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 as
ZipFile 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!!!
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 as
ZipFile 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!!!
Share Improve this question asked Nov 11, 2010 at 19:15 user405398user405398 2- can you please explain to me why i still the fakepath even after using dwr.util.getValue("file"). – Genjuro Commented Jul 18, 2012 at 14:09
- 1 Sorry, i am not able to understand you? What is fakepath means in terms of DWR. – user405398 Commented Jul 23, 2012 at 14:22
2 Answers
Reset to default 1Here you have 2 examples about creating a ZIP file:
http://www.java2s./Tutorial/Java/0180_File/0601_ZipOutputStream.htm
Here is an example about reading a ZIP file:
http://www.kodejava/examples/334.html
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-binary format. So, I then made the request in this Format.
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.
My code worked when I made a Base-64 encoded Request (i.e --data-binary)