如何使用 Java 上传 zip 文件？
我正在尝试上传 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 上传数据教程中看到的那样（不在他们的网站上。他们通过 dwr.rar 捆绑包提供），他们通过以下几行获取输入。
var image = dwr.util.getValue('uploadImage');
var file = dwr.util.getValue('uploadFile');
var color = dwr.util.getValue('color');
dwr.util.getValue() 是一个获取任何元素值的实用程序，在本例中是一个文件对象。//在教程中提到。
因此，我通过以下代码使用该实用程序获得了一个 zip 文件。
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;
}
此代码运行时没有错误。但输出是一个空的 zip 文件。我知道我做错了什么。我找不到那个。
实际上，我收到的是一个 zip 文件 输入流。
我应该如何写 输入流（zip 文件）到 zip.file 使用java？
如果我设置java会发生什么 方法参数作为
ZipFile 文件？我 没试过，但因为，我是 仍在寻找一个好的教程 了解一下。
任何建议或链接将更加感激！！！！ 提前致谢！！！
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() 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.
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>
The Java Code is:
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. But the output is a Empty zip file. I know i doing something wrong. I unable to find that.
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 asZipFile file? 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!!!!!
Thanks in Advance!!!
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这里有两个关于创建 ZIP 文件的示例：
http://www.java2s。 com/Tutorial/Java/0180_File/0601_ZipOutputStream.htm
以下是读取 ZIP 文件的示例：
http://www.kodejava.org/examples/334.html
Here you have 2 examples about creating a ZIP file:
http://www.java2s.com/Tutorial/Java/0180_File/0601_ZipOutputStream.htm
Here is an example about reading a ZIP file:
http://www.kodejava.org/examples/334.html
我还在 Java 中实现了相同类型的后端代码，并且我面临着制作 Zip 文件的相同问题，但其内容为空。
后来我发现我向 API 发出的请求，因为我附加的文件不是
--data-binary格式。所以，我随后以这种格式提出了请求。我不确定您在
multipart/form-data或Base-64编码中发出什么请求格式。当我发出 Base-64 编码请求（即
--data-binary）时，我的代码有效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.
Later I found that the Request I was making to API, in that the file I was Attaching was not in
--data-binaryformat. So, I then made the request in this Format.I am not sure what request format you are making either in
multipart/form-dataorBase-64encoded.My code worked when I made a Base-64 encoded Request (i.e
--data-binary)