URL url = new URL("http://localhost:8080/Work/images/abt.jpg");
InputStream in = new BufferedInputStream(url.openStream());
ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
int n=0;
while (-1!=(n=in.read(buf)))
{
out.write(buf, 0, n);
}
out.close();
in.close();
byte[] response1 = out.toByteArray();
FileOutputStream fos = new FileOutputStream("C://abt.jpg");
fos.write(response1);
fos.close();
在此代码中,最后3行有一些错误
SEVERE:servlet的Servlet.service()ImageDownloadServlet抛出了异常java.io. FileNotFoundException :C:/abt.jpg(没有这样的文件或目录)
我该如何解决?
答案 0 :(得分:3)
可能会尝试将C:/abt.jpg
切换为C:\\abt.jpg
答案 1 :(得分:3)
尝试使用File.pathSeparator而不是斜杠。
答案 2 :(得分:3)
String filePath = request.getParameter("action");
System.out.println(filePath);
// URL url = new
// URL("http://localhost:8080/Works/images/abt.jpg");
response.setContentType("image/jpeg");
response.setHeader("Content-Disposition", "attachment; filename=icon" + ".jpg");
URL url = new URL(filePath);
URLConnection connection = url.openConnection();
InputStream stream = connection.getInputStream();
BufferedOutputStream outs = new BufferedOutputStream(response.getOutputStream());
int len;
byte[] buf = new byte[1024];
while ((len = stream.read(buf)) > 0) {
outs.write(buf, 0, len);
}
outs.close();
}
答案 3 :(得分:2)
答案 4 :(得分:0)
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException,
IOException, MalformedURLException {
String filePath = request.getParameter("action");
// String filename = "abt.jpg";
System.out.println(filePath);
URL url = new URL("http://localhost:8080/Works/images/abt.jpg");
InputStream in = new BufferedInputStream(url.openStream());
ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
int n = 0;
while (-1 != (n = in.read(buf))) {
out.write(buf, 0, n);
}
out.close();
in.close();
byte[] response1 = out.toByteArray();
FileOutputStream fos = new FileOutputStream("/home/user/Downloads/abt.jpg");
fos.write(response1);
fos.close();
}
此图片将位于下载文件夹