警告：mysql_query（）期望参数2是资源，[关闭]

Question

i am trying to run this PHP Function:

function makeString() {
    // Create our random string
    $string = "";
    $characters = array('a', 'b', 'c', 'd');

    for ($i = 0; $i < 4; $i++)
    {
        $string .= $characters[mt_rand(0, 4)];
    }

    $sql="SELECT COUNT(*) FROM urls WHERE short_url = '{$string}'";
    $rs=mysql_query($sql,$conn);
    $result=mysql_fetch_array($rs);

    if($result['COUNT(*)'] > 0)
    {
        // if it already exists, do it again
         makeString();
    }
    else
    {
        return $string;

        //insert the data
    }
}

makeString();

but im getting these errors:

Warning: mysql_query() expects parameter 2 to be resource, null given in /home/charlie/public_html/short/shorten.php on line 15

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in /home/charlie/public_html/short/shorten.php on line 16

what do they mean?

all the table/database columns seem okay and there are no reserved words etc...

Answer 1

$conn is not available inside scope of function makeString() you can either remove it because it's not required if you already connected, or you can pass it as function argument like:

makeString($conn);