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.

Imbricate

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? – Navicert

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. – Imbricate

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? – Navicert

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. – Imbricate

Answer 1

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

Navicert

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

@fujy any updates for 2023? – Fley

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

Answer 2

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

Continence

may you explain it in some more details – Imbricate

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. #1092288 – Continence