(PHP - MYSQL) 我要数"likes"。当他们点击一个按钮时，它会添加 1 个赞

Question

这是我的dao文件

<?php

require_once( __DIR__ . '/DAO.php');

class insertVoteDAO extends DAO {

  public function insertVote1() {
      $sql = "UPDATE `ic_items` SET `flames`+1 WHERE id=1";
      $stmt = $this->pdo->prepare($sql);
      $stmt->execute();
      return $stmt->fetch(PDO::FETCH_ASSOC);
      }
    }
?>

在我的 index.php 中，我这样做：

 <?php
require_once( __DIR__ . '/dao/insertVoteDAO.php');
$voteDAO = new insertVoteDAO();

      if (!empty($_POST)) {
      $insertVote = $voteDAO->insertVote1();
    }
     ?>

这是我得到的错误：

Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '+1 WHERE id=1' at line 1 in /Applications/MAMP/htdocs/backend/major-IC/dao/insertVoteDAO.php:9 Stack trace: #0 /Applications/MAMP/htdocs/backend/major-IC/dao/insertVoteDAO.php(9): PDO->prepare('UPDATE `ic_item...') #1 /Applications/MAMP/htdocs/backend/major-IC/collection.php(72): insertVoteDAO->insertVote1() #2 {main} thrown in /Applications/MAMP/htdocs/backend/major-IC/dao/insertVoteDAO.php on line 9

它说我在查询中的语法有问题。

希望有人能帮助我。

Answer 1

这一行在 flames+1:

处是错误的

$sql = "UPDATE ic_items SET flames+1 WHERE id=1";

应该是

$sql = "UPDATE ic_items SET flames = flames+1 WHERE id=1";

Answer 2

同意 BanNsS1，您可以查看以下 link 以供进一步参考：Increment a database field by 1

(PHP - MYSQL) 我要数"likes"。当他们点击一个按钮时，它会添加 1 个赞

(PHP - MYSQL) I want to count "likes". When they click on a button it adds 1 like

php

mysql

dao