我有一个java servlet,它接受用户上传到我的web应用程序的图像.
我有另一台服务器(运行php),它将托管所有图像.如何从我的jsp服务器获取图像到我的php服务器？流程将是这样的:
public class ServletImgUpload extends HttpServlet
{
public void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException
{
// get image user submitted
// try sending it to my php server now
// return success or failure message back to user
}
}
谢谢
首先，为什么不直接将表单提交到 PHP 脚本呢？
<form action="http://example.com/upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit">
</form>
如果这不是一个选项，并且您确实需要将表单提交到 servlet，那么首先在 JSP 中创建一个 HTML 表单，如下所示：
<form action="upload" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit">
</form>
在侦听 of 的 servlet 中url-pattern，/upload您有 2 个选项来处理请求，具体取决于 PHP 脚本的处理方式。
如果 PHP 脚本采用相同的参数，并且可以按照 HTML 表单指示 servlet 执行的相同方式处理上传的文件（我仍然宁愿让表单直接提交给 PHP 脚本，但无论如何），那么您可以让 servlet 充当透明代理，它只是立即将字节从 HTTP 请求传输到 PHP 脚本。APIjava.net.URLConnection在这方面很有用。
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
HttpURLConnection connection = (HttpURLConnection) new URL("http://example.com/upload.php").openConnection();
connection.setDoOutput(true); // POST.
connection.setRequestProperty("Content-Type", request.getHeader("Content-Type")); // This one is important! You may want to check other request headers and copy it as well.
// Set streaming mode, else HttpURLConnection will buffer everything in Java's memory.
int contentLength = request.getContentLength();
if (contentLength > -1) {
connection.setFixedLengthStreamingMode(contentLength);
} else {
connection.setChunkedStreamingMode(1024);
}
InputStream input = request.getInputStream();
OutputStream output = connection.getOutputStream();
byte[] buffer = new byte[1024]; // Uses only 1KB of memory!
for (int length = 0; (length = input.read(buffer)) > 0;) {
output.write(buffer, 0, length);
}
output.close();
InputStream phpResponse = connection.getInputStream(); // Calling getInputStream() is important, it's lazily executed!
// Do your thing with the PHP response.
}
如果 PHP 脚本采用不同或更多参数（同样，我宁愿只是相应地更改 HTML 表单，以便它可以直接提交到 PHP 脚本），那么您可以使用Apache Commons FileUpload来提取上传的文件和Apache HttpComponents Client将上传的文件提交到 PHP 脚本，就像从 HTML 表单提交一样。
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
InputStream fileContent = null;
String fileContentType = null;
String fileName = null;
try {
List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
for (FileItem item : items) {
if (!item.isFormField() && item.getFieldName().equals("file")) { // <input type="file" name="file">
fileContent = item.getInputStream();
fileContentType = item.getContentType();
fileName = FilenameUtils.getName(item.getName());
break; // If there are no other fields?
}
}
} catch (FileUploadException e) {
throw new ServletException("Parsing file upload failed.", e);
}
if (fileContent != null) {
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/upload.php");
MultipartEntity entity = new MultipartEntity();
entity.addPart("file", new InputStreamBody(fileContent, fileContentType, fileName));
httpPost.setEntity(entity);
HttpResponse phpResponse = httpClient.execute(httpPost);
// Do your thing with the PHP response.
}
}