I have created a web service which will query the data base and save the file to a specific location. I have tested the code in the local system and its working fine. But when I am trying to put the code in the server it is showing file not found exception.
Below is the piece of code that I am running:
FileInputStream fileInputStream = new FileInputStream("D:\\ram_sahu/Hello.csv");
Could you guys please help me in this.
Thanks,
Ram
The problem is that you have an absolute path for the file which 'points' to the location of the file locally.
You need to put the path according to where it will be in the server.
Related
I need some help i am working on backup and restoring sqlite.db file , In this process i store a copy of db file in my device storage and restore it from there which is working perfectly fine. But when i send this file through any email server like gmail and download it into my storage it add timestamp to its name i handled that in code but it says file not found exception ,i guess gmail is not downloading it properly in storage or not giving access to other apps .
Whenever i rename that file just replacing any alphabet from that name than i can restore that file please help me out i am stuck here ,Thanks in advance
I have found the solution :
My problem was when a file is downloaded from gmail or any other server when we try to get its path it does not give its actual path it give path like /downloads/#2386
so when i try to open it, it shows file not found exception so to overcome this problem i have to get actual file path from that path, after getting that path and try to open it, it works perfectly fine.
I am uploading an excel file to the tomcat server. Which is saving inside my eclipse directory D:\workspace_Eclipse\.metadata\.plugins\org.eclipse.wst.server.core\tmp1\wtpwebapps\StatusPortal\Job_doc\abc.xls
When ever i am accessing this file its giving me file not found Exception \Job_doc\abc.xls.
Its could not able to find the path which is i am giving while accessing the file like
\Job_doc\abc.xls
I am giving the path \Job_doc\abc.xls while accessing.
This is because you are using a relative path. Eclipse will use the current working directory to be a temp location for deploying the webapp. So the file is uploaded to the folder relative to this path (This happens when you start the app from eclipse Run On Server. Define your paths as static constants(May be you can use absolute paths for testing). After testing you can use the relative paths on production deployment.
Still, you can do alternate way. Dont use the integrated tomcat server of Eclipse. Use a standalone server, use the descriptor file to link the webapp in workspace to tomcat. After the save, just reload the app in tomcat manager and try.
Try reading your file using ClassLoader as below:
InputStream inputStream =
getClass().getClassLoader().getResourceAsStream("/Job_doc/abc.xls");
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream ));
If you want to get the File object, then try as below:
URI uri = getClass().getClassLoader().getResource("/Job_doc/abc.xls").toURI();
File file = new File(uri);
I have a small problem calling a path(that has the python file, that I need to run) in the following code:
Process p = Runtime.getRuntime().exec(callAndArgs,env,
new java.io.File("C:\\Users\\Balkishore\\Documents\\NetBeansProjects\\Testinstrument_Rest\\build\\web"));//excuting python file
As it can be seen from the above code, the python file is called using the path specified in java.io.file function. But it is very specific, as it can be run only in my computer. How can i make it generic, so that it is possible to run this piece of code in any computer?
Any help would be very much appreciated.
Put your python script to the location relative to your working directory and use relative path. Alternatively use configuration file or property to read the path from.
If this file is already exist in the app then you need to do
ServletContext.getRealPath("/");
which will give you the path to web root now from here you need to move relatively to reach to your file
If this is an external file
put it in ${user.home}/appname/
String filePath = System.getProperty("user.home")+File.separator+"APP_NAME"
and instruct your users to put the file in this path, or read the path from some configuration file (.properties, .conf)
I had problems while finding the path of file(s) in Netbeans..
Problem is already solved (checked answer).
Today I noticed another problem: When project is finished,
I have to execute the generated .jar to launch the program, but it doesn't work because an error occurs: NullPointer (where to load a file) when accessing/openning jar outside Netbeans.
Is it possible to open a file with the class file in Java/Netbeans which works in Netbeans and even in any directory?
I've found already some threads about my problem in site but none was helpful.
Code:
File file = new File(URLDecoder.decode(this.getClass().getResource("file.xml").getFile(), "UTF-8"));
The problem you have is that File only refer to files on the filesystem, not files in jars.
If you want a more generic locator, use a URL which is what getResource provides. However, usually you don't need to know the location of the file, you just need its contents, in which case you can use getResourceAsInputStream()
This all assumes your class path is configured correctly.
Yes, you should be able to load a file anywhere on your file system that the java process has access to. You just need to have the path explicitly set in your getResource call.
For example:
File file = new File(URLDecoder.decode(this.getClass().getResource("C:\\foo\\bar\\file.xml").getFile(), "UTF-8"));
I'm new to help file creation in java. I have created a help file "sample.chm" with a 3rd party tool, added it to a java program with package name as "help" calling with runtime class and build the jar. When I run the jar file it is giving me an error that the "file cannot be found, null pointer Exception". I have given a relative path to identify the file like "../help/sample.chm" still it is not working and I tried with various classes to ientify the path. But still the same error.
Request you to please help me in fixing it.
The jar can be placed in different systems and should open this help file with out any issues.
I hope my explanation is sufficient you to identify the problem.
Regards,
Chandu
If you have a file inside a jar, you can't access it as you normally would. You can access it like this:
URL helpFile=Thread.currentThread().getContextClassLoader().getResource("help/sample.chm");
The method used above (getResource) will return a URL; if you want, you can get it as an InputStream as well by using getResourceAsStream instead.
At least a workaround unless a better solution pops up. Use the this.getClass().getClassLoader().getResource way to get an inputstream to the help file inside the jar.
Copy the bytes to a new help file in the target systems temp folder and use this extracted file with the external help file viewer.