这是我的应用程序中的代码：
1）我创建了一个接受HTTP请求的类：
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.commons.fileupload.FileItemIterator;
import org.apache.commons.fileupload.FileItemStream;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
public class FileUpload extends HttpServlet{
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
ServletFileUpload upload = new ServletFileUpload();
try{
FileItemIterator iter = upload.getItemIterator(request);
while (iter.hasNext()) {
FileItemStream item = iter.next();
String name = item.getFieldName();
InputStream stream = item.openStream();
// Process the input stream
ByteArrayOutputStream out = new ByteArrayOutputStream();
int len;
byte[] buffer = new byte[8192];
while ((len = stream.read(buffer, 0, buffer.length)) != -1) {
out.write(buffer, 0, len);
}
int maxFileSize = 10*(1024*1024); //10 megs max
if (out.size() > maxFileSize) {
throw new RuntimeException("File is > than " + maxFileSize);
}
}
}
catch(Exception e){
throw new RuntimeException(e);
}
}
}
2）然后在我的web.xml中添加了以下几行：
<servlet>
<servlet-name>fileUploaderServlet</servlet-name>
<servlet-class>com.testapp.server.FileUpload</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>fileUploaderServlet</servlet-name>
<url-pattern>/testapp/fileupload</url-pattern>
</servlet-mapping>
3）对于form.action这样做：
form.setAction(GWT.getModuleBaseURL()+"fileupload");
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