如何使用Java上传zip文件？
[英]How to upload a zip file using Java?
问题描述
I trying to upload a zip file. 
我试图上传一个zip文件。 In my project i am using DWR in the client side and Java in server side. 在我的项目中，我在客户端使用DWR，在服务器端使用Java。 As i have seen in DWR tutorials for uploading data(Its not in their website. They are providing it with dwr.rar bundle) they getting input by the below lines. 正如我在DWR教程中看到的上传数据（它不在他们的网站上。他们提供dwr.rar包）他们通过以下行获得输入。
var image = dwr.util.getValue('uploadImage');
var file = dwr.util.getValue('uploadFile');
var color = dwr.util.getValue('color');
dwr.util.getValue() is a utility to get the value of any element, in this case a file object.//Mentioned in the tutorial.
So, i get a zip file using that utility by the below code. 所以，我通过以下代码使用该实用程序获得了一个zip文件。
Javascript: 使用Javascript：
function uploadZip(){
var file = dwr.util.getValue("uploadFile");
dwr.util.setValue("uploadFile", null);
DataUpload.uploadData(file, function(data){
if(data != null){
$("#zipURL").html("<p>Upload Completed!!!</p>");
$("#zipURL").append("Location: "+data.path2);
}
});
}
HTML: HTML：
<html>
<head>ZIP Uploader
</head>
<body>
<table>
<tr><td>Select File: </td><td><input type="file" id="uploadFile" /></td>
<tr><td><input type="button" value="Upload" onclick="uploadZip()" /></td></tr> </table>
<div id="result"><span id="imgURL"></span>
<span id="zipURL"></span></div>
</body>
</html>
The Java Code is: Java代码是：
public class DataUpload {
private static String DATA_STORE_LOC = "D:/BeenodData/Trials/";
public Path uploadData(InputStream file) throws IOException{//In the tutorial the
//parameters are in type of BufferedImage & String.
//They used it for image and text file respectively.
//In an another example(out of DWR site) they used InputStream for receiving
//image
try {
byte[] buffer = new byte[1024];
int c;
File f2 = new File(DATA_STORE_LOC+dat+".zip");
path.setPath2(DATA_STORE_LOC+dat+".zip");
FileOutputStream fos = new FileOutputStream(f2);
c = file.read();
System.out.println(c);
while ((c = file.read()) != -1) {
fos.write(c);
}
file.close();
fos.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return path;
}
This code runs without error. 此代码运行没有错误。 But the output is a Empty zip file. 但输出是一个空的zip文件。 I know i doing something wrong. 我知道我做错了什么。 I unable to find that. 我找不到。
Actually, i am receiving a zip file as InputStream.
How should i have to write a InputStream(a zip file) to a zip.file using java?
What will happen if i set the java method parameter as
ZipFile file?ZipFile file会发生什么？ I didnt tried it, yet because, i am still searching a good tutorial to learn about it.
Any Suggestion or Links would be more appreciative!!!!! 任何建议或链接都会更加赞赏!!!!! Thanks in Advance!!! 提前致谢！！！
2 个解决方案
解决方案1
2 已采纳 2010-11-12 13:30:29
Here you have 2 examples about creating a ZIP file: 这里有2个关于创建ZIP文件的示例：
http://www.java2s.com/Tutorial/Java/0180_ File/0601 _ZipOutputStream.htm http://www.java2s.com/Tutorial/Java/0180_ 文件/ 0601 _ZipOutputStream.htm
Here is an example about reading a ZIP file: 这是一个关于读取ZIP文件的示例：
http://www.kodejava.org/examples/334.html http://www.kodejava.org/examples/334.html
解决方案2
0 2019-02-15 14:12:56
I have also implemented the Same kind of backend Code in Java, and I was facing the same Issue of Zip file being made, but its content being empty. 我还在Java中实现了相同类型的后端代码，我遇到了相同的Zip文件问题，但其内容为空。
Later I found that the Request I was making to API, in that the file I was Attaching was not in --data-binary format. 后来我发现我正在向API发出请求，因为我正在附加的文件不是--data-binary格式。 So, I then made the request in this Format. 所以，我接着以这种格式提出了请求。
curl --data-binary @"/mnt/c/checknew.zip" http://localhost/api/upload
I am not sure what request format you are making either in multipart/form-data or Base-64 encoded. 我不确定您在multipart/form-data或Base-64编码中使用的请求格式。
My code worked when I made a Base-64 encoded Request (ie --data-binary ) 当我制作Base-64编码请求（即--data-binary ）时，我的代码工作--data-binary
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