java 如何使用 Apache Commons 以多部分形式读取其他参数
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How can I read other parameters in a multipart form with Apache Commons
提问by pkaeding
I have a file upload form that is being posted back to a servlet (using multipart/form-data encoding). In the servlet, I am trying to use Apache Commons to handle the upload. However, I also have some other fields in the form that are just plain fields. How can I read those parameters from the request?
我有一个文件上传表单，它被回发到一个 servlet（使用 multipart/form-data 编码）。在 servlet 中，我尝试使用 Apache Commons 来处理上传。但是，我在表单中还有一些其他字段，它们只是普通字段。如何从请求中读取这些参数？
For example, in my servlet, I have code like this to read in the uplaoded file:
例如，在我的 servlet 中，我有这样的代码要在上传的文件中读取：
// Create a factory for disk-based file items
FileItemFactory factory = new DiskFileItemFactory();
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
// Parse the request
Iterator /* FileItem */ items = upload.parseRequest(request).iterator();
while (items.hasNext()) {
FileItem thisItem = (FileItem) items.next();
... do stuff ...
}
回答by stian
You could try something like this:
你可以尝试这样的事情：
while (items.hasNext()) {
FileItem thisItem = (FileItem) items.next();
if (thisItem.isFormField()) {
if (thisItem.getFieldName().equals("somefieldname") {
String value = thisItem.getString();
// Do something with the value
}
}
}
回答by Edward Verenich
Took me a few days of figuring this out but here it is and it works, you can read multi-part data, files and params, here is the code:
我花了几天的时间才弄明白这个问题，但在这里它确实有效，您可以读取多部分数据、文件和参数，这是代码：
try {
ServletFileUpload upload = new ServletFileUpload();
FileItemIterator iterator = upload.getItemIterator(req);
while(iterator.hasNext()){
FileItemStream item = iterator.next();
InputStream stream = item.openStream();
if(item.isFormField()){
if(item.getFieldName().equals("vFormName")){
byte[] str = new byte[stream.available()];
stream.read(str);
full = new String(str,"UTF8");
}
}else{
byte[] data = new byte[stream.available()];
stream.read(data);
base64 = Base64Utils.toBase64(data);
}
}
} catch (FileUploadException e) {
e.printStackTrace();
}
回答by Edward Verenich
Did you try request.getParam() yet?
你试过 request.getParam() 了吗？
