note 125026 deleted from function.preg-replace by crell

Note Submitter: Anonymous ---- Note that the replacement string using the ${n} notation for a group identifier (rather than the $n notation) must be in a SINGLE quoted string: $replacement = '${1}1,$3'; // Correct - Single Quoted $replacement = "${1}1,$3"; // Incorrect - Double Quoted $replacement = "$1 one , $3"; // Correct - Double Quoted but with $n notation. This might sneak up on you, because it is not an error to use "${1}". The double quotes allow a string that contains variable interpolation, and the braces make the variable a PHP "Variable-Variable". In the above example the undefined ${1} will evaluate to NULL rather than the replaced string. Valid PHP, but not what you wanted.