Commands out of sync; you can't run this command now

[cmartino]

Hello All,

I have the following code that gets log entries and compiles them into an array for viewing later:

function get_risk_logs($risk_no, $group_id) {
        // get dates first
        $stmt = mysqli_prepare($this->db_conn, "select date from risk_logs where risk_no = ? and group_id = ? order by date desc");
        if (!$stmt) {
            echo("Error preparing the 'get_risk_logs: 1' statement: " . mysqli_error($this->db_conn));
            exit();
        }
        mysqli_stmt_bind_param($stmt, "ii", $risk_no, $group_id);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_bind_result($stmt, $date);
        while (mysqli_stmt_fetch($stmt)) {
            $dates[] = $date;
        }
        mysqli_stmt_close($stmt);
        
        $dates = array_unique($dates);
        $dates = array_values($dates);

        /*
         * for each of the dates in the array above,
         * put all log information in an array corresponding to that date
         */
        $logs = array();
        for ($i = 0; $i < sizeof($dates); $i++) {
            $tmp_date = strval($dates[$i]);
            $logs[$i] = array('date'=>$tmp_date, 'info'=>array());
            $stmt = mysqli_prepare($this->db_conn, "select risk_no, date, time, user, details from risk_logs where date = ? and risk_no = ? and group_id = ?");
            if (!$stmt) {
                echo("Error preparing the 'get_risk_logs: 2' statement: " . mysqli_error($this->db_conn));
                exit();
            }
            mysqli_stmt_bind_param($stmt, "sii", $tmp_date, $risk_no, $group_id);
            mysqli_stmt_execute($stmt);
            mysqli_stmt_bind_result($stmt, $risk_no, $date, $time, $user_number, $details);
            while (mysqli_stmt_fetch($stmt)) {
                $count = 0;
                $user_name = $this->get_user_name($user_number);
                $string = "<b>User:</b>&nbsp;$user_name<p/><b>Time:</b>&nbsp;$time<p/><b>Details:</b>&nbsp;$details<p/>";
                $logs[$i]['info'][$count] = $string;
                $count++;
            }
            mysqli_stmt_close($stmt);
        }
        return $logs;
    }

When the above function is called, I get an error message that says "Error preparing the 'prepare mysql' statement: Commands out of sync; you can't run this command now". I have been trying to figure out why I am getting this error message but I am coming up short. Can someone help me out? Is this a PHP problem or a MySQL problem?

Thanks!

[cmartino]

Update:

I changed the second sql statement line to

$stmt = mysqli_prepare($this->db_conn, "select risk_no, date, time, user, details from risk_logs where date = '?' and risk_no = ? and group_id = ?");

Notice the 's surrounding the question mark after "date = ".

By doing this, I manage to eliminate the error message, but my array is not populating properly. Any idea why?