React Native [Android]
Samsung Phone
Libraries :
react-native-document-picker [ returns our URI]
react-native-get-real-path [ converts URI to real path]
Able to :
Get a URI from local files and get real path including images
Able to get URI from Google Drive when I select a file
Unable :
Convert Google Drive URI to real path
DocumentPicker.show({filetype: [DocumentPickerUtil.allFiles()],},(error,res) => {
RNGRP.getRealPathFromURI(path).then(function(androidPath){
console.log('AndroidPath : ', androidPath);
})
}
my URI from google drive is like so :
content://com.google.android.apps.docs.storage/document/acc%3D2%3Bdoc%3D1
Fixed bug to get absolute path of Google Drive File.
So it turns out that we cannot directly get the absolute path from the URI that has been returned by selecting Google Drive File. Hence we need to apply some sort of hack to solve the problem.
What I did was, I forked the react-native-get-real-path repo into our own and then changed few things in GRP.java file.
I basically created InputStream from obtained google drive file's URI and then, using that stream, copied the file into the app's cache directory and returned the absolute path to that file and voila.
Here is the code snippet for the solution:
input = context.getContentResolver().openInputStream(uri);
/* save stream to temp file */
/* displayName is obtained from the URI */
File file = new File(context.getCacheDir(), displayName);
OutputStream output = new FileOutputStream(file);
byte[] buffer = new byte[4 * 1024]; // or other buffer size
int read;
while ((read = input.read(buffer)) != -1) {
output.write(buffer, 0, read);
}
output.flush();
final String outputPath = file.getAbsolutePath();
return outputPath;
You can clone the git repository. Reference of https://github.com/Wraptime/react-native-get-real-path/pull/8.
Related
I'm writing a web application that, among other things, allows users to upload files to my server. In order to prevent name clashes and to organize the files, I rename them once they are put on my server. By keeping track of the original file name I can communicate with the file's owner without them ever knowing I changed the file name on the back end. That is, until they go do download the file. In that case they're prompted to download a file with a unfamiliar name.
My question is, is there any way to specify the name of a file to be downloaded using just HTML? So a user uploads a file named 'abc.txt' and I rename it to 'xyz.txt', but when they download it I want the browser to save the file as 'abc.txt' by default. If this isn't possible with just HTML, is there any way to do it?
When they click a button to download the file, you can add the HTML5 attribute download where you can set the default filename.
That's what I did, when I created a xlsx file and the browser want to save it as zip file.
Download
Download Export
Can't find a way in HTML. I think you'll need a server-side script which will output a content-disposition header. In php this is done like this:
header('Content-Disposition: attachment; filename="downloaded.pdf"');
if you wish to provide a default filename, but not automatic download, this seems to work.
header('Content-Disposition: inline; filename="filetodownload.jpg"');
In fact, it is the server that is directly serving your files, so you have no way to interact with it from HTML, as HTML is not involved at all.
just need to use HTML5 a tag download attribute
codepen live demo
https://codepen.io/xgqfrms/full/GyEGzG/
my screen shortcut.
update answer
whether a file is downloadable depends on the server's response config, such as Content-Type, Content-Disposition;
download file's extensions are optional, depending on the server's config, too.
'Content-Type': 'application/octet-stream',
// it means unknown binary file,
// browsers usually don't execute it, or even ask if it should be executed.
'Content-Disposition': `attachment; filename=server_filename.filetype`,
// if the header specifies a filename,
// it takes priority over a filename specified in the download attribute.
download blob url file
function generatorBlobVideo(url, type, dom, link) {
var xhr = new XMLHttpRequest();
xhr.open('GET', url);
xhr.responseType = 'arraybuffer';
xhr.onload = function(res) {
// console.log('res =', res);
var blob = new Blob(
[xhr.response],
{'type' : type},
);
// create blob url
var urlBlob = URL.createObjectURL(blob);
dom.src = urlBlob;
// download file using `a` tag
link.href = urlBlob;
};
xhr.send();
}
(function() {
var type = 'image/png';
var url = 'https://cdn.xgqfrms.xyz/logo/icon.png';
var dom = document.querySelector('#img');
var link = document.querySelector('#img-link');
generatorBlobVideo(url, type, dom, link);
})();
https://cdn.xgqfrms.xyz/HTML5/Blob/index.html
refs
https://developer.mozilla.org/en-US/docs/Web/HTML/Element/a#download
https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/Content-Disposition
https://developer.mozilla.org/en-US/docs/Web/HTTP/Basics_of_HTTP/MIME_types#important_mime_types_for_web_developers
Sometimes #Mephiztopheles answer won't work on blob storages and some browsers.
For this you need to use a custom function to convert the file to blob and download it
const coverntFiletoBlobAndDownload = async (file, name) => {
const blob = await fetch(file).then(r => r.blob())
const url = URL.createObjectURL(blob)
const a = document.createElement('a')
a.style.display = 'none'
a.href = url
a.download = name // add custom extension here
document.body.appendChild(a)
a.click()
window.URL.revokeObjectURL(url)
}
Same code as #Hillkim Henry but with a.remove() improvement
This forces the document to remove the a tag from the body and avoid multiple elements
const coverntFiletoBlobAndDownload = async (file, name) => {
const blob = await fetch(file).then(r => r.blob())
const url = URL.createObjectURL(blob)
const a = document.createElement('a')
a.style.display = 'none'
a.href = url
a.download = name // add custom extension here
document.body.appendChild(a)
a.click()
window.URL.revokeObjectURL(url)
// Remove "a" tag from the body
a.remove()
}
Well, #Palantir's answer is, for me, the most correct way!
If you plan to use that with multiple files, then i suggest you to use (or make one) PHP Download Manager.
BUT, if you want to make that to one or two files, I will suggest you the mod_rewrite option:
You have to create or edit your .htaccess file on htdocs folder and add this:
RewriteEngine on
RewriteRule ^abc\.txt$ xyz.txt
With this code, users will download xyz.txt data with the name abc.txt
NOTE: Verify if you have already the "RewriteEngine on " on your file, if yes, add only the second for each file you wish to redirect.
Good Luck ;)
(Sorry for my english)
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);
It is fairly common to allow users to download a file via having some path modifier in the URL
//MVC Action to download the correct file From our Content directory
public ActionResult GetFile(string name) {
string path = this.Server.MapPath("~/Content/" + name);
byte[] file = System.IO.File.ReadAllBytes(path);
return this.File(file, "html/text");
}
quoted from http://hugoware.net/blog/dude-for-real-encrypt-your-web-config
An application I'm working with has liberal path downloads ( directory based ) sprinkled throughout the application, hence it is super vulnerable to requests like "http://localhost:1100/Home/GetFile?name=../web.config" or ( ..%2fweb.config )
Is there an easy way to restrict access to the config file - do I need to provide a custom Server.MapPath with whitelisted directories - or is there a better way.
How do you secure your file downloads - are path based downloads inherently insecure?
A simple option, assuming that all files in the ~/Content directory are safe to download would be to verify that the path is actually under (or in) the ~/Content directory and not up from it, as ~/Content/../web.config would be. I might do something like this:
// MVC Action to download the correct file From our Content directory
public ActionResult GetFile(string name) {
// Safe path
var safePath = this.Server.MapPath("~/Content");
// Requested path
string path = this.Server.MapPath("~/Content/" + name);
// Make sure requested path is safe
if (!path.StartsWith(safePath))
// NOT SAFE! Do something here, like show an error message
// Read file and return it
byte[] file = System.IO.File.ReadAllBytes(path);
return this.File(file, "html/text");
}
using (var repo = new Repository("path/to/your/repo"))
{
LibGit2Sharp.PullOptions options = new LibGit2Sharp.PullOptions();
options.FetchOptions = new FetchOptions();
options.FetchOptions.CredentialsProvider = new CredentialsHandler(
(url, usernameFromUrl, types) =>
new UsernamePasswordCredentials()
{
Username = USERNAME,
Password = PASSWORD
});
repo.Network.Pull(new LibGit2Sharp.Signature(USERNAME, EMAIL, new DateTimeOffset(DateTime.Now)), options)
}
i do not konw how to set arguments,when i use it,one error will show-----Unsupported URL protocol.could you tell me how to set arguments?
It depends on the url you are using.
For instance, issue 649 clearly states:
git.git supports relative URLs in remote configurations and resolves them relative to the working directory.
libgit2 currently fails with "Unsupported URL protocol".
It expects paths to be absolute.
So if your url is actually a local path, make sure it is an absolute path (and not a relative one).
As commented by 崔重阳, using an https instea of an sssh url is supported.
I'm using the following function to convert paths into a valid Virtual Path:
public string GetFullPath(string path)
{
Ensure.Argument.NotNullOrEmpty(path, "path");
if (path[0] == '~') // a virtual path e.g. ~/assets/style.less
{
return path;
}
if (VirtualPathUtility.IsAbsolute(path)) // an absolute path e.g. /assets/style.less
{
return VirtualPathUtility.ToAppRelative(path,
HostingEnvironment.IsHosted ? HostingEnvironment.ApplicationVirtualPath : "/");
}
// otherwise, assume relative e.g. style.less or ../../variables.less
return VirtualPathUtility.Combine(VirtualPathUtility.AppendTrailingSlash(currentFileDirectory), path);
}
This passes all my tests other than for when the input path is a relative path, above the website directory.
For example given a currentFileDirectory of ~/foo/bar and a relative path of ../../../ I want to detect this and attempt to fix the path.
Server.MapPath is an easy way to validate either virtual or dos style paths. Validation is restricted for security reasons from validating any physical paths that are outside the web site. See my comment above.
To elaborate on Mike's answer we can validate whether a relative path attempts to go outside of the physical web site directory by:
Getting the combined path of the current path and relative path using Path.GetFullPath
Getting the file system path of the attempted path using Server.MapPath
Getting the file system path of the root of the application
Validating that the attempted path exists within the root path
Example:
var currentFileDirectory = "~/foo/bar";
var relativePath = "../../../";
var attemptedPath = Path.GetFullPath(Path.Combine(Server.MapPath(currentFileDirectory), relativePath)); // 1 + 2
var rootPath = Server.MapPath("~/"); // 3
if (attemptedPath.IndexOf(rootPath, StringComparison.InvariantCultureIgnoreCase) == -1) // 4
{
throw new Exception(string.Format("Path {0} is outside path {1}",
rootPath, HostingEnvironment.ApplicationPhysicalPath));
}