JSP如何在JSP中获取文件的标准名称
[英]JSP how to get the fully qualified name of a file in JSP
问题描述
I am trying to upload a file in JSP using the 
我正在尝试使用JSP在JSP中上传文件
<form action="EdgeWarUpload" method="post"
enctype="multipart/form-data">
<input type="file" name="file" size="50" />
<br />
<input type="submit" value="Upload File" />
where EdgeWarUpload is a servlet.User is browsing and selecting a file to upload.I want the fully qualified path with filename (path name + file name) in the servlet EdgeWarUpload to create a BufferedInputStream.But I am not able to get it.Kindly check and reply. EdgeWarUpload是Servlet，用户正在浏览并选择要上传的文件。我希望Servlet EdgeWarUpload中具有文件名（路径名+文件名）的完全限定路径创建BufferedInputStream，但是我无法获取它。检查并回复。
3 个解决方案
解决方案1
1 2013-06-19 08:35:44
<html>
<header></header>
<body>
<form method="POST" action="upload.do">
escolha o arquivo para fazer upload:
<input type="file" name="ctrupload"><br>
<input type="submit" name="submit" value="enviar...">
</form>
</body>
</html>
try this 尝试这个
public class Uploader extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
ServletInputStream sis = request.getInputStream();
byte[] b = new byte[request.getContentLength()];
System.out.println(request.getContentLength());
sis.read(b,0,b.length);
FileOutputStream fos = new FileOutputStream("Teste.jpg");
fos.write(b);
sis.close();
fos.flush();
fos.close();
}
}
you really dont required fully qualified path 您真的不需要完全限定的路径
解决方案2
0 2013-06-19 08:37:39
Mostly Not Possible. 通常是不可能的。
Browsers wont send the full path because its considered a security risk because it might tell thing about the client system, so most modern browsers support that. 浏览器不会发送完整路径，因为它被认为存在安全风险，因为它可能会告知客户端系统信息，因此大多数现代浏览器都支持。
on server side no use of that I guess . 在服务器端，我猜没用。 just use filename . 只需使用filename 。
解决方案3
0 2013-06-19 08:44:37
here you can use this code in servlet for the file name: 在这里，您可以在servlet中使用以下代码作为文件名：
DataInputStream in = new DataInputStream(request.getInputStream());
int formDataLength = request.getContentLength();
byte dataBytes[] = new byte[formDataLength];
int byteRead = 0;
int totalBytesRead = 0;
while (totalBytesRead < formDataLength) {
byteRead = in.read(dataBytes, totalBytesRead, formDataLength);
totalBytesRead += byteRead;
}
String file = new String(dataBytes);
saveFile = file.substring(file.indexOf("filename=\"") + 10);
saveFile = saveFile.substring(0, saveFile.indexOf("\n"));
saveFile = saveFile.substring(saveFile.lastIndexOf("\\") + 1, saveFile.indexOf("\""));
System.out.println("filename3: "+ saveFile); // name of the file
int lastIndex = contentType.lastIndexOf("=");
String boundary = contentType.substring(lastIndex + 1, contentType.length());
int pos;
pos = file.indexOf("filename=\"");
pos = file.indexOf("\n", pos) + 1;
pos = file.indexOf("\n", pos) + 1;
pos = file.indexOf("\n", pos) + 1;
int boundaryLocation = file.indexOf(boundary, pos) - 4;
int startPos = ((file.substring(0, pos)).getBytes()).length;
int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length;
FileOutputStream fileOut = new FileOutputStream(saveFile);
fileOut.write(dataBytes, startPos, (endPos - startPos));
fileOut.flush();
fileOut.close();
FileInputStream fis = new FileInputStream(saveFile);
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
for (int readNum; (readNum = fis.read(buf)) != -1;) {
bos.write(buf, 0, readNum);
}
byte[] bytes = bos.toByteArray();
bos.close();
fis.close();
in.close();
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