Bash script with variable from a different file in array

Question

I have a config.txt file with the following:

VAR=('a' 'b' 'c')
database='mydb'

and I want to use the VAR variable in the script (myscript.sh) to check if any of the data in it exists in a search that will be made on the database.

The search I am doing is like this:

output=$(mysql $database -u $user -p$pass -se "select name from my_table where word = '$VAR'")

You define VAR as an array but use it as a "scalar". What actual value are you wanting to use in that select statement? Do you want to iterate over the array? — glenn jackman
– glenn jackman, Commented Jul 20, 2017 at 12:47
I want to search the database for each one of the data stored in the variable VAR. One of them should return a result. — Mike
– Mike, Commented Jul 20, 2017 at 13:25

Olli K · Accepted Answer · 2017-07-20 11:36:09Z

1

You can read files into your script with the source command (aliased simply as .) like so:

. config.txt
# or
source config.txt

To get a comma separated string, check for example here

answered Jul 20, 2017 at 11:36

Olli K

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2 Comments

chepner Over a year ago

That won't quite work; it doesn't appear from its use that VAR is supposed to be a shell array.

Mike Over a year ago

VAR is supposed to have more than one data, therefore an array, and i want to use it to search the database for each one of them. How can i do that?

2 revs · Accepted Answer · 2017-07-20 14:17:22Z

0

To expand on @Olli's answer:

source config.txt

# SQL injection
# values=$(printf "'%s'\n" "${VAR[@]}" | paste -sd,)
values=$(printf "'%s'\n" "${VAR[@]//\'/\'\'}" | paste -sd,)

select="select name from my_table where word in ($values)"
output=$(mysql "$database" -u "$user" -p"$pass" -se "$select")

edited Jul 20, 2017 at 14:17

community wiki

2 revs
glenn jackman

4 Comments

Mike Over a year ago

each time i use source in the script i get: line 19: mysql: command not found line 21: ls: command not found

glenn jackman Over a year ago

that sounds like you need to ask a separate question, and you have to show the code.

Mike Over a year ago

I created a script to test your solution, the only thing i have more is at the beginning the #!/bin/sh, the rest is the same as you wrote above. i got GNU is because of that?

glenn jackman Over a year ago

Because of what? I don't understand your question.

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