I want to make the following curl call to get an access token:
curl -k -XPOST --user "{user}" {url}/access_token -d 'grant_type=authorization_code&code={code}&redirect_uri={uri}'
However, I get the following error statement:
'code' is not recognized as ane internal or external command, operable program or batch file.
'redirect_uri' is not recognized as ane internal or external command, operable program or batch file.
When I check the trace, I see that the body only contains this: 'grant_type=authorization_code
What can I do to make sure all the parameters get passed?
curl -X "POST" "{url}/access_token" \
--user "{user}" \
-H "Content-Type: application/x-www-form-urlencoded" \
--data-urlencode "code={code}" \
--data-urlencode "redirect_uri={uri}" \
--data-urlencode "grant_type=authorization_code"
I hope this helps
It seems that the &-character is interpreted as "execute as background job". Try escaping each &-character with a backslash - like \&
Alternatively you could write your content into a file and reference this file via the parameter -d #file
The #-character is important for curl to know that file is a file name.
Related
I'm attempting to execute the following CURL call via commandline:
curl -i -H "Content-Type:application/json" \
-u "m19389#dev.acp.co.com:qE2P/N7y1k.\(" \
-X GET -d "https://wefd.it.co.com:3905/events/com.co.mpm.dev.29160-wgdhfgd-v1/dsd/dsdfds-0-0-1-7c49768976-2g7kq"
I've added quotes around the arguments and all, and I'm definitely inclusing the curl url, so I'm not sure what's going on here. Where did I go wrong?
I am using Marklogic 9 and try to ingest data from external source into MarkLogic. I made an REST API on port 8031. When I try to execute the following curl command:
curl --anyauth --user admin:admin -i -X POST -d https://services7.arcgis.com/21GdwfcLrnTpiju8/arcgis/rest/services/Geluidsbelasting/FeatureServer/0/query?where=1%3D1&outFields=*&outSR=4326&f=json
-H "Content-type: application/json" -H "Accept: application/json" \
'http://localhost:8031
After executing this statement I receive the error:
Curl: URL is not specified
Can you please help me out!
Many thanks
Erik
Your -d parameter has special characters that are not escaped. Try putting quotes around your -d url. It will prevent your command from getting truncated and misinterpreted at & signs..
HTH!
I am trying to follow the documentation from timekit.io. I am attempting to do something as simple as send a curl request to do basic authentication as seen in this section of the docs. I have replaced the Timekit-App:name-of-app with the name of my app which I found in the app-settings of my timekit account. I also replaced the email & password with the one's I use to login into my account.
I simply copied and pasted the curl command as is into my terminal and I get a response that says {"error":"Content-type should be json!"} I am not sure if I am not supposed to copy and paste it as is, or what I may be doing wrong, but my understanding is I am supposed to get a json response with a email and a api token among some other data.
Here is my curl command.
curl -X POST \
-H 'Timekit-App: jl-fit' \
-d '{
"email": "email#email.com",
"password": "password"
}' \
https://api.timekit.io/v2/auth
Looks like you have discovered a bug in their docs/examples.
The API you're connecting to expects JSON content type, but curl by default (for POSTing data) uses application/x-www-form-urlencoded. You can fix it by adding the header field explicitly with: -H 'Content-Type: application/json'.
Also, when you use the -d/--data option, method defaults to POST, so it doesn't have to be specified explicitly.
All put together, this should work:
curl \
-H 'Content-Type: application/json' \
-H 'Timekit-App: jl-fit' \
-d '{"email": "email#email.com", "password": "password"}' \
"https://api.timekit.io/v2/auth"
When having multiple arguments, it can be convenient to keep them in an array (no need to escape newlines). For example:
args=(
-H 'Content-Type: application/json'
-H 'Timekit-App: jl-fit'
-d '{"email": "email#email.com", "password": "password"}'
"https://api.timekit.io/v2/auth"
)
curl "${args[#]}"
I'm trying to import this example into postman
curl -s --user 'api:YOUR_API_KEY' \
https://api.mailgun.net/v3/YOUR_DOMAIN_NAME/messages \
-F from='Excited User <mailgun#YOUR_DOMAIN_NAME>' \
-F to=YOU#YOUR_DOMAIN_NAME \
-F to=bar#example.com \
-F subject='Hello' \
-F text='Testing some Mailgun awesomness!'
Please help me understand the -s, --user, -F, what is that? And then when I try to import, I get this error: option-less arguments found. How can I fix this?
If you're trying to execute curl commands in Postman, select Import, then Paste Raw Text and then copy the command, but first remove all backslashes.
I have the following curl command which works perfectly fine:
curl --silent --location -cookie "$COOKIE" --cookie-jar "$COOKIE" \
--form 'username=xxx' --form 'password=yyy' 'http://example.com'
It logs into site http://example.com posting a form with variable names username and password.
Problem is: I do not want to pass the password in clear.
I have also tried to save the password in a file called passwd in the working directory (chmod 600 passwd) and used the following curl command (this is why I used --form instead of --data, which would have been worked fine with the password in clear), however, it does not work:
curl --silent --location -cookie "$COOKIE" --cookie-jar "$COOKIE" \
--form 'username=xxx' --form 'password=<passwd' 'http://example.com'
Any suggestion about how to solve this?
Regards,
Stefano
Hans Z. answer to use environmental variable is correct in my opinion. Though I might just add that in bash you could use read command, which would prompt for password and not make it visible in history.
So the usage would look like this
$ read -s PASSWD # -s is there so the characters you type don't show up
$ curl --silent --location -cookie "$COOKIE" --cookie-jar "$COOKIE" \
--form 'username=xxx' --form "password=${PASSWD}" 'http://example.com'
UPDATE:
Another solution found in the comments is to use curl's --data-urlencode name#filename argument.
Quoting the manpage:
name#filename
This will make curl load data from the given file (including any newlines), URL-encode that data and pass it on in the POST.
And the final command looked like
$ curl --silent --location -cookie "$COOKIE" --cookie-jar "$COOKIE" \
--data 'username=xxx' --data-urlencode "password#passwd" 'http://example.com'
Set the password in an environment variable, e.g. export PASSWD=secret and use it in --form "password=${PASSWD}".
Passing secrets in command line is insecure as they are available in process list.
You should use the --data-urlencode name#filename approach
This will make curl load data from the given file
(including any newlines), URL-encode that data and
pass it on in the POST.
or -K, --config <file>
Specify a text file to read curl arguments from. The
command line arguments found in the text file will be used
as if they were provided on the command line.
See also this anwser