If in Loop in bash - mysql

Question

This should be an incredibly easy question but I am not very familiar with bash and I am taking way longer than I should to figure it out.
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re= $(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")
if [ $re -eq 0 ]; then
echo sucess
fi
done
This is an example of what I am trying to do, I have an id array and I want to send a query to my db so I can get a flag in the row with a certain id and then do something based on that. But I keep getting unexpected token errors and I am not entirely sure why
Edit: While copying the code and deleting some private information somehow I deleted the then, it was present in the code I was testing.

Answer 1

Based on what you described and the partial script, I am not certain I can completely create what you are trying to do but the token error messages you are experiencing usually have to do with the way bash handles whitespace as a delimiter. A few comments based on what you posted:
You need to remove the space around the equal sign in declaring an variable, so the space after the equal sign in re= needs to removed.
Because bash will is sensitive to whitespace, you need to quote variables declarations that might contain a space. To be safe, quotes need to be around the sub-shell $( )
You were missing the then in the if statement
It is important that variables in the test brackets, that is single [ ]s, must be quoted. Using an unquoted string with -eq, or even just the unquoted string alone within test brackets normally works, however, this is an unsafe practice and can give unpredictable results.
So, taking into account the items noted, the updated script would look something like:
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re="$(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")"
if [ "$re" -eq "0" ]; then
echo "success"
fi
done
Can you try working the edits mentioned into your script and see if you are able to get it working? Remember, it will be helpful for you to use a site like ShellCheck to learn more about potential pitfalls or the uniquenesses of bash syntax. This will help to ensure you are working toward a solution to your specific need rather then getting trapped by some tricky syntax.
After you have worked through those edits, can you report back your experience?
EDIT
Based on your comments there is a good chance you are not running your script with bash despite the including #!/bin/bash at the top of your script. When you run the script as sh scriptname.sh you are forcing the script to be run by sh not bash. Try running your script like this /bin/bash scriptname.sh then report back on your experience.
For more information on the differences between various shells, see Unix/Linux : Difference between sh , csh , ksh and bash Shell

Answer 2

Your problem with your if statement is that you do not have the then keyword. A simple fix is:
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re= $(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")
if [ $re -eq 0 ]; then
echo sucess
fi
done
Also here is a great reference on if statements in bash