无法使用过程PHP方法从SELECT预准备语句中检索结果?

时间:2019-12-20 13:52:06

标签: php mysqli

我在从SELECT语句检索结果时遇到问题,我尝试调试,但似乎没有达到我的第二个if语句,该语句在一段时间内未显示结果。我不太确定自己是否在正确地构造它,或者我使用的方法有点过于复杂。

$sql ="SELECT t1.customerID, t1.date_of_activity, t1.number_of_tickets, t2.activity_name, t2.price, t2.location    
FROM booked_activities as t1 INNER JOIN activities as t2 ON t1.activityID = t2.activityID WHERE t1.customerID = ?";

  if($result = mysqli_prepare($conn, $sql)){

    mysqli_stmt_bind_param($stmt, "i", $customerID);

    if (mysqli_stmt_execute($stmt)) {

        if(mysqli_stmt_num_rows($result) > 0){

          while($row = mysqli_fetch_array($result)){
            echo "<div class=\"card\">\n";
            echo "
            <h2 class=\"description\">Number of tickets: " . $row['number_of_tickets'] . "</h2>
            <p class=\"price\">&euro;" . $row['price'] . "<br />" . $row['activity_name'] . "</p>
            <p style=\"text-align: center;\" class=\"description\">Activity Date: " . $row['date_of_activity'] . "</p>\n";
            echo "<p><button>View Booking</button></p>";
            echo "</div>\n"; 
          }

          mysqli_free_result($result);
      } else{
          echo "No bookings were found.";
      }
    }
     mysqli_stmt_close($stmt);
  }
 mysqli_close($conn);

0 个答案:

没有答案