CODEWITHSUNDEEP
bashvariables
user1819769
user1819769

Reputation: 85

How to use user input to select a variable in bash?

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?

Upvotes: 1

Views: 1965

Answers (3)

Darren Smith
Darren Smith

Reputation: 2488

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

Upvotes: 1

F. Hauri - Give Up GitHub
F. Hauri - Give Up GitHub

Reputation: 70742

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.

Upvotes: 4

stephanmg
stephanmg

Reputation: 766

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}".

Upvotes: 0

Related Questions