Passing a variable as part of an argument of a command

Question

I am trying to pass a variable as part of an argument to a command. Specifically, I want mysqldump to return elements whose ID is above some number. For example, this line works and returns all elements with ID above 30: mysqldump --user=MYUSER -p -B DB --tables sometable --where='id>30'.

I have a script that determines this number and assigns it to variable LASTID. I would like to pass the value of this variable into the argument (in place of "30"). I run this:

mysqldump --user=MYUSER -p -B DB --tables sometable --where='id>"${LASTID}"'

But it returns the entire database, instead of using the correct id. It is as if it is ignoring what I am passing to LASTID.

I also tried this:

eval $(mysqldump --user=MYUSER -p -B DB --tables sometable --where='id>"${LASTID}"') and eval echo $(mysqldump --user=MYUSER -p -B DB --tables sometable --where='id>"${LASTID}"').

score 2 · Accepted Answer · answered Jul 09 '22 at 18:04

2

It's:

mysqldump --user=MYUSER -p -B DB --tables sometable --where="id>${LASTID}"

Research the difference between single and double quotes in shell. Check your scripts with shellcheck. Eval is one letter from evil - do not use eval.

answered Jul 09 '22 at 18:04

KamilCuk

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Last sentence made my day – Mad Physicist Jul 09 '22 at 18:07
Of course! Silly me. I did learn the difference, but I added them around the variable instead of the whole argument. Thanks so much! – George B. Y. Jul 09 '22 at 18:12
1

And wow I found the script spellcheck. It looks invaluable. Although I suspect you are gonna complain about invaluable being too close to evaluable or containing all the letters of evil. – George B. Y. Jul 09 '22 at 18:14

Passing a variable as part of an argument of a command

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