Interpret / Parse / Replace variable in string with value

Php-mysql Course

To replace variables in string with their values defined in php, you can use the strtr() or preg_replace_callback() function.

Example with strtr()

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<?php
// array with data for the variables added in string
$data = array(
  '{$site}'=>'https://coursesweb.net/',
  '{$year}'=>date('Y')
);
$str = 'Code-snippets from {$site} , added in: {$year}.';

$str2 = strtr($str, $data);
echo $str2;      // Code-snippets from https://coursesweb.net/ , added in: 2014.

Example with preg_replace_callback()

- Click to select it.

<?php
// array with data for the variables added in string
$data = array(
  'id'=>'test2',
  'domain'=>'CoursesWeb.net'
);
$str = '<div id="{$id}">Parse this string, interpret {$domain} </div>';


$str2 = preg_replace_callback('/{\$([^}]+)}/', function($m) {
  GLOBAL $data;
  return $data[$m[1]];
}, $str);

echo $str2;      // <div id="test2">Parse this string, interpret CoursesWeb.net </div>

- It is important to define data for variables into an Array with the keys that match the form you added the variable in string. Interpreting variables in string can also be useful to create and use template page; see this tutorial: Create and Use in PHP a Simple Template Page.

<table><tr>
  <th>Title 1</th>
  <th>Title 2</th>
</tr></table>

.some_class {
  line-height: 150%;
}

document.getElementById("id_button").onclick = function(){
  window.open("http://coursesweb.net/");
}

$ar_dir = scandir("dir_name");
var_export($ar_dir);