-2

尝试运行此 mysql 查询/准备好的语句时遇到问题。

  <?php

  if (!empty($_POST['var1'])&&!empty($_POST['var2'])
    &&!empty($_POST['var3'])&&isset($_POST['var4'], 
    $_POST['var5'], $_POST['var6'])) {

  require_once 'connect.inc.php'; 

  $query = "INSERT INTO tablename (var1, var2, var3, var4, var5, var6)
  VALUES (?,?,?,?,?,?)";

  $stmt = mysqli_prepare($link, $query);

  mysqli_stmt_bind_param($stmt, "ssssss", $var1, $var2, $var3, $var4, $var5, $var6);

  $var1 = $_POST['var1'];
  $var2 = $_POST['var2'];
  $var3 = $_POST['var3'];
  $var4 = $_POST['var4'];
  $var5 = $_POST['var5'];
  $var6 = $_POST['var6'];

  mysqli_stmt_execute($stmt);

   if (mysqli_stmt_affected_rows($stmt)==1) {
   echo 'Thank you for your submission.';
      } else {
      mysqli_stmt_close($stmt);
      mysqli_close($link);  }

 } else {
echo 'We were unable to process your information. Please ensure all required fields 
        were filled out.'.mysqli_stmt_error($stmt);
} 

?>

当我运行代码时，我收到以下错误消息：

 Notice: Undefined variable: stmt in ...on line 62
  Warning: mysqli_stmt_error() expects parameter 1 to be mysqli_stmt, null given in ... on        line 62

第 62 行是包含mysqli_stmt_error($stmt)在末尾的行。据我所知，我已经$stmt正确设置了。（$link来自connect.inc.php页面。）我不确定为什么会收到消息null given，因为我确实有$stmt参数 1。

有谁知道我做错了什么？

4

1 回答 1

-1

您正在尝试捕获查询错误，但在显示 $_POST var 错误的地方使用它。查看改进的代码：

  <?php

  if (!empty($_POST['var1'])&&!empty($_POST['var2'])
    &&!empty($_POST['var3'])&&isset($_POST['var4'], 
    $_POST['var5'], $_POST['var6'])) {

    require_once 'connect.inc.php'; 

    $query = "INSERT INTO tablename (var1, var2, var3, var4, var5, var6)
    VALUES (?,?,?,?,?,?)";

    $stmt = mysqli_prepare($link, $query);

    $var1 = $_POST['var1'];
    $var2 = $_POST['var2'];
    $var3 = $_POST['var3'];
    $var4 = $_POST['var4'];
    $var5 = $_POST['var5'];
    $var6 = $_POST['var6'];

    mysqli_stmt_bind_param($stmt, "ssssss", $var1, $var2, $var3, $var4, $var5, $var6);
    mysqli_stmt_execute($stmt);

     if (mysqli_stmt_affected_rows($stmt)==1) {
          echo 'Thank you for your submission.';
        } else {
          mysqli_stmt_close($stmt);
          mysqli_close($link);  

          // do something with your statement error
          echo mysqli_stmt_error($stmt);
        }

} else {
  echo 'We were unable to process your information. Please ensure all required fields were filled out.';
} 
?>
于 2013-07-28T13:20:22.940 回答