加入2表，其中customer_id = $ username [重复]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

I'm developing a laundry service system, using php. I have 2 table, table customer and table ticket (with customer_id as the foreign key here).

so i'm trying to display a list of ticket after the user logged in, consists of data from table customer and table ticket.

i'm trying to join 2 table with customer_id as the link join.. for the logged in username(using $session to store username for logged in customer). here is my code:

$customerusername = $_SESSION['customerusername'];
$result = mysqli_query($conn,"SELECT * from TICKET INNER JOIN CUSTOMER ON TICKET.CUSTOMER_ID=CUSTOMER.CUSTOMER_ID 
                                        WHERE CUSTOMER_ID=(SELECT CUSTOMER_ID from CUSTOMER WHERE CUSTOMER_USER='$customerusername')");

however i got this error Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in this first line below:

while($row = mysqli_fetch_array($result)){   //Creates a loop to loop through results
    $found_row = true;
    echo " <tr>
            <td>".$row['TICKET_ID']."</td>
            <td>".$row['TICKET_PICKUP']."</td>
            <td>".$row['TICKET_PTIME']."</td>
            <td>".$row['TICKET_DELIVERY']."</td>
            <td>".$row['TICKET_DTIME']."</td>
            <td>".$row['TICKET_STATUS']."</td>
            <td>RM ".$row['TICKET_PRICE']."</td>
            <td>".$row['PAYMENT_STATUS']."</td>
            </tr>";
    }
    if ($found_row == false) {
            echo "No Ticket Submitted";
        }

what went wrong?

</div>

Answer 1

You have to use aliases on tables to specify columns that have same names and avoid ambiguous errors.

Your CUSTOMER_ID is present in your two tables. So MySQL cannot know which column you are talking about.

You could try something like this:

$customerusername = $_SESSION['customerusername'];
$result = mysqli_query($conn,
    "SELECT * from TICKET t 
     INNER JOIN CUSTOMER c ON t.CUSTOMER_ID=c.CUSTOMER_ID 
     WHERE t.CUSTOMER_ID=(SELECT CUSTOMER_ID from CUSTOMER WHERE CUSTOMER_USER='$customerusername')");