filenotfoundexception

Java - Access is denied java.io.FileNotFoundException [duplicate]

耗尽温柔 提交于 2019-11-27 03:56:19
This question already has an answer here: java.io.FileNotFoundException: (Access is denied) 4 answers I have the following code: List<FileItem> items = uploadHandler.parseRequest(request); for (FileItem item : items) { if (!item.isFormField()) { File file = new File("D:/Data"); } } When I am trying to save a file, I am getting the following error java.io.FileNotFoundException: D:\Data (Access is denied.) What could be the reason and how can I resolve this? I do have read and write permission on this folder. Julien When you create a new File , you are supposed to provide the file name, not only

java.io.FileNotFoundException (permission denied) despite chmod 777

末鹿安然 提交于 2019-11-27 02:41:07
问题 I have faced strange poblem while writing Grails application deployed on Tomcat. After creating simple test controller I want to write test contents in package com package com.domain.controller import java.io.File; import java.io.PrintWriter; class TestController { def index() { // test try { PrintWriter writer = new PrintWriter("/home/user/domains/domain.com/public_html/the-file-name.txt"); writer.println("The first line"); writer.println("The second line"); writer.close(); } catch

Getting all the time “permission denied” or “no such file or directory” by trying to save Bitmap image. What should i do?

元气小坏坏 提交于 2019-11-27 02:13:48
I`m trying to save Bitmap image by this code: File sdcard = Environment.getExternalStorageDirectory(); String filename = "test"; File folder = new File(sdcard, "/Download"); Log.v("ImageStorage1", "EXiST?: " + folder.exists()); folder.mkdirs(); Log.v("ImageStorage2", "EXIST!: " + folder.exists()); Log.v("ImageStorage", "Folder: " + folder); File file = new File(folder, filename + ".jpg"); try { FileOutputStream out = new FileOutputStream(file.getAbsoluteFile()); result.compress(Bitmap.CompressFormat.JPEG, 90, out); out.flush(); out.close(); } catch (Exception e) { e.printStackTrace(); } I`m

FileNotFoundException while getting the InputStream object from HttpURLConnection

一世执手 提交于 2019-11-27 00:45:36
I am trying to send a post request to a url using HttpURLConnection (for using cUrl in java). The content of the request is xml and at the end point, the application processes the xml and stores a record to the database and then sends back a response in form of xml string. The app is hosted on apache-tomcat locally. When I execute this code from the terminal, a row gets added to the db as expected. But an exception is thrown as follows while getting the InputStream from the connection java.io.FileNotFoundException: http://localhost:8080/myapp/service/generate at sun.net.www.protocol.http

java.io.FileNotFoundException: (Access is denied)

这一生的挚爱 提交于 2019-11-26 22:42:54
I am trying to read the files inside a folder, but when I run the program it throws this exception. I tried with some other folders also. It throws the same exception. Exception in thread "main" java.io.FileNotFoundException: C:\backup (Access is denied) at java.io.FileInputStream.open(Native Method) at java.io.FileInputStream.<init>(Unknown Source) You cannot open and read a directory, use the isFile() and isDirectory() methods to distinguish between files and folders. You can get the contents of folders using the list() and listFiles() methods (for filenames and File s respectively) you can

Can't Access Resources In Executable Jar

馋奶兔 提交于 2019-11-26 22:05:27
问题 Can someone please point out what I'm doing wrong here. I have a small weather app that generates and sends an HTML email. With my code below, everything works fine when I run it from Eclipse. My email gets generated, it's able to access my image resources and it sends the email with the included attachment. However, when I build the executable jar by running mvn install and run the jar using java -jar NameOfMyJar.jar I get java.io.FileNotFound Exceptions for my image resource. I know that I

java.io.FileNotFoundException when writing uploaded file to disk via getRealPath()

♀尐吖头ヾ 提交于 2019-11-26 21:07:05
Glassfish seems to be adding extra to the path I want to save my image file too, is there some way to use only the absolute path my servlet gets with String appPath = request.getServletContext().getRealPath(""); ? I have spent days trying different methods to upload an image file and have a servlet save it to disk. I used this example: http://www.codejava.net/java-ee/servlet/how-to-write-upload-file-servlet-with-servlet-30-api Using debug I can see the file name and path information is collected correctly but the code fails at `part.write(savePath + File.separator + fileName);`` And glassfish

Python's “open()” throws different errors for “file not found” - how to handle both exceptions?

家住魔仙堡 提交于 2019-11-26 19:48:42
问题 I have a script where a user is prompted to type a filename (of a file that is to be opened), and if the file doesn't exist in the current directory, the user is prompted again. Here is the short version: file = input("Type filename: ") ... try: fileContent = open(filename, "r") ... except FileNotFoundError: ... When I tested my script on my MacOS X in Python 3.3x it worked perfectly fine when I type the wrong filename on purpose (it executes the suite under "expect"). However, when I wanted

Android - How to get Uri from raw file?

醉酒当歌 提交于 2019-11-26 18:31:29
问题 I am trying to get the Uri from a raw file I have included in the project in the raw folder. But I am getting a FileNotFoundException , no matter what. The file is a .wav file, also tried it with a .mp4 , also doesn't work. Playing both files with MediaPlayer DOES work. The Uri returns: mark.dijkema.android.eindopdracht/2130968576 My Code: package mark.dijkema.android.eindopdracht; import java.io.DataInputStream; import java.io.File; import java.io.FileInputStream; import java.io

FileNotFoundException for HttpURLConnection in Ice Cream Sandwich

徘徊边缘 提交于 2019-11-26 17:40:17
问题 I have an Android app that works fine with Android 2.x and 3.x, but it fails when run on Android 4.x. The problem is in this section of code: URL url = new URL("http://blahblah.blah/somedata.xml"); HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection(); urlConnection.setRequestMethod("GET"); urlConnection.setDoOutput(true); urlConnection.connect(); InputStream inputStream = urlConnection.getInputStream(); When the application is running on Android 4.x, the getInputStream()