mysql command line sql execution mystery, only part of the string is considered

Question

from command line from my Mac terminal:

mysql --host=127.0.0.1 --port=3306 -uroot -p"mypass" wordpress -e "update users set user_pass = '$1$Hat7oFty$mA.L2vsQdD3MxvxAuDFKp0';"

completes successfully...

however.... only .L2vsQdD3MxvxAuDFKp0 is written to the user_pass field in every row. Whiskey Tango Foxtrot?

Needless to say when I issue update users set user_pass = '$1$Hat7oFty$mA.L2vsQdD3MxvxAuDFKp0'; directly to the DB from an application like DataGrip it takes the whole string correctly....

It's been a long time since I used bash. Could it be thinking $1$Hat7oFty$mA is a variable it should be substituting with? — Uueerdo
– Uueerdo, Commented Jul 30, 2018 at 18:40
hmmm didn't think of that. I guess I have to 'escape' it someohow — JasonGenX
– JasonGenX, Commented Jul 30, 2018 at 18:50

Progman · Accepted Answer · 2018-07-30 20:43:24Z

2

The $ are part of shell variables, which are unintentionally get replaced. You have to escape the $ character to keep it in the string as a literal $.

$ echo "$1$Hat7oFty$mA.L2vsQdD3MxvxAuDFKp0"
.L2vsQdD3MxvxAuDFKp0
$ echo "\$1\$Hat7oFty\$mA.L2vsQdD3MxvxAuDFKp0"
$1$Hat7oFty$mA.L2vsQdD3MxvxAuDFKp0

answered Jul 30, 2018 at 20:43

Progman

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