Echo variable inside Array PHP

Question

Hello I would like to be able to define a URL inside my config file and echo the url inside array:

$options = array( 'results_per_page' => 30,
                  'url' => 'http://localhost/page.php?page=*VAR*',
                  'db_handle' => $dbh
                );

In this case let's say I would like to define a variable:

$url = 'http://localhost/';

or

define( "SERVER_PATH", "http://localhost/" );

and being able to echo it inside the 'url' => "$url" page.php?page=*VAR*'

Any help will be welcome.

why do you want to echo it inside an array? practically it's not possible. — Muhammad Sumon Molla Selim
– Muhammad Sumon Molla Selim, Commented Jul 14, 2016 at 19:23
because I have this line in different php files and want to change the url in one place instead of each of them. — John Siniger
– John Siniger, Commented Jul 14, 2016 at 19:24
Create a temp $tmpUrl = $url then use that 'url' => "$tmpUrl" . "page.php?page=*VAR*'" I don't know I fully understand the issue but it seems like you do not want to overwrite something? — nerdlyist
– nerdlyist, Commented Jul 14, 2016 at 19:26

Muhammad Sumon Molla Selim · Accepted Answer · 2016-07-14 19:25:15Z

2

You can do this way:

Using variable:

$options = array( 'results_per_page' => 30,
                  'url' => $url . 'page.php?page=*VAR*',
                  'db_handle' => $dbh
                );

Using constant:

$options = array( 'results_per_page' => 30,
                  'url' => SERVER_PATH . 'page.php?page=*VAR*',
                  'db_handle' => $dbh
                );

answered Jul 14, 2016 at 19:25

Muhammad Sumon Molla Selim

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Comments

nerdlyist · Accepted Answer · 2016-07-14 19:35:01Z

2

You can also append the variable to the end of the array value by using the concatenation operator

$option['url'] = $option ['url'] . $var;

Or in less words

$option['url'] .= $var;

edited Jul 14, 2016 at 19:35

nerdlyist

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answered Jul 14, 2016 at 19:30

dinnertoast

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