formDecodeMultipart

Decodes the "multipart/form-data" request in multipart.

this is a convenience method for the case when you have a single file upload control in a form. (Or when you don't have any file upload controls, but are still using "multipart/form-data" anyway.) Pass the name of the file upload control in file_control_name, and funcform_decode_multipart will extract the uploaded file data into filename, content_type, and file. All of the other form control data will be returned (as strings, as with funcform_decode in the returned glib.hash_table.HashTable.

You may pass null for filename, content_type and/or file if you do not care about those fields. funcform_decode_multipart may also return null in those fields if the client did not provide that information. You must free the returned filename and content-type with funcGLib.free, and the returned file data with methodGlib.Bytes.unref.

If you have a form with more than one file upload control, you will need to decode it manually, using soup.multipart.Multipart.newFromMessage and soup.multipart.Multipart.getPart.

string[string]
formDecodeMultipart

Parameters

multipart soup.multipart.Multipart

a #SoupMultipart

fileControlName string

the name of the HTML file upload control

filename string

return location for the name of the uploaded file

contentType string

return location for the MIME type of the uploaded file

file glib.bytes.Bytes

return location for the uploaded file data

Return Value

Type: string[string]

a hash table containing the name/value pairs (other than file_control_name) from msg, which you can free with glib.hash_table.HashTable.destroy. On error, it will return null.