Hi All,

I am trying to write a script that will execute a mysql query based upon user input. I am running into issues with the following:

VERIFYQUERY=`mysql -u$DBUSER -p$DBPW -D$DB -e"select user_name from user where user_id='$REMOVEID'";`

VERIFYUSER=`echo $VERIFYQUERY | cut -f2`

The $VERIFYQUERY variable is setting properly and the output is something like "01 username".

All I need to do is cut the last field out to strip away the userid portion of the output but I am doing something wrong.

Any input is appreciated, I don't do much BASH scripting and am sure that this is an easy fix for someone with a bit more experience.

Thank You!

I had a typo in my script and have fixed it. I thought I was going crazy for a minute there..

Please use [code][/code] tags around your code and data, to preserve formatting and to improve readability. Please do not use quote tags, colors, or other fancy formatting.


No need to use an external tool for this. Unless your shell is very primitive, there are parameter substitution patterns that can do the job nicely.


More bash string manipulations.

In addition:

1)
$(..) is highly recommended over `..`

2)
Environment variables are generally all upper-case. So while not absolutely necessary, it's good practice to keep your own user variables in lower-case or mixed-case, to help differentiate them.

3)
QUOTE ALL OF YOUR VARIABLE SUBSTITUTIONS. You should never leave the quotes off a variable expansion unless you explicitly want the resulting string to be word-split by the shell. This is a vitally important concept in scripting, so train yourself to do it correctly now. You can learn about the exceptions later.

http://mywiki.wooledge.org/Arguments
http://mywiki.wooledge.org/WordSplitting
http://mywiki.wooledge.org/Quotes