php pdo：update + insert然后選擇return null

Question

出於某種原因，這個執行的PHP代碼返回NULL ... cud any1請幫助糾正它？

public function like($pid)
        {
             $uid = escape($_SESSION['user']);
             $sql = $this->_db->prepare("UPDATE postsinitial SET likes = likes+1 WHERE pid = :m;INSERT IGNORE INTO userlikedposts (ulp_userid,ulp_postid) VALUES (:k, :m)");
             $sql->bindValue(':k', $uid);
             $sql->bindValue(':m', $pid);
             $sql->execute();

             $query = $this->_db->prepare("SELECT likes FROM postsinitial WHERE pid = :n");
             $query->bindParam(':n', $pid);
             $query->execute();
             while($rows = $query->fetch())
             {
                return $rows['likes'];
             }

         }

但是，當我分別運行查詢的兩個部分，即注釋掉$ sql批處理代碼並單獨運行$ query批處理時，它工作並返回一個值..，它工作正常..但沒有按照說明組合..那我怎么按原樣運行呢？

我也嘗試過這個模型用於選擇查詢bt仍然是相同的結果：

$query = $this->_db->prepare("SELECT likes FROM postsinitial WHERE pid = :n");
             $query->bindParam(':n', $pid);
             $query->execute();
             while($rows = $query->fetch(PDO::FETCH_ASSOC))
             {
                return $rows[0]['likes'];
             }

Answer 1

答案很簡單：

您應該逐個運行查詢，而不是將它們全部填充到一個調用中。 運行你的插入查詢分離ROM更新，你會沒事的。

     public function like($pid)
     {
         $sql = "UPDATE postsinitial SET likes = likes+1 WHERE pid = ?";
         $this->_db->prepare($sql)->execute($_SESSION['user']);

         $sql = "INSERT IGNORE INTO userlikedposts (ulp_userid,ulp_postid) VALUES (?, ?)";
         $this->_db->prepare($sql)->execute([$_SESSION['user'], $pid]);

         $stmt = $this->_db->prepare("SELECT likes FROM postsinitial WHERE pid = ?");
         $stmt->execute([$pid]);
         return $stmt->fetchColumn();
     }