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Jan 15, 2022 at 1:04 comment added mFlorin defaults-extra-file didn't worked for me, but defaults-file worked ok. Thx, solved my problem (not wanting to put secrets in cmd).
Jun 3, 2021 at 13:46 comment added Mladen Jablanović @DavidTonhofer this should be an answer on its own.
Jun 16, 2019 at 21:05 comment added vinkomlacic The --defaults-extra-file must be the first option when executing the mentioned command. For me, it didn't work otherwise.
Mar 21, 2018 at 6:51 vote accept Kokizzu
Mar 1, 2018 at 18:02 comment added David Tonhofer 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
Feb 3, 2018 at 1:50 comment added Geoffrey Wiseman 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).
Oct 19, 2017 at 5:01 comment added Wildcard @Kusalananda, yes, but as per comments on unix.stackexchange.com/q/385339/135943, that does not mean it's safe!
Feb 1, 2017 at 0:52 comment added ChuckCottrill 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.
Jan 23, 2017 at 13:40 comment added Kusalananda 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/…
May 23, 2015 at 5:48 history answered garethTheRed CC BY-SA 3.0