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I need to load a db into a mysql server and the following command works from the shell:
From shell

mysql -h "172.17.0.2" -u "root"  -p"mypasswd" -Bse "create database mydb;"

But not from the script, in which the ip and the password are variables passed as arguments:

The script:

#!/bin/bash

set -e
sqlname=$1
sqlpass=$2
sqlip=$3
...  
set -x
echo "Creating the database..."
mysql -h $sqlip -u "root" -p$sqlpass -Bse "create database mydb;"

The result of the script: 

./myscript.sh mysql1 mypasswd 172.17.0.2

Creating the database...
+ mysql -h 172.17.0.2 -u root -pmypasswd -Bse 'create database mydb;'
ERROR 2003 (HY000): Can't connect to MySQL server on '172.17.0.2' (111)

-- begin edit
Same result for the variables inside double quotes: 

#!/bin/bash  
...  
mysql -h "$sqlip" -u root -p"$sqlpass" -Bse "create database mydb;"

+ echo 'Creating the database...' Creating the database...
+ mysql -h 172.17.0.2 -u root -pmypasswd -Bse 'create database mydb;' ERROR 2003 (HY000): Can't connect to MySQL server on '172.17.0.2' (111)

-- end edit

Double quotes and singles quotes with variables inside the script confuse me.

I am missing something obvious here about using variables inside and with quotes.

Any hint?

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  • Done. I am using a very simple password for testing, as simple as "mypasswd" Commented Apr 18, 2018 at 15:51

2 Answers 2

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The use of quotes is to stop the shell interpreting certain punctuation characters, such as space (parameter separator), > (output redirection), * (wild-card in file mask), etc.

Single quotes stop all interpretation of the quoted text, while double quotes allow $ expansion (and some others). The command inside a script is no different from that typed to an interactive prompt.

The IP address has no shell-specific characters, nor has root, so they do not need to be quoted. If your password has shell-specific characters, such as a space, it will need to be quoted, and because it requires $ expansion they need to be double quotes. The -Bse parameter has both space and ;, so it needs quoting, but either will do, since there is no $ expansion.

The variables you are expanding $sqlip and $sqlpass need to be defined within the subshell which executes the script, so they need to be set in the subshell or set as export in the calling shell.

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  • Hi @afh. In the script I use the shebang that will run the script in a subshell (edited the question). Is that what yo meant? Commented Apr 18, 2018 at 15:11
  • @AJN - No. Scripts always run in a subshell, unless invoked with the source / . command. Commented Apr 18, 2018 at 15:23
  • Where is 172.17.0.2? Is it the same machine as you're running on? Is it on your intranet? Commented Apr 18, 2018 at 15:41
  • It is a container, and there is no issue of connectivity. I tested the command (mentioned in the beginning of the question) from the host console and it works. Commented Apr 18, 2018 at 15:53
  • 2
    Searching for "sql error 2003 HY000 111" gives a lot of solutions, including this one and the following answer. But your command-line and script commands appear to be the same, so I don't see how this can vary in the script, unless you have unexported configuration variables, such as MYSQL_TCP_PORT. Commented Apr 18, 2018 at 15:57
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It’s possible that there are some characters in the password variable which are treated specially by the shell.

As a rule, it’s best practice to always quote shell variables to avoid unwanted side effects such as token splitting and wildcard expansion. See When is double-quoting necessary? from Unix and Linux Stack Exchange.

mysql -h "$sqlip" -u root -p"$sqlpass" -Bse "create database mydb;"
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  • Inside the script, I wrapped the variables in double quotes, but gives the same result (Look at the edited question). Commented Apr 18, 2018 at 15:05

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