remote mysql login using username / password not working when adding password to command line ...
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remote mysql login using username / password not working when adding password to command line
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I am running the following command to connect to a remove mysql server (it is part of a bash script that will connect automatically)
mysql -h slave1 -u monitor -pvP!KnK4*
And I receive the following response within the terminal window
-bash: !KnK4: event not found
When I run mysql -h slave1 -u monitor -p without adding the passing in the command line and use the same password when prompted to enter it in this works fine.
What am I doing wrong in this instance?
PS the password above is not the real password (it is much longer just made it smaller for stackexchange purposes)
bash terminal mysql
asked 14 hours ago
ZabsZabs
1275
add a comment |
I am running the following command to connect to a remove mysql server (it is part of a bash script that will connect automatically)
mysql -h slave1 -u monitor -pvP!KnK4*
And I receive the following response within the terminal window
-bash: !KnK4: event not found
When I run mysql -h slave1 -u monitor -p without adding the passing in the command line and use the same password when prompted to enter it in this works fine.
What am I doing wrong in this instance?
PS the password above is not the real password (it is much longer just made it smaller for stackexchange purposes)
bash terminal mysql
asked 14 hours ago
ZabsZabs
1275
apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
did you try quotes?
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago
add a comment |
I am running the following command to connect to a remove mysql server (it is part of a bash script that will connect automatically)
mysql -h slave1 -u monitor -pvP!KnK4*
And I receive the following response within the terminal window
-bash: !KnK4: event not found
When I run mysql -h slave1 -u monitor -p without adding the passing in the command line and use the same password when prompted to enter it in this works fine.
What am I doing wrong in this instance?
PS the password above is not the real password (it is much longer just made it smaller for stackexchange purposes)
bash terminal mysql
asked 14 hours ago
ZabsZabs
1275
I am running the following command to connect to a remove mysql server (it is part of a bash script that will connect automatically)
mysql -h slave1 -u monitor -pvP!KnK4*
And I receive the following response within the terminal window
-bash: !KnK4: event not found
When I run mysql -h slave1 -u monitor -p without adding the passing in the command line and use the same password when prompted to enter it in this works fine.
What am I doing wrong in this instance?
PS the password above is not the real password (it is much longer just made it smaller for stackexchange purposes)
bash terminal mysql
bash terminal mysql
asked 14 hours ago
ZabsZabs
1275
asked 14 hours ago
ZabsZabs
1275
edited 13 hours ago
Zabs
asked 14 hours ago
ZabsZabs
1275
asked 14 hours ago
ZabsZabs
1275
asked 14 hours ago
ZabsZabs
1275
1275
apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
did you try quotes?
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago
add a comment |
apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
did you try quotes?
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago
apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
did you try quotes?
– user907860
13 hours ago
did you try quotes?
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago
add a comment |
1 Answer
1
active
oldest
votes
If you are going to use characters that are special to the shell (e. g. !, *) as a portion of a parameter, you must present it in such a manner as to not be interpreted by the shell before it is passed to the command you're invoking.
Rather than:
$ mysql -h slave1 -u monitor -pvP!KnK4*
Instead use:
$ mysql -h slave1 -u monitor -p'vP!KnK4*'
And then, when you have a moment, change your MySQL user's password if this is your actual password because you have put it on the internet for all to see.
(As a further aside, you should be aware that anyone who is able to read your user's shell history file also has access to this password.)
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you are going to use characters that are special to the shell (e. g. !, *) as a portion of a parameter, you must present it in such a manner as to not be interpreted by the shell before it is passed to the command you're invoking.
Rather than:
$ mysql -h slave1 -u monitor -pvP!KnK4*
Instead use:
$ mysql -h slave1 -u monitor -p'vP!KnK4*'
And then, when you have a moment, change your MySQL user's password if this is your actual password because you have put it on the internet for all to see.
(As a further aside, you should be aware that anyone who is able to read your user's shell history file also has access to this password.)
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
add a comment |
If you are going to use characters that are special to the shell (e. g. !, *) as a portion of a parameter, you must present it in such a manner as to not be interpreted by the shell before it is passed to the command you're invoking.
Rather than:
$ mysql -h slave1 -u monitor -pvP!KnK4*
Instead use:
$ mysql -h slave1 -u monitor -p'vP!KnK4*'
And then, when you have a moment, change your MySQL user's password if this is your actual password because you have put it on the internet for all to see.
(As a further aside, you should be aware that anyone who is able to read your user's shell history file also has access to this password.)
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
add a comment |
If you are going to use characters that are special to the shell (e. g. !, *) as a portion of a parameter, you must present it in such a manner as to not be interpreted by the shell before it is passed to the command you're invoking.
Rather than:
$ mysql -h slave1 -u monitor -pvP!KnK4*
Instead use:
$ mysql -h slave1 -u monitor -p'vP!KnK4*'
And then, when you have a moment, change your MySQL user's password if this is your actual password because you have put it on the internet for all to see.
(As a further aside, you should be aware that anyone who is able to read your user's shell history file also has access to this password.)
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
If you are going to use characters that are special to the shell (e. g. !, *) as a portion of a parameter, you must present it in such a manner as to not be interpreted by the shell before it is passed to the command you're invoking.
Rather than:
$ mysql -h slave1 -u monitor -pvP!KnK4*
Instead use:
$ mysql -h slave1 -u monitor -p'vP!KnK4*'
And then, when you have a moment, change your MySQL user's password if this is your actual password because you have put it on the internet for all to see.
(As a further aside, you should be aware that anyone who is able to read your user's shell history file also has access to this password.)
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
answered 13 hours ago
DopeGhotiDopeGhoti
47k56190
47k56190
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
add a comment |
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
will try that (the pwd itself is different just similar syntax although much longer)
– Zabs
13 hours ago
1
1
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
Glad to hear this is not your actual password.
– DopeGhoti
13 hours ago
add a comment |
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apparently bash treats everyting before the ! sign, namely "pvP" in your case, as options. Try to add a space after "-p". And better don't use the "-p" option on a live server, since the password remains in the bash history and can be stolen
– user907860
14 hours ago
tried the space and it still doesn't seem to work :(
– Zabs
14 hours ago
did you try quotes?
– user907860
13 hours ago
characters like ! and * can well be special to bash. So use them in single quotes
– user907860
13 hours ago