asked on

bash dynamic variable assignment

I need to assign values to dynamic variables

EX "$time"ofpurchase=$var

andy tips?

I keep getting "command not found error"

thanks
jculkincys

Duncan Roe

help eval

SOLUTION

Duncan Roe

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jculkincys

ASKER

I almost got it

I just need a little more hand holding

I have dyname name on each side of the equation

eval "$time"_u_staff=$staff # doesnt work

I am trying to get more info

does the above help at all?

Gabriel Orozco

in order to better separate your bash variables, the preferred method is

eval ${time}_u_staff=${staff}

jculkincys

ASKER

Hmm I am stil getting errors

Portion of Code:

eval ${time}_u_faculty=${faculty}
echo "${time}_u_faculty --> $(${time}_u_faculty)"

Output of script with set -xv:

+ eval 00_u_faculty=0
00_u_faculty=0
++ 00_u_faculty=0
./analyzelogsdev.sh: 00_u_faculty=0: command not found

ASKER CERTIFIED SOLUTION

Arty K

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jculkincys

ASKER

Thanks for the help

any tips on reading a dynamic variable?

for example while in a for loop I want to pull all the values back

Ex.

for time in 1 2 3
do
echo ${u_faculty_${time}}

done

I am stuck with the following

+ time=01
echo `\${u_staff_${time}}`
${u_staff_${time}}
./analyzelogsdev.sh: ${u_staff_${time}}: bad substitution
++ echo
+ echo 'PLEASE WORK: '
PLEASE WORK:
+ exit 0

Arty K

you cannot use embedded braces {}, use this syntax:

eval echo "\$u_faculty_${time}"