I'm having trouble uploading an image in my website. Sometimes, it uploads. But sometimes it also says can’t create cache file. How do I fix it?
Here's my code:
File file = new File(imagePath);
BufferedImage bi;
try{
bi = ImageIO.read(file);
}catch(javax.imageio.IIOException e){
if(request.getParameter("fi") != null){
file = new File(context.getInitParameter("ImgPath") + "placeholder/150x80.png");
}else if (request.getParameter("li") != null){
file = new File(context.getInitParameter("ImgPath") + "placeholder/150x80.png");
}
bi = ImageIO.read(file);
}
OutputStream outImg = response.getOutputStream();
File cacheDir = new File(context.getInitParameter("ImgPath") + "cache");
try {
ImageIO.setCacheDirectory(cacheDir);
ImageIO.write(bi, "png", outImg);
} catch (Exception ex) {
}
outImg.close();
Problem: Your tomcat installation is failed to create temp folder on startup or temp folder is not writable.
Solution: Make sure temp folder exists under \temp and it is writable.
You can create it manually.
Or
You can override default temp folder location of Tomcat by setting the value for CATALINA_TMPDIR environment variable in catalina.bat (windows) or catalina.sh (linux).
# CATALINA_TMPDIR (Optional) Directory path location of temporary directory
# the JVM should use (java.io.tmpdir). Defaults to
# $CATALINA_BASE/temp.
In which security context are your code in? In some implementation of 'multiple bundle on single virtual machine' framework, there can exist two types of bundles, one has permission to write temp folder, and the other does not. If the former would access ImageIO package first, ImageIO package determined that it would have permission to write temp folder, and trying to use it all of subsequent calls, but, in such frameworks, the latter can call ImageIO as well, and it will fail, since that bundle does not have access to the temp file. The behavior would change either the former would access first or vice versa, and if the latter would call ImageIO first, it won't ever use cache directory for that virtual machine instance and you don't see any troubles.
If your code does not have access to the cache directory and someone which has access to there calls ImageIO first, your code will fail.
And, it seems that your code would call ImageIO.setCacheDirectory() with your local data folder. In such frameworks, there are many cases that other bundles running on the same virtual machine would not have access to the local folder of your bundle. If so, they would have to throw IOException, if your setCacheDirectory() call was successful and you have given the directory only your code can access.
If your code successfully set the cache directory to your local data folder which other bundles cannot access to, your code might work fine, but other bundles would fail when trying to use ImageIO.
Related
I have been searching for a way to get a file object from a file, in the resources folder. I have read a lot of similar questions on this website but non fix my problem exactly.
Link already referred to
how-to-get-a-path-to-a-resource-in-a-java-jar-file
that got really close to answering my question:
String path = this.getClass().getClassLoader().getResource(<resourceFileName>)
.toExternalForm()
I am trying to have a resource file that I can write data into and then bring that file object to another part of my program, I know I can technically create a temp file that, I then write data into then pass it into a part of my program, the problem with this approach is that I think it can take a lot of system recourses, my program will need to create a lot of these temp files.
Is there any way, I can reuse one file in the resource folder? all I need is to get it's path (and it needs to work in a jar).I have tried this snipper of code i created for testing, i don't really know why it returns false, because in the ide it returns true.
public File getFile(String fileName) throws FileNotFoundException {
//Getting file from the resources folder
ClassLoader classLoader = getClass().getClassLoader();
URL fileUrl = classLoader.getResource(fileName);
if (fileUrl == null)
throw new FileNotFoundException("Cannot find file " + fileName);
System.out.println("before: " + fileUrl.toExternalForm());
final String result = fileUrl.toExternalForm()
.replace("jar:" , "")
.replace("file:" , "");
System.out.println("after: " + result);
return new File(result);
}
Output:
before: jar:file:/C:/Users/%myuser%/Downloads/Untitlecd.jar!/Recording.wav
after: /C:/Users/%myuser%/Downloads/Untitlecd.jar!/Recording.wav
false
i have been searching for a way to get a file object from a file in the resources folder.
This is flat out impossible. The resources folder is going to end up jarred into your distribution, and you can't edit jar files, they are read only (or at least, you should consider them so. Non-idiotic deployments will generally mark their own code files (which includes those jars) as read-only to the running process. Even if not, editing jar files is extremely heavy and not something you want to do. Even if you do, on windows, open files can't be edited/replaced like this without significant headaches).
The 'resources' folder simply isn't designed for files that are meant to be modified.
The usual strategy is to make a directory someplace (for example, the user's home dir, accessing via System.getProperty("user.home"), and then make/edit files within that dir. If you wish, you can put templates in your resources folder and use those to 'initialize' that dir hanging off the user's home dir with a skeleton version.
If you have a few ten thousand files to make, whatever process needs this needs to be adjusted to not need this. For example, by using a database (H2, perhaps, if you want to ship it with your java app and have it be as low impact as possible).
Is there a way in Java to construct a File instance on a resource retrieved from a jar through the classloader?
My application uses some files from the jar (default) or from a filesystem directory specified at runtime (user input). I'm looking for a consistent way of
a) loading these files as a stream
b) listing the files in the user-defined directory or the directory in the jar respectively
Edit: Apparently, the ideal approach would be to stay away from java.io.File altogether. Is there a way to load a directory from the classpath and list its contents (files/entities contained in it)?
I had the same problem and was able to use the following:
// Load the directory as a resource
URL dir_url = ClassLoader.getSystemResource(dir_path);
// Turn the resource into a File object
File dir = new File(dir_url.toURI());
// List the directory
String files = dir.list()
ClassLoader.getResourceAsStream and Class.getResourceAsStream are definitely the way to go for loading the resource data. However, I don't believe there's any way of "listing" the contents of an element of the classpath.
In some cases this may be simply impossible - for instance, a ClassLoader could generate data on the fly, based on what resource name it's asked for. If you look at the ClassLoader API (which is basically what the classpath mechanism works through) you'll see there isn't anything to do what you want.
If you know you've actually got a jar file, you could load that with ZipInputStream to find out what's available. It will mean you'll have different code for directories and jar files though.
One alternative, if the files are created separately first, is to include a sort of manifest file containing the list of available resources. Bundle that in the jar file or include it in the file system as a file, and load it before offering the user a choice of resources.
Here is a bit of code from one of my applications...
Let me know if it suits your needs.
You can use this if you know the file you want to use.
URL defaultImage = ClassA.class.getResource("/packageA/subPackage/image-name.png");
File imageFile = new File(defaultImage.toURI());
A reliable way to construct a File instance on a resource retrieved from a jar is it to copy the resource as a stream into a temporary File (the temp file will be deleted when the JVM exits):
public static File getResourceAsFile(String resourcePath) {
try {
InputStream in = ClassLoader.getSystemClassLoader().getResourceAsStream(resourcePath);
if (in == null) {
return null;
}
File tempFile = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
tempFile.deleteOnExit();
try (FileOutputStream out = new FileOutputStream(tempFile)) {
//copy stream
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
}
return tempFile;
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
Try this:
ClassLoader.getResourceAsStream ("some/pkg/resource.properties");
There are more methods available, e.g. see here:
http://www.javaworld.com/javaworld/javaqa/2003-08/01-qa-0808-property.html
This is one option: http://www.uofr.net/~greg/java/get-resource-listing.html
I'm trying to create a directory hierarchy and then copy a file from my assets into the newly created directory. I've used the code samples offered here on stackoverflow but I'm not getting anywhere. Essentially my code goes like this:
InputStream in = getAssets.open(mydb.sqlite);
File dir = new File("/data/data/my.package/databases/");
dir.mkdirs();
out = new FileOutputStream("/data/data/my.package/databases/mydb.sqlite");
while ((len = in.read(buffer)) > 0)
{
out.write(buffer, 0, len);
}
out.flush();
out.close();
in.close
This is all done within the main activity on first launch and of course should only be done once. Before the code runs, the only directory that exists is /data/. Yes, it is a database file but that shouldn't matter at this point since I'm just copying the file and haven't yet tried to access it with the database code.
The code doesn't throw any exceptions, and dir.mkdirs() returns true. Regardless, the directories are not created and the file is not copied. I have added the permissions line for WRITE_EXTERNAL_STORAGE to my manifest as has been suggested in several places, FWIW. I can step through the code and everything appears to be fine. But of course it's not.
What do I need to be looking for here? File permissions problems? Why does the code think everything is there (ie, when it gets to the copy code, it doesn't throw FileNotFoundException) but in reality there's nothing on the file system?
Edit: Here's what I've learned. getDir() creates the directory in /data/data/my.package/app_*. Not sure why it uses this prefix on the directory name. This won't work if you need to put something in a standard location such as databases/ or files/.
Use openFileOutput instead as /data/data is internal storage and you should only use the Android APIs to get a file handle there. See: http://developer.android.com/guide/topics/data/data-storage.html#filesInternal
Edit:
/data/data/my.package is handled by openFileOutput/getDir so you would create your directory with getDir('databases',MODE_PRIVATE) and create your file there.
Edit2:
From getDir documentation:
You can use the returned File object to create and access files in this directory.
I am trying to learn how to use the Simple XML Framework as detailed in this thread : Best practices for parsing XML.
I am using the following code :
public class SimpleXMLParserActivity extends Activity {
/** Called when the activity is first created. */
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
//setContentView(R.layout.main);
Serializer serializer = new Persister();
Example example = new Example("Example message", 123);
File result = new File("example.xml");
try {
Log.d("Start", "Starting Serializer");
serializer.write(example, result);
} catch (Exception e) {
// TODO Auto-generated catch block
Log.d("Self", "Error");
e.printStackTrace();
}
}
}
I am having a problem understanding the line
File result = new File("example.xml");
1) Does this line create a new file in my app called example.xml ? If so where is this file located.
2) Or does this line look for an existing file called example.xml and then add to it ? If so where should the example.xml file be placed in my app bundle so that it can be found. I do notice at the moment I am getting an error message :
java.io.FileNotFoundException: /example.xml (Read-only file system)
Thank you.
File result = new File("example.xml")
This line will just store the filename "example.xml" in a new File object. There is no check if that file actually exists and it does not try to create it either.
A file without specifying an absolute path (starting with / like new File("/sdcard/example.xml")) is considered to be in the current working directory which I guess is / for Android apps (-> /example.xml (Read-only file system))
I guess serializer.write(example, result); tries to create the actual file for your but fails since you can't write to '/'.
You have to specify a path for that file. There are several places you can store files, e.g.
Context#getFilesDir() will give you a place in your app's home directory (/data/data/your.package/files/) where only you can read / write - without extra permission.
Environment#getExternalStorageDirectory() will give you the general primary storage thing (might be /sdcard/ - but that's very different for devices). To write here you'll need the WRITE_EXTERNAL_STORAGE permission.
there are more places available in Environment that are more specialized. E.g. for media files, downloads, caching, etc.
there is also Context#getExternalFilesDir() for app private (big) files you want to store on the external storage (something like /sdcard/Android/data/your.package/)
to fix your code you could do
File result = new File(Environment.getExternalStorageDirectory(), "example.xml");
Edit: either use the provided mechanisms to get an existing directory (preferred but you are limited to the folders you are supposed to use):
// via File - /data/data/your.package/app_assets/example.xml
File outputFile = new File(getDir("assets", Context.MODE_PRIVATE), "example.xml");
serializer.write(outputFile, result);
// via FileOutputStream - /data/data/your.package/files/example.xml
FileOutputStream outputStream = openFileOutput("example.xml", Context.MODE_PRIVATE);
serializer.write(outputStream, result);
or you may need to create the directories yourself (hackish way to get your app dir but it should work):
File outputFile = new File(new File(getFilesDir().getParentFile(), "assets"), "example.xml");
outputFile.mkdirs();
serializer.write(outputFile, result);
Try to avoid specifying full paths like "/data/data/com.simpletest.test/assets/example.xml" since they might be different on other devices / Android versions. Even the / is not guaranteed to be /. It's safer to use File.separatorChar instead if you have to.
2 solutions to do it cleanly :
use openFileOutput to write a private file in the application private directory (which could be located in the internal memory or the external storage if the app was moved there). See here for a snippet
or use the File constructor to create the File anywhere your app has write access. This is if you want to store the file on the SDCard for example. Instantiating a file doesn't create it on the file system, unless you start writiung to it (with FileOutputStream for example)
I'd recommend approach 1, it's better for users because these files get erased when your app is uninstalled. If the file is large, then using the External Storage is probably better.
What I read on the Android pages, I see it creates a file with that name:
File constructor
I think it writes it to the /data/data/packagname directory
edit: the 'packagename' was not shown in the tekst above. I put it between brackets. :s
Try saving to /sdcard/example.xml.
Is there a way in Java to construct a File instance on a resource retrieved from a jar through the classloader?
My application uses some files from the jar (default) or from a filesystem directory specified at runtime (user input). I'm looking for a consistent way of
a) loading these files as a stream
b) listing the files in the user-defined directory or the directory in the jar respectively
Edit: Apparently, the ideal approach would be to stay away from java.io.File altogether. Is there a way to load a directory from the classpath and list its contents (files/entities contained in it)?
I had the same problem and was able to use the following:
// Load the directory as a resource
URL dir_url = ClassLoader.getSystemResource(dir_path);
// Turn the resource into a File object
File dir = new File(dir_url.toURI());
// List the directory
String files = dir.list()
ClassLoader.getResourceAsStream and Class.getResourceAsStream are definitely the way to go for loading the resource data. However, I don't believe there's any way of "listing" the contents of an element of the classpath.
In some cases this may be simply impossible - for instance, a ClassLoader could generate data on the fly, based on what resource name it's asked for. If you look at the ClassLoader API (which is basically what the classpath mechanism works through) you'll see there isn't anything to do what you want.
If you know you've actually got a jar file, you could load that with ZipInputStream to find out what's available. It will mean you'll have different code for directories and jar files though.
One alternative, if the files are created separately first, is to include a sort of manifest file containing the list of available resources. Bundle that in the jar file or include it in the file system as a file, and load it before offering the user a choice of resources.
Here is a bit of code from one of my applications...
Let me know if it suits your needs.
You can use this if you know the file you want to use.
URL defaultImage = ClassA.class.getResource("/packageA/subPackage/image-name.png");
File imageFile = new File(defaultImage.toURI());
A reliable way to construct a File instance on a resource retrieved from a jar is it to copy the resource as a stream into a temporary File (the temp file will be deleted when the JVM exits):
public static File getResourceAsFile(String resourcePath) {
try {
InputStream in = ClassLoader.getSystemClassLoader().getResourceAsStream(resourcePath);
if (in == null) {
return null;
}
File tempFile = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
tempFile.deleteOnExit();
try (FileOutputStream out = new FileOutputStream(tempFile)) {
//copy stream
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
}
return tempFile;
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
Try this:
ClassLoader.getResourceAsStream ("some/pkg/resource.properties");
There are more methods available, e.g. see here:
http://www.javaworld.com/javaworld/javaqa/2003-08/01-qa-0808-property.html
This is one option: http://www.uofr.net/~greg/java/get-resource-listing.html