What is the right way to POST multipart form-data using curl

Question

I used this syntax to post a file along with some parameters   curl -v -include --form  key1 value1  --form upload localfilename URL   The file is around 500K in size  First of all  I see content length to be 254 on the transmit side  Later the server response s content length is 0   Where am I going wrong    Here is the complete trace of the command     Couldn t find host xxx xxx xxx xxx in the  netrc file  using defaults   About to connect   to xxx xxx xxx xxx port yyyy   0      Trying xxx xxx xxx xxx      Adding handle  conn  0x4b96a0   Adding handle  send  0   Adding handle  recv  0   Curl addHandleToPipeline  length  1   - Conn 0  0x4b96a0  send pipe  1  recv pipe  0   Connected to xxx xxx xxx xxx  xxx xxx xxx xxx  port yyyy   0    POST  zzzzzz UploadFile HTTP 1 1   User-Agent  curl 7 32 0   Host  xxx xxx xxx xxx yyyy   Accept        Content-Length  254   Expect  100-continue   Content-Type  multipart form-data  boundary ------------------------948a6137eef50079     HTTP 1 1 100 Continue   HTTP 1 1 100 Continue    HTTP 1 1 200 OK   HTTP 1 1 200 OK   Server Apache-Coyote 1 1 is not blacklisted   Server  Apache-Coyote 1 1   Server  Apache-Coyote 1 1   Added cookie JSESSIONID  C1D7DD042E250211D9DEA82688876F88  for domain xxx xxx xxx xxx  path  zzzzz   expire 0   Set-Cookie  JSESSIONID C1D7DD042E250211D9DEA82688876F88  Path  zzzzzz     HttpOnly   Set-Cookie  JSESSIONID C1D7DD042E250211D9DEA82688876F88  Path  zzzzzz   HttpOnly   Content-Type  text html charset ISO-8859-1 Content-Type  text html charset ISO-8859-1   Content-Length  0   Content-Length  0   Date  Tue  01 Oct 2013 11 54 24 GMT   Date  Tue  01 Oct 2013 11 54 24 GMT   Connection  0 to host xxx xxx xxx xxx left intact

User · Answer

On Windows 10  curl 7 28 1 within powershell  I found the following to work for me    filePath    c  temp dir with spaces myfile wav   curlPath     myfilename       filePath  curl -v -F  curlPath URL

User · Answer

to upload a file using curl in Windows I found that the path requires escaped double quotes  e g   curl -v -F  upload    C  myfile txt    URL

User · Answer

The following syntax fixes it for you   curl -v -F key1 value1 -F upload  localfilename URL

User · Answer

I had a hard time sending a multipart HTTP PUT request with curl to a Java backend  I simply tried curl -X PUT URL      --header  Content-Type  multipart form-data  boundary ---------BOUNDARY       --data-binary  file  and the content of the file was -----------BOUNDARY Content-Disposition  form-data  name  quot name1 quot  Content-Type  application xml version 1 0 charset UTF-8   lt xml gt content lt  xml gt  -----------BOUNDARY Content-Disposition  form-data  name  quot name2 quot  Content-Type  text plain  content -----------BOUNDARY--  but I always got an error that the boundary was incorrect  After some Java backend debugging I found out that the Java implementation was adding a  r n-- as a prefix to the boundary  so after changing my input file to                            lt -- here s the CRLF -------------BOUNDARY        lt -- added  --  at the beginning     -------------BOUNDARY        lt -- added  --  at the beginning     -------------BOUNDARY--      lt -- added  --  at the beginning  everything works fine  tl dr Add a newline  CRLF  r n  at the beginning of the multipart boundary content and -- at the beginning of the boundaries and try again  Maybe you are sending a request to a Java backend that needs this changes in the boundary

User · Answer

This is what worked for me curl --form file   filename  URL   It seems when I gave this answer  4  years ago   I didn t really understand the question  or how form fields worked  I was just answering based on what I had tried in a difference scenario  and it worked for me  So firstly  the only mistake the OP made was in not using the   symbol before the file name  Secondly  my answer which uses file     only worked for me because the form field I was trying to do the upload for was called file  If your form field is called something else  use that name instead  Explanation From the curl manpages  under the description for the option --form it says   This enables uploading of binary files etc  To force the  content  part to be a file  prefix the file name with an   sign  To just get the content part from a file  prefix the file name with the symbol  lt   The difference between   and  lt  is then that   makes a file  get attached in the post as a file upload  while the  lt  makes a text field and just get the contents for that text field from a file   Chances are that if you are trying to do a form upload  you will most likely want to use the   prefix to upload the file rather than  lt  which uploads the contents of the file  Addendum Now I must also add that one must be careful with using the  lt  symbol because in most unix shells   lt  is the input redirection symbol  which coincidentally will also supply the contents of the given file to the command standard input of the program before  lt    This means that if you do not properly escape that symbol or wrap it in quotes  you may find that your curl command does not behave the way you expect  On that same note  I will also recommend quoting the   symbol   You may also be interested in this other question titled  application x-www-form-urlencoded or multipart form-data  I say this because curl offers other ways of uploading a file  but they differ in the content-type set in the header  For example the --data option offers a similar mechanism for uploading files as data  but uses a different content-type for the upload  Anyways that s all I wanted to say about this answer since it started to get more upvotes  I hope this helps erase any confusions such as the difference between this answer and the accepted answer  There is really none  except for this explanation

[forms] What is the right way to POST multipart/form-data using curl?

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