I'm trying to populate a directory from the contents of a bundle built into my plug-in. The following code works when the bundle is a file-system, but fails when the bundle is a JAR.
What is the best way to test if a URL is a directory? Or is there a completely different, better approach for creating a file structure from a resource bundle?
static private void bundleCopy(String dir, String destination) throws IOException {
Bundle bundle = com.mds.apg.Activator.getDefault().getBundle();
for (#SuppressWarnings("unchecked")
Enumeration<URL> en = (Enumeration<URL>) bundle.findEntries(dir, "*", true);
en.hasMoreElements();) {
URL url = en.nextElement();
String toFileName = destination + url.getPath().substring(dir.length());
File toFile = new File(toFileName);
InputStream in;
try {
in = url.openStream();
} catch (FileNotFoundException e) {
// this exception get thrown for file system directories but not for jar file ones
if (!toFile.mkdir()) {
throw new IOException("bundleCopy: " + "directory Creation Failed: "
+ toFileName);
}
continue;
}
FileCopy.coreStreamCopy(in, toFile);
}
}
I found an answer:
The key point is that the Enumeration entries for directories end in a '/'.
The following correctly distinguishes between directories and files for both JARs and file systems:
static private void bundleCopy(String dir, String destination)
throws IOException, URISyntaxException {
Bundle bundle = com.mds.apg.Activator.getDefault().getBundle();
Enumeration<URL> en = bundle.findEntries(dir, "*", true);
while (en.hasMoreElements()) {
URL url = en.nextElement();
String pathFromBase = url.getPath().substring(dir.length()+1);
String toFileName = destination + pathFromBase;
File toFile = new File(toFileName);
if (pathFromBase.lastIndexOf('/') == pathFromBase.length() - 1) {
// This is a directory - create it and recurse
if (!toFile.mkdir()) {
throw new IOException("bundleCopy: " + "directory Creation Failed: " + toFileName);
}
} else {
// This is a file - copy it
FileCopy.coreStreamCopy(url.openStream(), toFile);
}
}
}
You could try something like new File(FileLocator.resolve(url).toUri()) to convert from the Eclipse-specific URL to one using a native Java protocol.
Related
I'm working on zip files (android-studio-java)
and I tried below solutions:
android.permission.READ_EXTERNAL_STORAGE ---> failed (no read access for zip file)
Access documents and other files ---> failed (no read access for zip file)
android.permission.MANAGE_EXTERNAL_STORAGE ---> passed
My question is: Is there any lower level of permission for zip files or I have to use MANAGE_EXTERNAL_STORAGE?
Thank you,
Actually you can use ACTION_OPEN_DOCUMENT, no need for MANAGE_EXTERNAL_STORAGE
Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT);
intent.addCategory(Intent.CATEGORY_OPENABLE);
// you can set here the type you need setType(application/zip)
intent.setType("*/*");
intent.addFlags(Intent.FLAG_GRANT_WRITE_URI_PERMISSION);
intent.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
startActivityForResult(intent, ZIP_READ_CODE);
OnActivityResult
if (resultCode == RESULT_OK) {
case ZIP_READ_CODE: {
Uri uri = data.getData();
File file = null;
try {
file = read_uri_to_file(uri);
} catch (IOException e) {
//handle error
}
}
}
Read the File:
private File read_uri_to_file(Uri uri) throws IOException {
String displayName = "";
Cursor cursor = getContentResolver().query(uri, null, null, null, null);
if(cursor != null && cursor.moveToFirst()){
try {
displayName = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
}finally {
cursor.close();
}
}
//create temp file to copy and handle selected file
File file = File.createTempFile(
FilenameUtils.getBaseName(displayName),
"."+FilenameUtils.getExtension(displayName)
);
//copy selected file, to a temp file
InputStream inputStream = getContentResolver().openInputStream(uri);
FileUtils.copyInputStreamToFile(inputStream, file);
return file;
}
Here I am trying to add Help .html files on Cntl+space. Its working on my laptop because I have source code but not on server. No error but not returning path on server.
Here is my code:
Bundle bundle=Platform.getBundle("my plugin ID");
URL url = FileLocator.find(bundle, new Path("/doc/myfile/file/"), null);
//Here in file folder my html files.
String fPath = "";
try {
URL fileURL = FileLocator.toFileURL(url);
fPath=fileURL.getPath();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
String fullPath = filePath + name+"."+ myfileextension;
return fullPath;//returning myfile path
You must give FileLocator.toFileURL the full name of the file you want, you can't use just the folder name.
This is because FileLocator.find may have to extract the file from the plug-in jar to a temporary location to make it available.
So:
URL url = FileLocator.find(bundle, new Path("/doc/myfile/file/" + name + "." + myfileextension), null);
Also check your build.properties file. You must include all the folders you want to use in the plugin in this.
I tried to checkout a file using sharpsvn from remote repository,but i found sharpsvn can not to checkout single file only checkout folder,please help me to know how to checkout a file?Thx.
My code
SvnUpdateResult result;
SvnCheckOutArgs checkoutArgs = new SvnCheckOutArgs();
string target = txtRepository.Text.Trim();
SvnUriTarget url = new SvnUriTarget(target);
string fileName = url.FileName;
string path = folder + "\\" + fileName;
using (SvnClient client = new SvnClient())
{
try
{
client.CheckOut(url,txtLocalFilePath.Text.Trim(),out result);//.Update(path,updateArgs,out result);
if (result != null)
{
WriteCheckOutTime(txtRepository.Text.Trim(), result.Revision);
MessageBox.Show("Check out success!", "Info", MessageBoxButtons.OK, MessageBoxIcon.Information);
return;
}
}
catch (SvnException svnException)
{
MessageBox.Show(svnException.Message + "Check out error!", "Error", MessageBoxButtons.OK, MessageBoxIcon.Warning);
}
catch (UriFormatException uriException)
{
MessageBox.Show(uriException.Message + "Check out error!", "Error", MessageBoxButtons.OK, MessageBoxIcon.Warning);
}
}
The smallest element you can check out with Subversion is a directory. It is not possible to check out a single file.
You can check out a directory, but leave it empty, via the Sparse Directories feature. Then update only the file you're interested in. But you must start with a directory.
FYI if you want to checkout empty use following syntax
//first define args
SvnCheckOutArgs args = new SvnCheckOutArgs();
// then for checkout only forlder empty
args.Depth = SvnDepth.Empty;
//checkout folder
client.CheckOut(url,txtLocalFilePath.Text.Trim(),args,out result)
So I am new to GWT and am not sure what the best programming practices are for what I am trying to do. In my web application the user will be able to upload a data file, my application needs to be able to access this file, do some stuff to it, and then let the user download the manipulated file.
So far I have been able to successfully upload a file with an upload servlet with this doPost method:
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
try {
DiskFileItemFactory fileItemFactory = new DiskFileItemFactory();
ServletFileUpload fileUpload = new ServletFileUpload(fileItemFactory);
fileUpload.setSizeMax(FILE_SIZE_LIMIT);
List<FileItem> items = fileUpload.parseRequest(req);
for (FileItem item : items) {
if (item.isFormField()) {
logger.log(Level.INFO, "Received form field:");
logger.log(Level.INFO, "Name: " + item.getFieldName());
logger.log(Level.INFO, "Value: " + item.getString());
} else {
logger.log(Level.INFO, "Received file:");
logger.log(Level.INFO, "Name: " + item.getName());
logger.log(Level.INFO, "Size: " + item.getSize());
}
if (!item.isFormField()) {
if (item.getSize() > FILE_SIZE_LIMIT) {
resp.sendError(HttpServletResponse.SC_REQUEST_ENTITY_TOO_LARGE, "File size exceeds limit");
return;
}
String fileName = item.getName();
if (fileName != null) {
fileName = FilenameUtils.getName(fileName);
}
fileName = getServletContext().getRealPath("/uploadedFiles/" + fileName);
byte[] data = item.get();
FileOutputStream fileOutSt = new FileOutputStream(fileName);
fileOutSt.write(data);
fileOutSt.close();
if (!item.isInMemory())
item.delete();
}
}
} catch (Exception e) {
logger.log(Level.SEVERE, "Throwing servlet exception for unhandled exception", e);
throw new ServletException(e);
}
}
When I look in my war folder, the uploadedFiles folder is created successfully and files are put there.
At this point I am a bit stuck, I have been researching but cannot seem to find a clear concise answer on what is the best way for me to access the uploaded files on the client side in order to manipulate them and then allow the user to download them. Maybe I am approaching this wrong, I am not sure. If someone could point me in the right direction or show me some good examples of the right way to do things that would be great, thanks.
To access the file in client side you need a new servlet or the same you are using with a doGet method.
The client should ask for the file via an Anchor or an Image depending on the file type but adding a parameter so as the server is able to identify the file. Normally you can use the name of the FileInput you used for uploading or maybe you could return a tag from the server.
I would recommend to you to take a try to gwt-upload, it would save a lot of time to you.
I solved my problem. When the file was successfully uploaded, I stored the file name. Later I used a RPC to access the file on the server. I passed the file name to the RPC so that it knows what file I am working on, then it looks for that file in the upload folder. So I can create the java file like this,
File file = new File((this.getServletContext().getRealPath("uploadedFiles") + File.separator + fileName));
and manipulate it how I see fit.
I built an assembly containing one js file.
I marked the file as Embedded Resource and added it into AssemblyInfo file.
I can't refernce the Assembly from a web site. It is in the bin folder but I don't see the reference to it.
It seems like not having at least a class inside the assembly I can't reference it.
I would include the js file into my pages from the assembly.
How should I do this?
Thanks
I do exactly the same thing in one of my projects. I have a central ScriptManager class that actually caches the scripts as it pulls them, but the core of extracting the script file from the embedded resource looks like this:
internal static class ScriptManager
{
private static Dictionary<string, string> m_scriptCache =
new Dictionary<string, string>();
public static string GetScript(string scriptName)
{
return GetScript(scriptName, true);
}
public static string GetScript(string scriptName, bool encloseInCdata)
{
StringBuilder script = new StringBuilder("\r\n");
if (encloseInCdata)
{
script.Append("//<![CDATA[\r\n");
}
if (!m_scriptCache.ContainsKey(scriptName))
{
var asm = Assembly.GetExecutingAssembly();
var stream = asm.GetManifestResourceStream(scriptName);
if (stream == null)
{
var names = asm.GetManifestResourceNames();
// NOTE: you passed in an invalid name.
// Use the above line to determine what tyhe name should be
// most common is not setting the script file to be an embedded resource
if (Debugger.IsAttached) Debugger.Break();
return string.Empty;
}
using (var reader = new StreamReader(stream))
{
var text = reader.ReadToEnd();
m_scriptCache.Add(scriptName, text);
}
}
script.Append(m_scriptCache[scriptName]);
if (encloseInCdata)
{
script.Append("//]]>\r\n");
}
return script.ToString();
}
}
EDIT
To provide more clarity, I've posted my ScriptManager class. To extract a script file, I simply call it like this:
var script = ScriptManager.GetScript("Fully.Qualified.Script.js");
The name you pass in it the full, case-sensitive resource name (the exception handler gets a list of them by calling GetManifestResourceNames()).
This gives you the script as a string - you can then put it out into a file, inject it into the page (which is what I'm doing) or whatever you like.
Assembly myAssembly = // Get your assembly somehow (see below)...
IList<string> resourceNames = myAssembly.GetManifestResourceNames();
This will return a list of all resource names that have been set as 'Embedded Resource'. The name is usually the fully qualified namespace of wherever you put that JS file. So if your project is called My.Project and you store your MyScript.js file inside a folder in your project called Resources, the full name would be My.Project.Resources.MyScript.js
If you then want to use that JS file:
Stream stream = myAssembly.GetManifestResourceStream(myResourceName);
Where myResourceName argument might be "My.Project.Resources.MyScript.js". To get that JS file in that Stream object, you'll need to write it as a file to the hard drive, and serve it from your website as a static file, something like this:
Stream stream = executingAssembly.GetManifestResourceStream(imageResourcePath);
if (stream != null)
{
string directory = Path.GetDirectoryName("C:/WebApps/MyApp/Scripts/");
using (Stream file = File.OpenWrite(directory + "MyScript.js"))
{
CopyStream(stream, file);
}
stream.Dispose();
}
And the code for CopyStream method:
private static void CopyStream(Stream input, Stream output)
{
byte[] buffer = new byte[8 * 1024];
int len;
while ((len = input.Read(buffer, 0, buffer.Length)) > 0)
{
output.Write(buffer, 0, len);
}
}
You might want to stick all this code in an Application_Start event in your Global.asax. You don't want it to run for each request
Now getting a reference to your Assembly is a different matter, there are many ways. One way is to include all the above code in your Assembly in question, then make sure you reference that Assembly from your main WebApp project in Visual Studio, then get a reference to the currently executing Assembly like so.
namespace My.Project
{
public class ResourceLoader
{
public static void LoadResources()
{
Assembly myAssembly = Assembly.GetExecutingAssembly();
// rest of code from above (snip)
}
}
}
Then call ResourceLoader.LoadResources() from your Application_Start event in your Global.asax.
Hope this helps
Fully working example (I hope):
namespace TestResources.Assembly
{
public class ResourceLoader
{
public static void LoadResources()
{
Assembly myAssembly = Assembly.GetExecutingAssembly();
Stream stream = myAssembly
.GetManifestResourceStream("TestResources.Assembly.CheckAnswer.js");
if (stream != null)
{
string directory = Path.GetDirectoryName("C:/WebApps/MyApp/Scripts/");
using (Stream file = File.OpenWrite(directory + "MyScript.js"))
{
CopyStream(stream, file);
}
stream.Dispose();
}
}
private static void CopyStream(Stream input, Stream output)
{
byte[] buffer = new byte[8 * 1024];
int len;
while ((len = input.Read(buffer, 0, buffer.Length)) > 0)
{
output.Write(buffer, 0, len);
}
}
}
}
Some caveats:
Change "C:/WebApps/MyApp/" to wherever your web app is located, maybe write some code to work this out dynamically
Make sure the /Scripts folder exists in your webapp root
I think it will overwrite the 'MyScript.js' file if it already exists, but just in case you might want to add some code to check for that file and delete it
Then stick a call to this code in your Global.asax file:
protected void Application_Start()
{
ResourceLoader.LoadResources();
}
Then the path for your web site will be /Scripts/MyScript.js eg:
<head>
<!-- Rest of head (snip) -->
<script type="text/javascript" src="/Scripts/MyScript.js"></script>
</head>