PHP 警告:mysqli_stmt::bind_param():变量数与准备语句中的参数数不匹配

Question

不确定为什么我会收到此 PHP 警告消息。准备好的语句中似乎有四个参数，bind_param() 中也有四个变量。谢谢你的帮助!

  if($stmt = $mysqli -> prepare("SELECT url, month, year, cover_image FROM back_issues ORDER BY year DESC, month DESC")) {
   $stmt -> bind_param("ssis", $url, $month, $year, $cover_image);

   $stmt -> execute();

   $stmt -> bind_result($url, $month, $year, $cover_image);

   $stmt -> fetch();

   while ($stmt->fetch()) {
     echo "<li class='item'><a href='$url'><img src='$cover_image' alt='$cover_image' width='' height='' /></a><br /><span class='monthIssue'>$month $year</span></li>";
   }

   $stmt -> close();
   $mysqli -> close();

 }

Answer 1

if($stmt = $mysqli -> prepare("SELECT url, month, year, cover_image FROM back_issues ORDER BY year DESC, month DESC")) {
   $stmt -> bind_param("ssis", $url, $month, $year, $cover_image);
   [...]
}

这些行没有任何意义!如果你想使用参数，你必须将它们用于 where 条件或类似条件，非常像:

$mysqli->prepare("SELECT * FROM back_issues WHERE url =? AND month =? AND year =? and cover_image = ?");
$stmt->bind_param("ssis", $url, $month, $year, $cover_image);

如果您没有任何占位符与参数绑定(bind)，bind_param() 方法将产生您遇到的错误。此外，该 IF 语句应该做什么？如果您想验证该查询是否产生任何结果，您必须先运行它，然后再进行验证。