如何使用纯SQL实现PHP条件？

我有以下表格结构：

// Posts
+----+------------+-----------------------+----------------+-------------+
| id |   title    |        content        |  money_amount  |  author_id  |
+----+------------+-----------------------+----------------+-------------+
| 1  | title 1    | content 1             | NULL           | 12345       |
| 2  | title 2    | content 2             | 25             | 42355       |
| 3  | title 3    | content 3             | 5              | 53462       |
| 4  | title 4    | content 4             | NULL           | 36346       |
| 5  | title 5    | content 5             | 15             | 13322       |
+----+------------+-----------------------+----------------+-------------+
//                                          ^^ NULL means this post is free


// Money
+---------+--------------+
| post_id | user_id_paid | 
+---------+--------------+
| 2       | 42355        |  // He is author of post
| 2       | 34632        |  // This row means besides author, this user 34632 can see this post too. Because he paid the money of this post.
| 3       | 53462        |  // He is author of post
| 5       | 13322        |  // He is author of post
| 3       | 73425        |  // This row means besides author, this user 34632 can see this post too. Because he paid the money of this post.
+---------|--------------+

注1:表中的所有

post\u id

（s）属于非免费的帖子

注2:在

Money

表中始终有一行属于post作者（非免费）

注3:

Money

表只是用来确定谁可以看到这样的帖子

---

现在，此用户

$\u会话['current\u user']='23421'

想要查看此帖子

id=2

。这是我的密码：

$stm = $this->dbh->prepare(SELECT * FROM Posts WHERE id = '2');
$stm->execute();
$result = $stm->fetch(PDO::FETCH_ASSOC);

if ( $result[money] == '') {   // money_amount is NULL in the Posts table
    this post is free and everybody can see it
} else {

    $stm = $this->dbh->prepare(SELECT count(1) FROM Money WHERE post_id = '2' and user_id = $_SESSION['current_user']);
    $num_rows = $stm->fetchColumn();

    if($num_rows){
        $paid = true;  // This means current user paid the cost of post and he can see it.
    } else {
        $paid = false; // this means current user didn't pay the cost of post and he cannot see it.
    }
}

---

我想知道，我是否可以在一个查询中实现这两个查询，并使用MySQL而不是PHP实现该条件？

您可以使用联接，下面的查询使用

左联接

SELECT * FROM Money
LEFT JOIN Posts ON Money.post_id = Posts.id
WHERE ((Posts.money_amount IS NOT NULL AND Money.user_id_paid = :userId)
      OR Posts.money_amount IS NULL) AND Posts.id = :postId

请注意，
：userId
是
PDO
参数化查询的占位符，您应该在执行前将参数绑定到占位符。比如：

$postId = 2;
$stmt->bindParam('userId', $_SESSION['current_user']);
$stmt->bindParam('postId', $postId);

还要注意，绑定占位符名称时不需要冒号。使用
右键联接
意味着您从
Posts
表中选择
，并联接
Money
表
 如果
和
存在
函数（MySql），下面是使用
的解决方案：

@frz3993实际上我需要检查
money\u amount
列的值是否为
NULL
然后
join
到
money
表并检查当前用户的id是否存在。我如何使用MySQL实现这个条件呢？试试这个查询，我写得很快，我不知道它是否正确。如果结果>0，用户可以看到帖子<代码>从货币、帖子（邮政ID＝2和POSS.MouyyAUMT为空）或（Muny.PasySID＝2和Muny.UsRyIdApLay=＝$Sale[ [ CurrutuxAudi] ]）/if > IF语句中的子查询>中选择计数（*）：
...
$stmt = $conn->prepare(" 
        SELECT IF(p.money_amount,1,0) as notfree, 
        EXISTS(SELECT * FROM `Money` WHERE `post_id` = ? AND`user_id_paid` = ?) as paid
        FROM `Posts` p WHERE p.id = ? ");

$stmt->execute([2, $_SESSION['current_user'], 2]);
$result = $stmt->fetch(\PDO::FETCH_ASSOC);

if (!$result['notfree']) {  // post is free
    // this post is free and everybody can see it
} else {
    $paid = ($result['paid'])? true : false; 
}