我想使用下面给出的代码上传pdf文件。它提供浏览功能但不上传文件。当我点击sendfile按钮时,显示uploadfile.html代码页。我怎样才能做到这一点??? 给定代码中的错误在哪里???
文件名-upload.html
<%@ page language="java" %>
<HTml>
<HEAD><TITLE>Display file upload form to the user</TITLE></HEAD>
<% // for uploading the file we used Encrypt type of multipart/
form-data and input of file type to browse and submit the file %>
<BODY> <FORM ENCTYPE="multipart/form-data" ACTION=
"uploadfile.html" METHOD=POST>
<br><br><br>
<center><table border="2" >
<tr><center><td colspan="2"><p align=
"center"><B>PROGRAM FOR UPLOADING THE FILE</B><center></td></tr>
<tr><td><b>Choose the file To Upload:</b>
</td>
<td><INPUT NAME="F1" TYPE="file"></td></tr>
<tr><td colspan="2">
<p align="right"><INPUT TYPE="submit" VALUE="Send File" ></p></td></tr>
<table>
</center>
</FORM>
</BODY>
</HTML>
文件名 - uploadfile.html
<%@ page import="java.io.*" %>
<%
//to get the content type information from JSP Request Header
String contentType = request.getContentType();
//here we are checking the content type is not equal to Null and
as well as the passed data from mulitpart/form-data is greater than or
equal to 0
if ((contentType != null) && (contentType.indexOf("multipart/
form-data") >= 0)) {
DataInputStream in = new DataInputStream(request.
getInputStream());
//we are taking the length of Content type data
int formDataLength = request.getContentLength();
byte dataBytes[] = new byte[formDataLength];
int byteRead = 0;
int totalBytesRead = 0;
//this loop converting the uploaded file into byte code
while (totalBytesRead < formDataLength) {
byteRead = in.read(dataBytes, totalBytesRead,
formDataLength);
totalBytesRead += byteRead;
}
String file = new String(dataBytes);
//for saving the file name
String saveFile = file.substring(file.indexOf("filename=\
"") + 10);
saveFile = saveFile.substring(0, saveFile.indexOf("\n"));
saveFile = saveFile.substring(saveFile.lastIndexOf("\\")
+ 1,saveFile.indexOf("\""));
int lastIndex = contentType.lastIndexOf("=");
String boundary = contentType.substring(lastIndex + 1,
contentType.length());
int pos;
//extracting the index of file
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;
// creating a new file with the same name and writing the
content in new file
FileOutputStream fileOut = new FileOutputStream(saveFile);
fileOut.write(dataBytes, startPos, (endPos - startPos));
fileOut.flush();
fileOut.close();
%><Br><table border="2"><tr><td><b>You have successfully
upload the file by the name of:</b>
<% out.println(saveFile); %></td></tr></table> <%
}
%>
答案 0 :(得分:37)
这显然是Roseindia代码段。首先,它是worst学习资源。不要使用它。它只教导不良做法。将该网站添加到您的黑名单。事实上,任何充满广告横幅和绝望过时的低质量代码片段的“教程”网站都由业余爱好者清楚地保留,主要关注广告收入而不是严肃的教学。这种垃圾“教程”网站的其他例子是javabeat,tutorialspoint,journaldev,javatpoint等。这些网站的常见之处在于它们起源于印度。
除了您错误地使用.html
文件扩展名而不是.jsp
这一事实(即使他们使用.jsp
扩展名正确地展示了他们的示例),代码仍有几个主要问题片段:
<font>
和<center>
标记。Content-Length
请求标头,该标头本身并不总是存在。如果此标头不存在,则代码会中断。String
。这可能会导致结果字节错误/损坏。DataInputStream
包装器是不必要的,代码没有任何好处。这太可怕了。
从JSP上传文件的正确方法是将表单提交到@MultipartConfig
带注释的servlet类,然后使用request.getPart()
来获取文件。您可以在此答案中找到一个代码段:How to upload files to server using JSP/Servlet?
在这个答案中详细阐述了学习Java EE的正确方法:Java EE web development, where do I start and what skills do I need?