How to create ZIP files using list of Input streams?

Question

In my case I have to download images from the resources folder in my web app. Right now I am using the following code to download images through URL.

url = new URL(properties.getOesServerURL() + "//resources//WebFiles//images//" + imgPath);

filename = url.getFile();               

is = url.openStream();
os = new FileOutputStream(sClientPhysicalPath + "//resources//WebFiles//images//" + imgPath);

b = new byte[2048];

while ((length = is.read(b)) != -1) {
    os.write(b, 0, length);
}

But I want a single operation to read all images at once and create a zip file for this. I don't know so much about the use of sequence input streams and zip input streams so if it is possible through these, please let me know.

so are you asking how to download all images in //resources//WebFiles//images// and zip them into one zip file? How does imgPath get assigned? — Shane Haw
– Shane Haw, Commented Aug 24, 2013 at 10:53
I already have imgPath.In my case i have list of imagepath that is located at //resources//WebFiles//images.I have to download all these images through URL at once. — Avyaan
– Avyaan, Commented Aug 24, 2013 at 11:00
Couldn't you just a ZipOuputStream docs.oracle.com/javase/6/docs/api/java/util/zip/… and do something like what is done here examples.javacodegeeks.com/core-java/util/zip/… by looping through each image? — Shane Haw
– Shane Haw, Commented Aug 24, 2013 at 11:07
thanx for your efforts,but i have already gone throgh that example.it is useful upto 70%.here they have used the physical path of the file but in my case i dont have aphysical path. i have rendered image on broser only. — Avyaan
– Avyaan, Commented Aug 24, 2013 at 11:34

fujy · Accepted Answer · 2017-04-14 18:11:16Z

8

The only way I can see you being able to do this is something like the following:

try {

    ZipOutputStream zip = new ZipOutputStream(new FileOutputStream("C:/archive.zip"));

    //GetImgURLs() is however you get your image URLs

    for(URL imgURL : GetImgURLs()) {
        is = imgURL.openStream();
        zip.putNextEntry(new ZipEntry(imgURL.getFile()));
        int length;

        byte[] b = new byte[2048];

        while((length = is.read(b)) > 0) {
            zip.write(b, 0, length);
        }
        zip.closeEntry();
        is.close();
    }
    zip.close();
}

Ref: ZipOutputStream Example

edited Apr 14, 2017 at 18:11

fujy

5,3015 gold badges33 silver badges51 bronze badges

answered Aug 24, 2013 at 12:38

Shane Haw

7239 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Shane Haw Over a year ago

@amitsingh it's a pleasure, glad I could help.

Viktor Fridman Over a year ago

@fujy any updates for 2023?

fujy Over a year ago

@ViktorFridman I am not the answer author/owner, I just edited it

Crickcoder · Accepted Answer · 2013-08-24 10:00:29Z

2

The url should return zip file. Else you have to take one by one and create a zip using your program

answered Aug 24, 2013 at 10:00

Crickcoder

2,1655 gold badges22 silver badges36 bronze badges

2 Comments

Avyaan Over a year ago

may you explain it in some more details

Crickcoder Over a year ago

You can make a list of Files. Then, create a zip file. Loop over the list and write each file to that zip. You can refer to this for creating zipfile. stackoverflow.com/questions/1091788/…

Collectives™ on Stack Overflow

How to create ZIP files using list of Input streams?

2 Answers 2

3 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related