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, Aug 24 '13 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, Aug 24 '13 at 11:00
Couldn't you just a ZipOuputStream http://docs.oracle.com/javase/6/docs/api/java/util/zip/ZipOutputStream.html and do something like what is done here http://examples.javacodegeeks.com/core-java/util/zip/create-zip-file-from-multiple-files-with-zipoutputstream/ by looping through each image? — Shane Haw, Aug 24 '13 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, Aug 24 '13 at 11:34

score 8 · Accepted Answer · edited Apr 14 '17 at 18:11

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 '17 at 18:11

fujy

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answered Aug 24 '13 at 12:38

Shane Haw

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@amitsingh it's a pleasure, glad I could help. – Shane Haw Aug 24 '13 at 13:30
@fujy any updates for 2023? – Viktor Fridman Aug 24 '23 at 13:59
@ViktorFridman I am not the answer author/owner, I just edited it – fujy Aug 24 '23 at 19:57

score 2 · Answer 2 · answered Aug 24 '13 at 10:00

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 '13 at 10:00

Crickcoder

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may you explain it in some more details – Avyaan Aug 24 '13 at 10:22
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. http://stackoverflow.com/questions/1091788/how-to-create-a-zip-file-in-java – Crickcoder Aug 24 '13 at 23:27

How to create ZIP files using list of Input streams?

2 Answers2