Array display troubles in PHP AJAX

Question

I am learning something new with checking for whether a username is used in the database and have created some code based on a tutorial I found online. I understand the logic, though not sure I have approached it the right way. In essence. Information is passed over from a form field. If what has been entered matches a field in the database then I want it to return / echo a result 'yes'. if it does not match, I need it to echo 'no'. Seems straight forward.

The tutorial is designed for predetermined values. i.e. $existing_users=array('test',''one','two',three');

Though I want the 'test',''one','two',three' to actually pull dynamically from a database.

So I went about adding the setting it up to do so, though the code I have written doesn't work when I try to place dynamic values in. My code is as below:

$existing_users = array();
mysql_select_db($database_db, $db);
$result = mysql_query('SELECT * FROM clients') or exit(mysql_error());
while ($row = mysql_fetch_assoc($result)) {$existing_users[] = $row['shortcode'];}

$arr = $existing_users;

$display = "'" . implode("', '", $arr) . "'";

    // THIS IS THE PROBLEM 
    // If the code is written out as:
    // $existing_users=array('test',''one','two',three'); 
    // It works.
    // When the script is coded as below. It doesn't work.
    // Note. If I echo $display is displays 'test',''one','two',three'

$existing_users=array($display);

//value received from the get method
$user_name=$_POST['user_name'];

//checking weather user exists or not in $existing_users array
if (in_array($user_name, $existing_users))
{

//user name is not available
echo "no";
} 
else
{ 
//user name is available 
echo "yes";
}

I am not sure whether I am approaching this the right way because I am hacking an online tute, there potentially could be a far easier way. Any thoughts would be greatly appreciated.

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Mihai Iorga
– Mihai Iorga, Commented Aug 26, 2012 at 0:08
You are adding a string to the $existing_users array. Why are you doing all that converting and copying of $existing_users to begin with? — PeeHaa
– PeeHaa, Commented Aug 26, 2012 at 0:11

Nicolas Brown · Accepted Answer · 2012-08-26 00:24:33Z

3

here's a faster way to do it:

$user_name=mysqli_real_escape_string($_POST['user_name']);

$result = mysql_query("SELECT * FROM clients where shortcode like '$user_name'") or exit(mysql_error());

if(mysql_num_rows($result)==0)
    echo 'no';
else
    echo 'yes'

i didnt validate the input from $_POST tho

what tutorial are you following anyway??

edited Aug 26, 2012 at 0:24

answered Aug 26, 2012 at 0:13

Nicolas Brown

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4 Comments

Nicolas Brown Over a year ago

i was just showing how to do it faster, even mentioned i didnt validate the input sigh

Nicolas Brown Over a year ago

i should probably get familiar with mysqli

PeeHaa Over a year ago

You should. Either mysqli_* or PDO.

uınbɐɥs Over a year ago

mysqli_real_escape_string. You forgot the i, if you are doing it the modern way...

Mihai Iorga · Accepted Answer · 2012-08-26 00:19:35Z

You don't need to remake $existing_users, as you are already creating that array from database query

$existing_users = array();
mysql_select_db($database_db, $db);
$result = mysql_query('SELECT * FROM clients') or exit(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
    $existing_users[] = $row['shortcode'];
}

$user_name=$_POST['user_name'];

if (in_array($user_name, $existing_users)){
    echo "no";
} else { 
    //user name is available 
    echo "yes";
}

and try to move your code to PDO

$db = new PDO('mysql:host=localhost;dbname='.$database_db, 'username', 'password', array(ATTR::PDO_EMULATE_PREPARES => false));
$stmt = $db->prepare("SELECT * FROM clients WHERE `shortcode`=:shortcode");
$stmt->execute(array(':shortcode' => $_POST['user_name']));

if($stmt->rowCount() == 1){
    echo 'no';
} else {
    echo 'yes';
}

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