How to read image as FileInputStream from source folder?

Question

I have been trying to upload image to database. So I made a browse button to go through directories and get the image. Suppose if no image is selected, a default image will be uploaded to the database. The image is stored in a folder named 'resources' under source folder.In the image, helper is the folder containting .java files and below is the resources folder I am uploading the image as FileInputStream and the code is

File pic=new File(imgloc);
FileInputStream fis=new FileInputStream(pic);

If the image is not selected to upload a default image my code is

File pic=new File(this.getClass().getResource("/resources/profile1.png").getFile());
FileInputStream fis=new FileInputStream(pic);

And my query to database is

    ps.setBinaryStream(1,fis ,(int) pic.length()); //While substituing for '?' in query

After build to jar it produce FileNotFoundException when trying to add default image. May be the question is asked several times I didn't find solution in that. I want read the image as file..Thanks in advance.

you cant read the source folder from runtime - the resource should be on you classpath — Scary Wombat, Jul 06 '17 at 05:01
I have created a `resources` folder under source folder packages — Previnkumar, Jul 06 '17 at 05:04
But i can able to see the `resources` folder when i opened the jar file with `winrar` — Previnkumar, Jul 06 '17 at 05:07
@ScaryWombat I think We can read file at runtime from source folder. — afzalex, Jul 06 '17 at 05:09
Are they in below directory containing your class files? see this answer https://stackoverflow.com/a/6608824/2310289 — Scary Wombat, Jul 06 '17 at 05:10
@afzalex Sure if you fully path it. The OP us using getResource. As per javadocs, *This method locates the resource through the system class loader (see getSystemClassLoader()).* If you can provide some code showing otherwise please do so. — Scary Wombat, Jul 06 '17 at 05:13
I have tried `ClassLoader` too.. That too didnt make favour. and @ScaryWombat It is saved as a directory below class files — Previnkumar, Jul 06 '17 at 05:16
@ScaryWombat I have described the path fully `/resources/profile1.png' . I thing thats enough path. If anything missed please tell what to add in my code — Previnkumar, Jul 06 '17 at 05:19
@ScaryWombat None answers worked from the question u marked as duplicate — Previnkumar, Jul 06 '17 at 05:32

score 1 · Accepted Answer · edited Jun 20 '20 at 09:12

1

to help illustrate I create a project with a source directory

src

below source I had a package starthere with a class Main

I create another source directory xxx and put in a file a.txt

I also created under xxx a directory called text and put in a file named b.txt

Also I created under src a directory called other and put in a file called c.txt

My code is

    URL url = Main.class.getResource("/a.txt");
    System.out.println(url);
    URL url2 = Main.class.getResource("/text/b.txt");
    System.out.println(url2);
    URL url3 = Main.class.getResource("/other/c.txt");
    System.out.println(url3);

and my output is

D:\temp>java -cp ab.jar starthere.Main

jar:file:/D:/temp/ab.jar!/a.txt

jar:file:/D:/temp/ab.jar!/text/b.txt

jar:file:/D:/temp/ab.jar!/other/c.txt

edit

As you are not wanting to use a File, but simply the InputStream, you can use

InputStream fis = Main.class.getResourceAsStream("/a.txt");

and then set using ps.setBinaryStream(1, fis);

edited Jun 20 '20 at 09:12

Community

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answered Jul 06 '17 at 05:49

Scary Wombat

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Yes bro, I tried this thing. After this code i used `File pic=new File(url.to URI);` because i need to load it to database Worked in code bro But not in jar. It showed exception `illegal argument exception: URI is not hierarchical`. – Previnkumar Jul 06 '17 at 05:58
I am not following you. The error is a completely different error. Also this code `File pic=new File(url.to URI);` does not make sense at all – Scary Wombat Jul 06 '17 at 06:02
Bro sorry for mistakes made by me, But my question itself is i want to get image as `file` Then only i can add it to database – Previnkumar Jul 06 '17 at 06:06
As per my code, are you getting an URL? This is the first step – Scary Wombat Jul 06 '17 at 06:08
yeah bro. I got `url`. Then kindly say what shld i do – Previnkumar Jul 06 '17 at 06:09
Working in code bro. But in jar `file:\C:\Users\Previn\Documents\NetBeansProjects\helper\dist\helper.jar!\resources\profile1.png java.io.FileNotFoundException: file:\C:\Users\Previn\Documents\NetBeansProjects\helper\dist\helper.jar!\resources\profile1.png (The filename, directory name, or volume label syntax is incorrect)`. This exception arises. – Previnkumar Jul 06 '17 at 06:17
Well as you don't really want the file, your just want the inputstream, use `getResourceAsStream` instead and then use https://docs.oracle.com/javase/7/docs/api/java/sql/PreparedStatement.html#setBinaryStream(int,%20java.io.InputStream) – Scary Wombat Jul 06 '17 at 06:20

afzalex · Answer 2 · 2017-07-06T06:21:26.770

Consider following package structure

/
/testing_package
/testing_package/Main.java
/testing_package/Image.png

Now Here is a program to get Image.png and put it in database

public class Main {
    public static void main(String...args) {
        System.out.println(Main.class.getResource("image.jpg"));
        //PreparedStatement as ps
        ps.setBinaryStream(1, Main.class.getResourceAsStream("image.jpg")));
    }
}

Thing to keep in mind
When using getResource(...) or getResourceAsStream(...) use relative location to the class instance whose function you are invoking.

Have a look at this answer too

How to read image as FileInputStream from source folder?

2 Answers2