未授予的 CAMERA 权限,则尝试使用此操作将导致 SecurityException。
EasyImage.openCamera(Activity activity, int type);
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
EasyImage.handleActivityResult(requestCode, resultCode, data, this, new DefaultCallback() {
@Override
public void onImagePickerError(Exception e, EasyImage.ImageSource source, int type) {
//Some error handling
}
@Override
public void onImagesPicked(List<File> imagesFiles, EasyImage.ImageSource source, int type) {
//Handle the images
onPhotosReturned(imagesFiles);
}
});
}
于 2018-01-11T09:12:39.427 回答
-3
试试这个代码
Intent photo= new Intent("android.media.action.IMAGE_CAPTURE");
startActivityForResult(photo, CAMERA_PIC_REQUEST);
于 2014-06-12T15:35:33.357 回答