Passing parameters along with a multipart/form-data upload form (Java Http Post Upload)
Solution 1
If isFormField
on FileItemStream
returns true it's a normal field. You can use openStream
and read the contents into a String.
Something like this:
FileItemStream item = iter.next();
if(item.isFormField()) {
// Normal field
String name = item.getFieldName();
String value = Streams.asString(item.openStream());
} else {
// File
}
Streams.asString
takes a second parameter which is the charset encoding to use, you might need to specify one that is suitable for your site.
Solution 2
To send a parameter with a FileUpload it just needs to be added in the URL within the setAction method
As follows:
formPanel.setAction("<ProjectURL>/<YourServletName>?<YourParameterName>="+parameter);
And in your servlet simply get the parameter as follows:
req.getParameter("<YourParameterName>");
Hope it helps ;-)
mona
Updated on July 13, 2022Comments
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mona almost 2 years
I have a code base which currently uploads file using Post and has enctype as multipart/form-data. Now I need to include some form items i.e. some parameters will also be passed along with the file upload. I have my html form created out but I cannot use request.getParameter because it is a multipart form. Could anyone suggest me how do I pass parameters along with my upload file. I am providing the codes below. Please suggest me how to get around based on compatibility of my codes
if (!ServletFileUpload.isMultipartContent(request)) { throw new CustomUploadException("Not a file upload request"); } ServletFileUpload upload = new ServletFileUpload(); FileItemIterator iter = upload.getItemIterator(request); while (iter.hasNext()) { FileItemStream item = iter.next(); if (item.isFormField() == false && item.getFieldName().equalsIgnoreCase("xmlfile")) { String fileName = item.getName(); myBean.setFileName(fileName ); } }