如何使用Java上传zip文件？ [英] How to upload a zip file using Java?
问题描述
我试图上传一个zip文件。在我的项目中，我在客户端使用DWR，在服务器端使用Java。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。
I trying to upload a zip file. In my project i am using DWR in the client side and Java in server side. 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.
var image = dwr.util.getValue('uploadImage');
var file = dwr.util.getValue('uploadFile');
var color = dwr.util.getValue('color');
dwr.util.getValue（）是一个获取值的实用程序任何元素，在本例中是一个文件对象。//在教程中提到。
dwr.util.getValue() is a utility to get the value of any element, in this case a file object.//Mentioned in the tutorial.
所以，我得到一个使用该文件的zip文件通过以下代码实用。
So, i get a zip file using that utility by the below code.
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>
<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>
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;
}
此代码运行无误。但输出是一个空的zip文件。我知道我做错了什么。我无法找到。
This code runs without error. But the output is a Empty zip file. I know i doing something wrong. I unable to find that.
实际上，我收到一个zip文件为
InputStream。
Actually, i am receiving a zip file as InputStream.
我应该如何使用java 将一个
的InputStream（一个zip文件）写入zip.file
？
How should i have to write a InputStream(a zip file) to a zip.file using java?
如果我将java
方法参数设置为 ZipFile文件，会发生什么？我b $ b没有尝试过，但因为，我是
仍然在寻找一个好的教程，以
了解它。
What will happen if i set the java
method parameter as ZipFile file? I
didnt tried it, yet because, i am
still searching a good tutorial to
learn about it.
任何建议或链接都会更加赞赏!!!!!
先谢谢!!!
Any Suggestion or Links would be more appreciative!!!!! Thanks in Advance!!!
推荐答案
这里有2个关于创建ZIP文件的例子：
Here you have 2 examples about creating a ZIP file:
http：//www.java2s。 com / Tutorial / Java / 0180_ File / 0601 _ZipOutputStream.htm
http://www.java2s.com/Tutorial/Java/0180_File/0601_ZipOutputStream.htm
以下是阅读ZIP文件的示例：
Here is an example about reading a ZIP file:
http://www.kodejava.org /examples/334.html
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