Curl Http Post File Upload Using Curl -Data in Linux Command Line

Using cURL to upload POST data with files

You need to use the -F option:

-F/--form <name=content> Specify HTTP multipart POST data (H)

Try this:

curl \
  -F "userid=1" \
  -F "filecomment=This is an image file" \
  -F "image=@/home/user1/Desktop/test.jpg" \
  localhost/uploader.php

cURL http post file upload using curl --data in linux command line

You usually can't simply pick -F or -d (--data) at your choice. The web server that will receive your post expects one of the formats. If the form you're trying to submit uses the type 'multipart/form-data', then and only then you must use the -F type. If not, you should use -d which then causes a posting with the type 'application/x-www-form-urlencoded'.

multipart/formposts with -F uses a special formatting of the post with Mime-headers separating the different parts and each part having its own set of headers.

-d is just raw data being sent for the server to interpret/decode.

Mentioned in the curl FAQ.

(But since you write the PHP code in this case, you just have to decide which POST method you want to accept and then make your curl command line use that.)

Send request to cURL with post data sourced from a file

You're looking for the --data-binary argument:

curl -i -X POST host:port/post-file \
  -H "Content-Type: text/xml" \
  --data-binary "@path/to/file"

In the example above, -i prints out all the headers so that you can see what's going on, and -X POST makes it explicit that this is a post. Both of these can be safely omitted without changing the behaviour on the wire. The path to the file needs to be preceded by an @ symbol, so curl knows to read from a file.

Correct curl command for uploading local file to url

There's more than one way to "upload a file to a URL", so we cannot actually know unless you give us more details.

But what's clear is that that you lack either a -d or a -F option on your command line, and you should drop the -X POST.

Multi-part formpost

If you upload with a multipart, which is how most "uploads" to HTTP works, it could be something like this:

curl https://waapi.pepipost.com/api/v2/media/upload/ -H 'Authorization: Bearer myAuthorizationToken' -F "file=@C:\Users\Slomil\Desktop\UserGuide.pdf"

Note that this command line sets the upload part to get the name file, which you should change to the name you want.

"regular" POST

If you just want to send the binary file "raw" in a POST (which your setting of Content-Type might indicate you want), use --data-binary like this:

curl https://waapi.pepipost.com/api/v2/media/upload/ -H 'Authorization: Bearer myAuthorizationToken' -H 'Content-Type: document' --data-binary "@C:\Users\Slomil\Desktop\UserGuide.pdf"

(I copied the Content-Type from the question, although it looks unusual and odd.)

Curl POST upload file test

This is where you are getting it wrong.

file=@demo2_test_imageinput.png

As per your function def, your multipart field name should be upload_file

async def create_upload_file(upload_file: UploadFile = File(...))

So your curl request must have the parameter

upload_file=@demo2_test_imageinput.png

http file upload using cURL in linux(command line)

Try

curl -i -F filename=image.jpg -F image=@/path/to/image.jpg http://localhost/xmlcreate/curlupload.php

posting a file with curl command?

tl;dr

The type option allows you to specify the Content-Type of a given part, in a multipart request.

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The -F option (and the more verbose --form) emulates a filled-in form in which a user has pressed the submit button. It will make curl to POST data using the Content-Type header with the multipart/form-data value.

In multipart requests, each part may have an optional Content-Type header, as stated in the RFC 7578:

4.4. Content-Type Header Field for Each Part

Each part MAY have an (optional) Content-Type header field, which defaults to text/plain. If the contents of a file are to be sent, the file data SHOULD be labeled with an appropriate media type, if known, or application/octet-stream.

While @ makes a file to get attached in the form as a file upload, the type parameter allows you to specify the Content-Type of a given part.

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

The following syntax fixes it for you:

curl -v -F key1=value1 -F upload=@localfilename URL

curl upload all files in a directory with a single curl command?

Use a loop to create an array of alternating -F and filename=@filename. Then substitute the array into the curl command.

files=()
while read -r filename; do
    files+=(-F "$filename=@$filename")
done < <(find . -name '*.txt' -type f)

curl -k -H "Accept: application/json" -H "Authorization: Bearer $bearertoken" "${$files[@]}" -F "deployment-name=${CIRCLE_SHA1:0:7}" -F "deployment-source=circleci" -F "enable-duplicate-filtering=false" -F "deploy-changed-only=true" "${$files[@]}" http:www.blah.com

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