I want to access a URL which requires a username/password. I'd like to try accessing it with curl. Right now I'm doing something like:
curl I get an error. I guess I need to specify a username and password along with the above command.
How can I do that?
20 Answers
Use the -u flag to include a username, and curl will prompt for a password:
curl -u username You can also include the password in the command, but then your password will be visible in bash history:
curl -u username:password It is safer to do:
curl --netrc-file my-password-file ...as passing a plain user/password string on the command line, is a bad idea.
The format of the password file is (as per man curl):
machine <example.com> login <username> password <password> Note:
- Machine name must not include
https://or similar! Just the hostname. - The words '
machine', 'login', and 'password' are just keywords; the actual information is the stuff after those keywords.
Or the same thing but different syntax
curl You can also just send the user name by writing:
curl -u USERNAME Curl will then ask you for the password, and the password will not be visible on the screen (or if you need to copy/paste the command).
To securely pass the password in a script (i.e. prevent it from showing up with ps auxf or logs) you can do it with the -K- flag (read config from stdin) and a heredoc:
curl --url url -K- <<< "--user user:password" Usually CURL command refer to as
curl -u USERNAME:PASSWORD if you don't have any password or want to skip command prompt to demand for password simple leave the password section blank.
i.e. curl -u USERNAME:
curl -X GET -u username:password {{ }} -O pretty easy, do the below:
curl -X GET/POST/PUT <URL> -u username:password Other answers have suggested netrc to specify username and password, based on what I've read, I agree. Here are some syntax details:
Like other answers, I would like to stress the need to pay attention to security regarding this question.
Although I am not an expert, I found these links insightful:
To summarize:
Using the encrypted versions of the protocols (HTTPS vs HTTP) (FTPS vs FTP) can help avoid Network Leakage.
Using netrc can help avoid Command Line Leakage.
To go a step further, it seems you can also encrypt the netrc files using gpg
With this your credentials are not "at rest" (stored) as plain text.
1To let the password least not pop up in your .bash_history:
curl -u user:$(cat .password-file) I had the same need in bash (Ubuntu 16.04 LTS) and the commands provided in the answers failed to work in my case. I had to use:
curl -X POST -F 'username="$USER"' -F 'password="$PASS"' "" Double quotes in the -F arguments are only needed if you're using variables, thus from the command line ... -F 'username=myuser' ... will be fine.
Relevant Security Notice: as Mr. Mark Ribau points in comments this command shows the password ($PASS variable, expanded) in the processlist!
3You can use command like,
curl -u user-name -p > new-file-name.ext Then HTTP password will be triggered.
2Plain and simply put the most secure way would be to use environment variables to store/retrieve your credentials. Thus a curl command like:
curl -Lk -XGET -u "${API_USER}:${API_HASH}" -b cookies.txt -c cookies.txt -- "" Would then call your restful api and pass the http WWW_Authentication header with the Base64 encoded values of API_USER and API_HASH. The -Lk just tells curl to follow http 30x redirects and to use insecure tls handling (ie ignore ssl errors). While the double -- is just bash syntax sugar to stop processing command line flags. Furthermore, the -b cookies.txt and -c cookies.txt flags handle cookies with -b sending cookies and -c storing cookies locally.
The manual has more examples of authentication methods.
2The safest way to pass credentials to curl is to be prompted to insert them. This is what happens when passing the username as suggested earlier (-u USERNAME).
But what if you can't pass the username that way? For instance the username might need to be part of the url and only the password be part of a json payload.
tl;dr: This is how to use curl safely in this case:
read -p "Username: " U; read -sp "Password: " P; curl --request POST -d "{\"password\":\"${P}\"}" unset P U read will prompt for both username and password from the command line, and store the submitted values in two variables that can be references in subsequent commands and finally unset.
I'm gonna elaborate on why the other solutions are not ideal.
Why are environment variables unsafe
- Access and exposure mode of the content of an environment variable, can not be tracked (ps -eww ) since the environment is implicitly available to a process
- Often apps grab the whole environment and log it for debugging or monitoring purposes (sometimes on log files plaintext on disk, especially after an app crashes)
- Environment variables are passed down to child processes (therefore breaking the principle of least privilege)
- Maintaining them is an issue: new engineers don't know they are there, and are not aware of requirements around them - e.g., not to pass them to sub-processes - since they're not enforced or documented.
Why is it unsafe to type it into a command on the command line directly Because your secret then ends up being visible by any other user running ps -aux since that lists commands submitted for each currently running process. Also because your secrte then ends up in the bash history (once the shell terminates).
Why is it unsafe to include it in a local file Strict POSIX access restriction on the file can mitigate the risk in this scenario. However, it is still a file on your file system, unencrypted at rest.
2In some API maybe it does not work (like rabbitmq).
there is alternative:
curl curl also the above format is usable in browsers.
You can use gpg encrypted netrc file with curl like so:
netrc_file="$HOME/netrc.gpg" curl --netrc-file <(gpg -d $netrc_file) This is MUCH more than the OP asked for, but since this is a top result for securely passing passwords to curl, I'm adding these solutions here for others who arrive here searching for that.
NOTE: -s arg for read command is not POSIX, and so is not available everywhere, so it won't be used below. We will use stty -echo and stty echo instead.
NOTE: All bash variables below could instead be declared as locals if in a function, rather than unsetting.
NOTE: perl is pretty generally available on all systems I've tried due to it being a dependency for many things, whereas ruby and python are not, so using perl here. If you can guarantee ruby/python where you're doing this, you can replace the perl command with their equivalent.
NOTE: Tested in bash 3.2.57 on macOS 10.14.4. Some small translation may be required for other shells/installs.
Securely prompt a user for a (reusable) password to pass to curl. Particularly useful if you need to call curl multiple times.
For modern shells, where echo is a built-in (check via which echo):
url=' printf "Username: " read username printf "Password: " stty -echo # disables echoing user input, POSIX equivalent for 'read -s' read pass printf "\n" # we need to move the line ahead stty echo # re-enable echoing user input echo ${pass} | sed -e "s/^/-u ${username}:/" | curl --url "${url}" -K- unset username unset pass For older shells, where echo is something like /bin/echo (where whatever it echos can be seen in the process list):
THIS VERSION CANNOT REUSE THE PASSWORD, see lower down instead.
url=' printf "Username: " read username printf "Password: " stty -echo # disables echoing user input, POSIX equivalent for 'read -s' perl -e ' my $val=<STDIN>; chomp $val; print STDERR "\n"; # we need to move the line ahead, but not send a newline down the pipe print $val; ' | sed -e "s/^/-u ${username}:/" | curl --url "${url}" -K- stty echo # re-enable echoing user input unset username If you happen to need to store the password temporarily to a file, to re-use it for multiple commands before clearing it (say because you're using functions for code re-use and don't want to repeat code and can't pass the value around via echo). (Yes, these are a little contrived looking in this form not being functions in different libraries; I tried to reduce them to the minimum code needed to show it.)
When echo is built-in (this is especially contrived, since echo is a built-in, but provided for completeness):
url=' filepath="$(mktemp)" # random path, only readable by current user printf "Username: " read username printf "Password: " stty -echo # disables echoing user input, POSIX equivalent for 'read -s' read pass echo "${pass}" > "${filepath}" unset pass printf "\n" # we need to move the line ahead stty echo # re-enable echoing user input cat "${filepath}" | sed -e "s/^/-u ${username}:/" | curl --url "${url}" -K- rm "${filepath}" # don't forget to delete the file when done!! unset username When echo is something like /bin/echo:
url=' filepath="$(mktemp)" # random path, only readable by current user printf "Username: " read username printf "Password: " stty -echo # disables echoing user input, POSIX equivalent for 'read -s' $(perl -e ' my $val=<STDIN>; chomp $val; open(my $fh, ">", $ARGV[0]) or die "Could not open file \"$ARGV[0]\" $\!"; print $fh $val; close $fh; ' "$filepath") printf "\n" # we need to move the line ahead stty echo # re-enable echoing user input cat "${filepath}" | sed -e "s/^/-u ${username}:/" | curl --url "${url}" -K- rm "${filepath}" # don't forget to delete the file when done!! unset username If you are on a system that has Gnome keyring app a solution that avoids exposing the password directly is to use gkeyring.py to extract the password from the keyring:
server=server.example.com file=path/to/my/file user=my_user_name pass=$(gkeyring.py -k login -tnetwork -p user=$user,server=$server -1) curl -u $user:$pass ftps://$server/$file -O In my case I needed a prompt where the user can enter their credentials.
The most simplistic way to get a prompt for username and password would be the following one-liner:
read -p "Username: " U ; curl -u "$U" <URL> ; unset U The read command prompts for the username. The -u "$U" parameter tells curl to try to authenticate with the username $U and prompt for the password.
As a bonus the password will only be visible to curl and won't end up in any log or history.
The man page of curl has more details about different authentication modes:
I also like the approach the user "iammyr" takes. It can cover cases where the authentication algorithm of cURL fails:
You should make sure what is the authentication type.
If it is digest authentication , http header is:
GET /xxxxxxxxxxx HTTP/1.1 Host: 192.168.3.142 User-Agent: Go-http-client/1.1 Authorization: Digest username="admin", realm="admin@51200304-49-test", nonce="5.1722929000077545", uri="/xxxxxxxxxxx", response="52d30eba90820f2c4fa4d3292a4a7bbc", cnonce="f11900fe0906e3899e0cc431146512bb", qop=auth, nc=00000001 Accept-Encoding: gzip you can use --digest option :
curl --digest -u 'username:password' ' 
