linux - Assign into named variable within a shell function -

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• indirect variable assignment in bash 3 answers

i have typical issue assign variable select statement, i've tried many options, none of them work..

i have function replace column name on upper(column name).

upper=(column1 column2)  # array variable  upper_function() { query="$1" ((i=0;i<${#upper[@]};i++)) query=$(echo "$query" | sed -e "s/\b"${upper[$i]}"\b/upper(${name[$i]})/g") done sqlquery="$query" } 

below have sql statement:

sqlquery="select column1,column2 table" 

now, change column1 , column2

upper(column1) ,  upper(column2).  

so run function:

upper_function "$sqlquery" 

solution above works fine, case instead of

sqlquery="$query" 

i make function more automatic , assigning name of select statement variable eg below:

$2="$query" 

and run function:

upper_function "$sqlquery" "sqlquery" 

but not working , have no idea why, how assign variable properly? thanks

this is, @ core, bashfaq #6.

printf -v can used indirect assignments in modern (3.1+) versions of bash:

upper=( column1 column2 )  # array variable  upper_function() {   local colname query=$1 varname=$2   colname in "${upper[@]}";     query=$(echo "$query" | sed -e "s/\b"${colname}"\b/upper($colname})/g")   done   printf -v "$varname" %s "$query" } 

...thereafter:

upper_function "$sqlquery" sqlquery 

---

on older shells, instead:

eval "$varname=\$query" 

...note need trust value of $varname not malicious, code injected variable run function. thus, never want use filename, table name, or other potentially-attacker-controlled variable name variable assigned in way.