I'm quite new to android programming and I have the following problem.
I want to be able to put an image om my server and then if I use my app it should use that image as a background.
From previous research I understand I cant save any files to the drawable file?
So is this even possible?
I am now this far:
URL url = new URL ("http://oranjelan.nl/oranjelan-bg.png");
InputStream input = url.openStream();
try {
String storagePath = Environment.getExternalStorageDirectory();
OutputStream output = new FileOutputStream (storagePath + "/oranjelangb.png");
try {
byte[] buffer = new byte[1000000];
int bytesRead = 0;
while ((bytesRead = input.read(buffer, 0, buffer.length)) >= 0) {
output.write(buffer, 0, bytesRead);
}
} finally {
output.close();
}
} finally {
input.close();
}
But I get the following error
# String storagePath = Environment.getExternalStorageDirectory();
The compiller says cannot convert file to string.
It should be possible. Simple steps may include :-
1) Download image file from server, Store it to SDcard or assets folder.
links for step 1 >> link1 link2
2) Create a Bitmap from the file you downloaded.
3) Set that bitmap as a Background image.
You can pick steps and search on SO there should be lots of answers available.
Related
I have been using this tutorial to make some face detections on picture. The problem is when I getting the file path that used on java
String xmlFile = "E:/OpenCV/facedetect/lbpcascade_frontalface.xml";
CascadeClassifier classifier = new CascadeClassifier(xmlFile);
How can I translate on android studio. I try put my lbpcascade_frontalface.xml on raw resources. CascadeClassifier is a class that opencv library provided. The only problem is they only loaded string path (on xmlfile).
this is my code.
String pathtoRes = getRawPathAtempt2(context);
CascadeClassifier cascadeClassifier = new CascadeClassifier();
cascadeClassifier.load(pathtoRes);
I translated to some method like this.
public String getRawPathAtempt2(Context context) {
return "android.resource://" + context.getPackageName() + "/raw/" + "lbpcascade_frontalface.xml";
}
I get the asertions error by opencv that tell me the file is null. Thats mean I had been wrong when I used file path on my method. how can I get file path on raw resources? help me please I have been stuck for several days now
how can i get file path on raw resources?
You can't. There is no path. It is a file on your development machine. It is not a file on the device. Instead, it is an entry in the APK file that is your app on the device.
If your library supports an InputStream instead of a filesystem path, getResources().openRawResource() on a Context to get an InputStream on your raw resource.
This is how i solved the problem with #CommonWare Answer
i'm using input stream to get file
is = context.getResources().openRawResource(R.raw.lbpcascade_frontalface);
so make the file will be opened and i will make a point to the File class
File cascadeDir = context.getDir("cascade", Context.MODE_PRIVATE);
mCascadeFile = new File(cascadeDir, "lbpcascade_frontalface.xml");
and then i make a search to the path file using getAbsolutePath like this
cascadeClassifier.load(mCascadeFile.getAbsolutePath());
take a look at my full code
try {
is = context.getResources().openRawResource(R.raw.lbpcascade_frontalface);
File cascadeDir = context.getDir("cascade", Context.MODE_PRIVATE);
mCascadeFile = new File(cascadeDir, "lbpcascade_frontalface.xml");
os = new FileOutputStream(mCascadeFile);
byte[] buffer = new byte[4096];
int bytesRead;
while ((bytesRead = is.read(buffer)) != -1) {
os.write(buffer, 0, bytesRead);
}
is.close();
os.close();
cascadeClassifier.load(mCascadeFile.getAbsolutePath());
} catch (IOException e) {
Log.i(TAG, "face cascade not found");
}
Uri video = Uri.parse("android.resource://com.test.test/raw/filename");
Using this you can access the file in raw folder, if you want to access the file in asset folder use this URL...
file:///android_asset/filename
Make sure you don't add the extension to the filename. E.g. "/movie" not "/movie.mp4".
Refer to this link:
Raw folder url path?
We can use this Alternative method for finding the path of Files in Raw Folder.
InputStream inputStreamdat=getResources().openRawResource(R.raw.face_landmark_model);
File model=getDir("model", Context.MODE_PRIVATE);
File modelFile=new
File(model,"face_landmark_model.dat");
FileOutputStream os1 = new
FileOutputStream(modelFile);
byte[] buffer1 = new byte[4096];
int bytesRead1;
while ((bytesRead1=inputStreamdat.read(buffer1))!=-1) {
os1.write(buffer1, 0, bytesRead1);
}
inputStreamdat.close();
os1.close();
You Can Use
modelFile.getAbsolutePath()); For getting the Path
my application will downloads non media files like PDF,TXT, XML etc. Using following code I am saving in storage.
FileOutputStream out = new FileOutputStream("Save locaation");
final int fileBufferSizee = 1024;
byte[] buffer = new byte[fileBufferSizee ];
int read;
while ((read = in.read(buffer)) != -1) {
out.write(buffer, 0, read);
}
in.close();
in = null;
out.flush();
out.close();
out = null;
working fine. Actually I need URI of saved file. for that I need to insert information in android media database.
How to insert non media files in android media database ? any suggestions ?
if u want to get URL from absolute path then do this
Uri.parse(new File("/sdcard/cats.jpg").toString()));
save this in one URI variable
I need to get image from the web and store it in the phone for later use.
I tryed this:
public Drawable grabImageFromUrl(String url) throws Exception
{
return Drawable.createFromStream((InputStream)new URL(url).getContent(), "src");
}
So this my function to grab image from Url, i just need a proccess to get the returned drawable and save.
How can i do that ?
see this complete example give here
http://android-example-code.blogspot.in/p/download-store-and-read-images-from.html
Based off here, you can actually download the image using a different method. Is it absolutely necessary that you store it as a drawable before saving it? Because I think you could save it first, and THEN open it, if need be.
URL url = new URL ("file://some/path/anImage.png");
InputStream input = url.openStream();
try {
//The sdcard directory e.g. '/sdcard' can be used directly, or
//more safely abstracted with getExternalStorageDirectory()
String storagePath = Environment.getExternalStorageDirectory();
OutputStream output = new FileOutputStream (storagePath + "/myImage.png");
try {
byte[] buffer = new byte[aReasonableSize];
int bytesRead = 0;
while ((bytesRead = input.read(buffer, 0, buffer.length)) >= 0) {
output.write(buffer, 0, bytesRead);
}
} finally {
output.close();
}
} finally {
input.close();
}
I am working on an API which returns me a zip file containing multiple XML files, which i have to parse individually after extracting the zip file.
Here is the link for that(will download the zip-file) :
http://clinicaltrials.gov/ct2/results?term=&recr=&rslt=&type=&cond=&intr=&outc=&lead=&spons=&id=&state1=&cntry1=&state2=&cntry2=&state3=&cntry3=&locn=&gndr=Female&age=0&rcv_s=&rcv_e=&lup_s=&lup_e=studyxml=true
Here is my current code to save the zip-file in sdcard:
File root = Environment.getExternalStorageDirectory();
String url= "http://clinicaltrials.gov/ct2/results?term=&recr=&rslt=&type=&cond=&intr=&outc=&lead=&spons=&id=&state1=&cntry1=&state2=&cntry2=&state3=&cntry3=&locn=&gndr=Female&age=0&rcv_s=&rcv_e=&lup_s=&lup_e=xml=true";
try {
HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection();
conn.setDoInput(true);
conn.setConnectTimeout(10000); // timeout 10 secs
conn.connect();
InputStream input = conn.getInputStream();
FileOutputStream fOut = new FileOutputStream(new File(root, "new.zip"));
int byteCount = 0;
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = input.read(buffer)) != -1) {
fOut.write(buffer, 0, bytesRead);
byteCount += bytesRead;
}
fOut.flush();
fOut.close();
} catch (Exception e) {
e.printStackTrace();
}
Problem :
New.zip File is getting created in sdcard but it seems nothing is downloading also the file size is 0kb.
Is my code proper or I have to use something else to handel zipfiles.
Edit Solved :
I am extremely sorry the api link is invalid ... it should be
http://clinicaltrials.gov/ct2/results?term=&recr=&rslt=&type=&cond=&intr=&outc=&lead=&spons=&id=&state1=&cntry1=&state2=&cntry2=&state3=&cntry3=&locn=&gndr=Female&age=0&rcv_s=&rcv_e=&lup_s=&lup_e=&studyxml=true
& is required before studtxml..
Thnx every 1 for quick response ..
There is something wrong either in your .zip file URL or in .zip file size(0 byte size) because if we download this .zip file (From URL given by you) from web browser then also its downloaded with 0 byte size.
Downloaded .zip file URL.
Your Url in the code String url=... is not giving me a zip file.
http://clinicaltrials.gov/ct2/results?term=&recr=&rslt=&type=&cond=&intr=&outc=&lead=&spons=&id=&state1=&cntry1=&state2=&cntry2=&state3=&cntry3=&locn=&gndr=Female&age=0&rcv_s=&rcv_e=&lup_s=&lup_e=xml=true
The link you provided is different
http://clinicaltrials.gov/ct2/results?term=&recr=&rslt=&type=&cond=&intr=&outc=&lead=&spons=&id=&state1=&cntry1=&state2=&cntry2=&state3=&cntry3=&locn=&gndr=Female&age=0&rcv_s=&rcv_e=&lup_s=&lup_e=studyxml=true
Looks like there's an error: lup_e=xml=true should be lup_e=studyxml=true
I am copying an image to a private directory like so:
FileChannel source = null;
FileChannel destination = null;
source = new FileInputStream(sourceFile).getChannel();
destination = new FileOutputStream(destFile).getChannel();
destination.transferFrom(source, 0, source.size());
source.close();
destination.close();
..but when I insert it back in to Gallery, untouched, at a later time:
private void moveImageToGallery(Uri inUri) throws Exception {
MediaStore.Images.Media.insertImage(getContentResolver(), ImageUtil.loadFullBitmap(inUri.getPath()), null, null);
}
..it apparently loses its Exif data. The rotation no longer works. Is there some way I can copy an image file and not lose that data? Thanks for any suggestions.
FileChannel, here, seems to actually read the data, decode it, reencode it, then write it; thus losing the EXIF data. Copying a file (byte-by-byte) does not alter its content. The only thing that can happen before/after a copy is a file access change (remember: Android is based on Linux, Linux being an UNIX => rwx permissions (see chmod)), eventually denying the read or write of the file. So it is clear FileChannel does something unwanted.
This code will do the work:
InputStream in = new FileInputStream(source);
OutputStream out = new FileOutputStream(dest);
byte[] buf = new byte[1024]; int len;
while ((len = in.read(buf)) > 0)
out.write(buf, 0, len);
in.close();
out.close();