Post Request with Wget

Post request with Wget?

Wget currently doesn't not support "multipart/form-data" data. --post-file is not for transmitting files as form attachments, it expects data with the form: key=value&otherkey=example. It is actually possible to post other formats (json) if you send the corresponding header.

--post-data and --post-file work the same way: the only difference is that --post-data allows you to specify the data in the command line, while --post-file allows you to specify the path of the file that contain the data to send.

Here's the documentation:

 --post-data=string
--post-file=file
Use POST as the method for all HTTP requests and send the specified data
in the request body. --post-data sends string as data, whereas
--post-file sends the contents of file. Other than that, they work in
exactly the same way. In particular, they both expect content of the
form "key1=value1&key2=value2", with percent-encoding for special
characters; the only difference is that one expects its content as a
command-line parameter and the other accepts its content from a file. In
particular, --post-file is not for transmitting files as form
attachments: those must appear as "key=value" data (with appropriate
percent-coding) just like everything else. Wget does not currently
support "multipart/form-data" for transmitting POST data; only
"application/x-www-form-urlencoded". Only one of --post-data and
--post-file should be specified.

Regarding your authentication token, it should either be provided in the header, in the path of the url, or in the data itself. This must be indicated somewhere in the documentation of the service you use. In a POST request, as in a GET request, you must specify the data using keys and values. This way the server will be able to receive multiple information with specific names. It's similar with variables.

Hence, you can't just send a magic token to the server, you also need to specify the name of the key. If the key is "token", then it should be token=YOUR_TOKEN.

wget --post-data 'user=foo&password=bar' http://example.com/auth.php

Also, you should consider using curl if you can because it is easier to send files using it. There are many examples on the Internet for that.

How to make http/https POST requests manually?

You can use wget for this. From the manual it supports digest authentication and can send POST requests.

There seems to be a GUI at wget::gui, but I don't know how reliable or complete it is.

How to send a GET request with Wget and output it on the screen

Instead of -O response_from_API_server.json try -O-. Instead of saving the response to a file it prints to stdout.

Send urlencoded POST request with wget

as suggested in the comments by @Roadowl, bash itself can urlencode data, the wget invocation could look like

wget --post-data=$(printf "url="; urlencode $u) http://example.com

using urlencode implementation from https://gist.github.com/cdown/1163649 ,

urlencode() {
# urlencode <string>
old_lc_collate=$LC_COLLATE
LC_COLLATE=C

local length="${#1}"
for (( i = 0; i < length; i++ )); do
local c="${1:i:1}"
case $c in
[a-zA-Z0-9.~_-]) printf "$c" ;;
*) printf '%%%02X' "'$c" ;;
esac
done

LC_COLLATE=$old_lc_collate
}

.. probably won't work will null bytes though, given that bash doesn't like null bytes and probably won't allow them in variables

Sending multiline command output via wget as post-data

You do not escape - you quote the usage. Check your scripts with shellcheck.

... --post-data="$(ls -lah)" ...

BASH - post array with wget --post-data

That wget command looks OK, and worked in a small test I did.

Your problem is probably on the server side.



Related Topics



Leave a reply



Submit