I have MySQL password saved on a file foo.php, for example P455w0rd, when I try to use it:

$ cat foo.php | grep '$dbpwd=' | cut -d '"' -f 2 | mysql -U root -p mydb -h friendserver
Enter password: (holds)

$ echo P455w0rd | mysql -u root -p mydb -h friendserver
Enter password: (holds)

Both option still ask for password, what's the correct way to send password from stdin?

  • 3
    must be no blank between -p and your password. May 23, 2015 at 7:10
  • mysql doesn't read the password from stdin, I wasn't able to figure it what it does read it from.
    – Omn
    Mar 19, 2018 at 0:20
  • 2
    The proper answer is don't put your passwords on the command line where anyone with access to /proc can trivially read them as long as the program is running. That's what a ~/.my.cnf is for, properly chmod'ed to 0600
    – Shadur
    Jun 16, 2019 at 21:29

6 Answers 6


The mysql client utility can take a password on the command line with either the -p or --password= options.

If you use -p, there must not be any blank space after the option letter:

$ mysql -pmypassword

I prefer the long options in scripts as they are self-documenting:

mysql --password=mypassword --user=me --host=etc
  • 13
    This is insecure because any user could view the password - either directly via /proc/$pid/cmdline or via the ps command. It's true that mysql overwrites the password in argv during startup but there is always a time window where another user could observe the password. Also, on some systems the argv overwriting might not work. Jul 14, 2019 at 14:49
  • Thanks for the hint that there must not be any blank space after the option letter. Saved my day. Oct 7, 2020 at 8:11
  • 8
    Of course this is insecure, but most people are doing something like this in a build process that will be deleted shortly afterwards...
    – Adam F
    Jun 29, 2021 at 22:41

You have to be very careful how you pass passwords to command lines as, if you're not careful, you'll end up leaving it open to sniffing using tools such as ps.

The safest way to do this would be to create a new config file and pass it to mysql using either the --defaults-file= or --defaults-extra-file= command line option.

The difference between the two is that the latter is read in addition to the default config files whereas with the former, only the one file passed as the argument is used.

Your additional configuration file should contain something similar to:


Make sure that you secure this file.

Then run:

mysql --defaults-extra-file=<path to the new config file> [all my other options]
  • 2
    MySQL will modify its argv to overwrite the parameters given to the -p flag. At least that's what's concluded here, together with other relevant info: unix.stackexchange.com/questions/78757/…
    – Kusalananda
    Jan 23, 2017 at 13:40
  • 1
    This is the best solution, as it provides a measure of security. You need to chmod go-rwx, and make sure this argument preceeds all other arguments. Feb 1, 2017 at 0:52
  • 3
    @Kusalananda, yes, but as per comments on unix.stackexchange.com/q/385339/135943, that does not mean it's safe!
    – Wildcard
    Oct 19, 2017 at 5:01
  • Might be worth adding a note about --login-path now that it's supported. It's not much better than this, but it is slightly less plain-text (even though the barrier to converting the contents to plaintext is low). Feb 3, 2018 at 1:50
  • 22
    Set MYSQL_PWD in the environment (export MYSQL_PWD=muhpassword) and execute your command without the -p. See MySQL Program Environment Variables. In spite of the manual's dire warnings, this is rather safe. Unless you start weird warez in the same shell later. So we run: MYSQL_PWD=$(cat foo.php etc) mysql -u foouser -h barhost Mar 1, 2018 at 18:02

create a file ~/.my.cnf, make it only accessible by yourself, permission 600.


Then you don't need type password any more.

   bash$ mysql -u myuser mydatabase

You can use a nifty Linux trick...

/dev/stdin can be used as a file resource for --defaults-file like so:

echo -e "[client]user=xxx\npassword=xxx" | mysql --defaults-file=/dev/stdin -e 'select user()

You can usually use - instead of /dev/stdin, but it's a convention that a lot of programs (including mysql) do not honour.

A little more about the /dev/stdin file resource:

$ ls -la /dev/stdin
lrwxrwxrwx 1 root root 15 Jun 19 08:35 /dev/stdin -> /proc/self/fd/0

This way you can also check empty password situations

  • 2
    you should use password instead of pass to not see the following error: Info: Using unique option prefix 'pass' is error-prone and can break in the future. Please use the full name 'password' instead.
    – Bash Stack
    Aug 30, 2020 at 10:45
  • 1
    Great, thank you! Aug 31, 2020 at 19:38
  • What's with the 'select user() bit? I couldn't get this to work. It just outputs user().
    – geoidesic
    Jan 24 at 17:56

As @maxschlepzig said in a comment,

Note that there is the MYSQL_PWD environment variable which is read by mysql if you don't specify -p.

So in bash:

read -s -p "Enter the mysql password for $DBUSER @ $DBNAME: " DBPASS 
mysqldump -u $DBUSER $DBNAME > dump.sql

If you want to start mysql with a password provided you have to fetch the password in a variable first:

MYSQLPASS=`cat foo.php | grep '$dbpwd=' | cut -d '"' -f 2`

Then you can start your mysql command with:

mysql -U root -p ${MYSQLPASS} mydb -h friendserver
  • 10
    This also jeopardizes your password security. Note that there is the MYSQL_PWD environment variable which is read by mysql if you don't specify -p. And under Linux this is a secure method because the environment of a user cannot be read by other unprivilleged users - in contrast to the argument vector. Jul 14, 2019 at 14:55

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