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I'd like to write bash script to parse data from a configuration file. I searched for this but without finding something I could modify to suit my needs.

Joomla! config file:

   public $access = '1';
   public $debug = '0';
   public $debug_lang = '0';
   public $dbtype = 'mysqli';
   public $host = 'localhost';
   public $user = 'template';
   public $password = 'template';
   public $db = 'template_druha';
   public $dbprefix = 'dsf1i_';
   public $live_site = '';
   public $secret = '2w9gHzPb4HfAs2Y9';
   public $gzip = '0';
   public $error_reporting = 'default';

I'd like to parse the database credentials on lines with "$user" and "$password" and store them in a variable. What is the best practice?

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1  
If your file is PHP, use PHP (or a PHP AST parser). – Chris Down Sep 16 '15 at 16:40
    
Parse the file using awk to extract the $user and $password values from this joomla config file. – steve Sep 16 '15 at 16:42
    
Have php read in and parse the file, then have print statements to return the values you want – roaima Sep 16 '15 at 16:48
up vote 2 down vote accepted

With GNU grep, you could do:

user=$(grep -oP "\\\$user.+?'\K[^']+" file)
pass=$(grep -oP "\\\$password.+?'\K[^']+" file)

The -P enables Perl Compatible Regular Expressions, which give us \K (ignore anything matched so far). The -o means "only print the matching portion of the line. Then, we search for $var (we need three \, to avoid expanding the variable and to avoid the $ being taken as part of the regex), a single quote and one or more non-' characters until the next '.

Alternatively, you could use awk:

user=$(awk -F"'" '/\$user/{print $2}' file)
pass=$(awk -F"'" '/\$password/{print $2}' file)

Here, we are setting the field delimiter to ', so the value of the variable will be the second field. The awk command prints the second field of matching lines.

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up vote 1 down vote

For your sample input:

$ cat /tmp/foo
public $access = '1';
public $debug = '0';
public $debug_lang = '0';
public $dbtype = 'mysqli';
public $host = 'localhost';
public $user = 'template-user';
public $password = 'template-pass';
public $db = 'template_druha';
public $dbprefix = 'dsf1i_';
public $live_site = '';
public $secret = '2w9gHzPb4HfAs2Y9';
public $gzip = '0';
public $error_reporting = 'default';

You could do:

user="$(grep '$user' /tmp/foo | sed -e 's/  *$//g' -e 's/;$//' | awk -F= '{ print $2}')"
pass="$(grep '$password' /tmp/foo | sed -e 's/  *$//g' -e 's/;$//' | awk -F= '{ print $2}')"
  • The grep searchs for the user or password lines in the given file
  • The first sed expression removes any trailing whitespace
  • The second sed expression removes the trailing ;
  • The awk uses = as the column delimiter, and prints the second column
  • The var=$(...) evaluates all that, takes it output and stores it in the variable
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up vote 1 down vote

Since this file is a PHP file, the most reliable way to extract data for it is to parse it with PHP. If you need to use the data in a shell script, write some PHP code that prints out variable assignments in shell syntax.

#!/bin/sh
eval "$(php -r '
    include $argv[1];
    $config = new JConfig();
    echo "joomla_user=\x27" . preg_replace("/\x27/", "\x27\\\x27\x27", $config->user) . "\x27\n";
    echo "joomla_password=\x27" . preg_replace("/\x27/", "\x27\\\x27\x27", $config->password) . "\x27\n";
' /path/to/configuration.php)"
echo "User is $joomla_user"
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