根据值在phpmyadmin上更新和设置SQL

想知道为什么我的SQL更新案例无法正常运行。 我已经有了价值观：

$usrcustomerid = 110;
$search = 'face';

正在查找集合中我的案例的修复程序，因为触发此SQL时没有更新

SQL更新代码

   public function UpdateUserSearch($db, $usrcustomerid, $search) {
   $stmt = $db->prepare(
     " UPDATE usr_customer
       SET search_counter = (CASE WHEN search_counter = 1 THEN 2
                                  WHEN search_counter = 2 THEN 3
                                  WHEN search_counter = 3 THEN 1 END),

           search1 = (CASE WHEN search_counter = 1 THEN $search END),
           search2 = (CASE WHEN search_counter = 2 THEN $search END),
           search3 = (CASE WHEN search_counter = 3 THEN $search END)
       WHERE usrcustomerid = $usrcustomerid "
   );

   $stmt->bindValue(':usrcustomerid', $usrcustomerid, PDO::PARAM_INT);
   $stmt->bindValue(':search', $search, PDO::PARAM_STR);
   $stmt->execute();
   $rowAffected = $stmt->rowCount();

   return $rowAffected;
 }

表格usr\U客户

在下图中，用户的搜索计数器设置为1，三个搜索列中没有值，因此根据我的逻辑，代码应该在“search1”中插入我的“$search”值，并将搜索计数器从1更新为2

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PS我知道SQL注入，请不要用提示回答，因为我正在尝试首先让我的代码工作。

这不是SQL问题，而是PHP问题。在SQL查询中替换的值应以：而不是$作为前缀

     " UPDATE usr_customer
       SET search_counter = (CASE WHEN search_counter = 1 THEN 2
                                  WHEN search_counter = 2 THEN 3
                                  WHEN search_counter = 3 THEN 1 END),

           search1 = (CASE WHEN search_counter = 1 THEN :search END),
           search2 = (CASE WHEN search_counter = 2 THEN :search END),
           search3 = (CASE WHEN search_counter = 3 THEN :search END)
       WHERE usrcustomerid = :usrcustomerid "

见：