1

I want to send a number of predefined queries to mysql. i will use the variable defined internally by using the externally inputted variable as below. But not good.

[testuser@testserver work]$ cat test1.sh 
#!/bin/sh

query1='select * from mysql.user limit 2;'
query2="select * from mysql.user limit 2;"

echo $1
echo "$1"
#mysql -uuser-p -e "$1"

this is result

[testuser@testserver work]$ sh test3.sh query1 
query1
query1

but i want result

[testuser@testserver work]$ sh test1.sh  query1 
select * From mysql.user limit 1

how to modify this bash scrpit?

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3
  • 1
    U could use echo "${!1}" syntax! Commented Nov 4, 2019 at 8:50
  • Question: what is your ultimate goal here? Indirection is generally not the way to go. If you just want to define the variables query1 and query2, you can use source test1.sh; echo query1, if you just want to retrieve it, I would use a case statement. Commented Nov 4, 2019 at 10:31
  • @user1819769 : $1 is the first parameter to your script. You pass query1 as parameter, so of course query1 is printed by the echo statement. Commented Nov 4, 2019 at 10:46

3 Answers 3

Reset to default
4

Try simply:

#!/bin/bash

query1='select * from mysql.user limit 2;'
query2="select * from mysql.user limit 2;"

echo "${!1}"
#mysql -uuser-p -e "${!1}"

But you could:

#!/bin/bash

queries=('select * from mysql.user limit 2;'
        "select * from mysql.user limit 2;")

echo "${queries[$1]}"
#mysql -uuser-p -e "${queries[$1]}"

Then run ./test 0 or ./test 1.

Or even

#!/bin/bash

declare -A queries=$'(
    [query1]=\047select * from mysql.user limit 2;\047
    [query2]="select * from mysql.user limit 2;")'

echo "${queries[$1]}"
#mysql -uuser-p -e "${queries[$1]}"

Note: Syntax $'....\047...\047...' is a trick for using simple quotes in quoted string.

Then run ./test3 query1.

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3 Comments

So $1 needs now to be numeric because we use an array?
@stephanmg In my second purpose, yes!
@F.Hauri: I'd wanted to suggest this too, because if he will have more queries this will somehow clutter the code.
1

Can also use a case statment:

#!/usr/bin/env bash


case "$1" in
    --query1)
        echo "select * from mysql.user limit 1;"
        ;;
    --query2)
        echo "select * from mysql.user limit 2;"
        ;;
    *)
        echo "unkown option"
        exit 1
        ;;
esac

then run something like: ./test --query1

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1 Comment

Double dash are not needed! case $1 in query1 ) ...;; query2) ...;;esac will work too!
0

You can add the following lines:

#!/bin/bash

query1='query 1'
query2='query 2'

eval var="$"$1
echo $var

Edit: This uses eval which should be avoided, which might be unsafe if you don't trust the data. There is for Bash scripts variable indirection via the mechanism "${!VARNAME}", so you can use "${!1}".

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2 Comments

Avoid using eval!!
Possibly interesting if not writing Bash but Sh scripts: unix.stackexchange.com/questions/425024/…

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