在同一脚本中重用包含变量的代码 bash

Question

I have a small block of code in bash like below

#!/bin/bash

query="select * from table where id = ${row_id}"

row_id=1
echo "$query"


row_id=3
echo "$query"


row_id=4
echo "$query"

expected output is below

select * from table where id = 1
select * from table where id = 3
select * from table where id = 5

But I am getting nothing as output

I know I am referencing variable before assigning it.

The idea here is to use reusable code instead of writing the same code at many places

How can I achieve what I want

Answer 1

you should be getting

select * from table where id =
select * from table where id =
select * from table where id =

and as you already mentioned the reason is

I know I am referencing variable before assigning it.

One way to implement this

$ for row_id in 1 3 5; 
  do 
    echo "select * from table where id = $row_id"; 
  done

select * from table where id = 1
select * from table where id = 3
select * from table where id = 5

UPDATE

Based on the comment

Here if row_id is a random variable I get as part of another query then how do I get the correct query as my output

which is different from the posted question, better to define a function

$ getquery() { echo "select * from table where id = $1"; }

$ getquery $RANDOM

select * from table where id = 12907

Answer 2

You can create a function and call the function at various place by assign variable to it

#!/bin/bash

# create a function with variable and write your command
# here your command is print the query 
my_function_name(){
arg1=$1
echo "select * from table where id = ${arg1}"
}

# assign varaible 
row_id=1

# print the ourput of function when above variable is assigned
query=$(my_function_name "$row_id")

echo $query


# assign varaible 
row_id=2

# print the ourput of function when above variable is assigned
query=$(my_function_name "$row_id")

echo $query

# assign varaible 
row_id=3

# print the ourput of function when above variable is assigned
query=$(my_function_name "$row_id")

echo $query