I want to echo the results of this mysql query

SELECT username, task_completed FROM accounts WHERE referrer_by = :referrer_by AND task_completed < 1

I have this code so far but this doesnt return anything and it should return the values which are less than 1

$ref_query=$db->prepare("SELECT username, task_completed FROM accounts WHERE referrer_by = :referrer_by AND task_completed < 1");
$ref_query->bindValue(":referrer_by", $user);

$result = $ref_query->execute();

if($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        echo "<br> Username: " . $row["username"] . " - Tasks: " . $row["task_completed"] . " <br>";
    }
}
Re: How to echo this Query 80 80

Sorry, I'm not familiar with the PHP framework you're using. Does bindValue() properly escape the $user variable?
Also, do you know if it's returning num_rows at all? What about echo'ing echo $result->num_rows; right above the if statement to see if it's even reaching the while loop.

Re: How to echo this Query 80 80

@Dani thank you for your reply. I did tried with echoing the num_rows but didnt return anythng also.

Re: How to echo this Query 80 80

Firstly, add thsi to the top of your page to see any errors returned, this only whilst in development - error_reporting(E_ALL); ini_set('display_errors', 1);

As we don't know what is in your data table, is there any records to be returnd that is < 1?
What do you gte when you echo out?

Re: How to echo this Query 80 80

For some or other reason I cannot edit my previous code, wanted to add the following code -

Re: How to echo this Query 80 80

Dani, pse check, I added first comment, wanted to add code but from there, code block shows with all my code attempts but does not want to post the code??

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