How to show content mysql in php dynamically

Question

Hi friends I want to display content in the database (mysql) dynamically in php For example: assuming that I have three links I want that when the user clicks on a link I show him the content of this link How to do please And thank you in advance

Gokhul Varman · Answer

your PHP code $sql = 'SELECT username FROM users' ; $run = mysql_query( $sql $link ); $users = array(); if ( $run && mysql_num_rows( $run ) ) { while ( ( $fetch = mysql_fetch_assoc( $run ) ) !== false ) { $users[] = $fetch[ 'username' ]; } } // return string echo $users = implode( ''  $users ); ?> your javascript code http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js > ../jquery.airport-1.1.source.js > text/javascript > var users = []; $. get ( 'getData.php'  function(data) { users = data.split( '' ); }); $(document).ready(function() { $( '#div1' ).airport( users ); });