Unix Shell脚本-在另一个变量内替换变量并将值连接到同一变量

Question

I have the below requirement to concatenate the value of a variable to its own variable and create one array:

Table=CRS
Job=JOBNAME
gen_job_$Table=GENERATEDJOB
gen_job_$TABLE=$Job,${gen_job_${Table}}
echo ${gen_job_${Table}} should give JOBNAME,GENERATEDJOB

I tried using the eval function also as below:

eval gen_job_$Table=$job,eval echo \$gen_job_$Table

However, I am not able to display the final result.

Answer 1

You perhaps just want to consider using an array for a situation like this (and I recommend reading this page , however, if you want to dynamically build your variable names it is possible. To set a variable whose name is stored in another variable use printf -v like

table=CRS
job=JOBNAME
holds_name="gen_job_$Table"
printf -v "$holds_name" 'GENERATEDJOB'

Then when you want to access the variable whose name is set in holds_name you can use indirection:

printf '%s,%s\n' "$job" "${!holds_name}"

Using braces around the variable name, if the first character inside the brace is ! then the rest of the word is treated as the name of a variable that is expanded to find the name of the actual variable whose value should be used.