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How to show content mysql in php dynamically

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How to show content mysql in php dynamically

ma ouml

9yAsked Aug 28, 2017, 5:41 AM1kPHP

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

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Gokhul Varman9y ago
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 );
});
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