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為什么MySQL INSERT語句無法正常工作

[英]Why is MySQL INSERT statement not working without error

運行良好的php / mySQL預訂功能突然停止將預訂條目插入數據庫,而無需更改代碼和正常運行的數據庫連接。

我運行工作在其他網站頁面的水貨版本; 兩者之間的唯一區別是,殘破的版本在php 5.6上運行,而功能正常的版本仍在5.4上運行。

即使表沒有更新,添加錯誤日志也不會帶來任何結果,並且我看不到php 5.4和5.6之間的任何不贊成使用的語句。

誰能找到我所缺少的問題?

 //If the confirm button has been hit:
if (isset($_POST['submit'])) {

//Create the foreach loop
  foreach ($_POST['class_id'] as $classes) {
  $class_id = (int)$classes;
  //UPDATE the bookings table **THIS PART IS NOT WORKING**:
  $query = "INSERT INTO bookings (user_id, booking_name, class_id, time_stamp) VALUES ('$user_id', '$username', '$class_id', NOW())";
  mysqli_query($dbc, $query);

}
foreach($_POST['class_id'] as $classes){
  $class_id = (int)$classes;
  //Change the booking numbers **THIS WORKS FINE**:
  $increase = "UPDATE classes SET online_bookings = (online_bookings + 1), total_bookings = (total_bookings + 1), free_spaces = (free_spaces - 1) WHERE class_id = $class_id";
  mysqli_query($dbc, $increase);
  }
   mysqli_close($dbc);

..以及提供$ _POST數據的表:

echo'<div class="container">';
echo'<div class="span8 offset1 well">';
echo'<p class="lead text-info">Do you want to reserve space at these classes?</p>';



//table header
echo '<table id="dancers" class="table table-bordered table-hover">';
  echo '<thead><tr><th>Date</th><th>Time</th><th>Venue</th><th>Who\'s going?</th></tr></thead>';
//create the form
echo '<form id="makebkg" method="post" action="' . $_SERVER['PHP_SELF'] . '">';


//Get the class IDs from the GET to use in the POST

    foreach ($_GET['sesh'] as $class_id) {

    $sql = "SELECT class_id, DATE_FORMAT(date, '%a, %d %b') AS new_date, DATE_FORMAT(time, '%H:%i') AS new_time, venue FROM classes WHERE class_id = '$class_id'";
    $data = mysqli_query($dbc, $sql);


//get table data
    while ($row = mysqli_fetch_array($data)) {
    $date = $row["new_date"];
    $time = $row["new_time"];
    $venue = $row["venue"];
    $class_id = $row["class_id"];
  }


//Show a table of the selected classes


      echo '<tr><td>' . $date . '</td>';
      echo '<td>' . $time . '</td>';
      echo '<td>' . $venue . '</td>';
      echo '<td>' . $username . '</td></tr>';
      echo '<input type="hidden" name="date[]" value="' . $date . '" />';
      echo '<input type="hidden" name="time[]" value="' . $time . '" />';
      echo '<input type="hidden" name="venue[]" value="' . $venue. '" />';
      echo '<input type="hidden" name="username[]" value="' . $username . '" />';
      echo '<input type="hidden" name="class_id[]" value="' . $class_id . '" />';



    }
 echo'</table>';



    //Go Back button
    echo '<a class="btn btn-link pull-left" href="classes.php"><i class="icon-arrow-left"></i> Go back</a>';

  // Make booking button - LIVE
      echo'<div id="confirmbtn">';
     echo '<input type="submit" id="confirm" name="submit" class="btn btn-large btn-primary pull-right" value="Confirm">';
      echo '</div>';

好的,我終於解決了問題。 事實證明,托管公司已將MySQL模式更改為“嚴格”。

此處的INSERT語句使某些表列留空,因此嚴格模式將拒絕整個插入。 與更新insert命令相比,在insert命令之前立即更改模式是解決問題的更快方法:

  // TURN OFF STRICT MYSQL MODE
  $strict = "SET sql_mode = ''";
  mysqli_query($dbc, $strict);

感謝您的建議和寬容的編碼器的容忍度。

您是否嘗試檢查查詢?

error_reporting(1);

$q = mysqli_query($dbc, $query);

if (!$q)
{
  echo 'Error' . mysqli_error($dbc);
}

對其他查詢執行相同的操作。

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