向我的查询添加另一个数据库连接导致看不到某些行

Question

对了，这个我不是特别熟练，这其实是我第一次join查询，所以要温柔。我将提供尽可能多的细节，因为它可能会像煎锅一样击中你们大多数人的脸，但它正在让我疯狂！

我在查询我尝试在 codeigniter 中编写的博客时遇到问题。我已经为帖子设置了一个包含 2 个连接的查询，它是使用三个表的类别：帖子、类别和 posts_categories 现在我也在尝试加入我的 cmets 表来进行计数。

这是我的模型中的代码，它显示了我编写的两个通用帖子：

            $this->db->select('posts.id,
                            posts.title,
                            posts.slug,
                            posts.content,
                            posts.author,
                            posts.date,
                            posts.time,
                            posts.tags,
                            posts.status,
                            GROUP_CONCAT(categories.name SEPARATOR \'-\') AS categories
                            ');
        $this->db->group_by(array('posts.id'));
        $this->db->from('posts');
        $this->db->join('posts_categories', 'posts_categories.blog_entry_id = posts.id');
        $this->db->join('categories', 'posts_categories.blog_category_id = categories.category_id');
        $query = $this->db->get();
        return $query->result_array();

这是结果：

(
[0] => Array
    (
        [id] => 1
        [title] => My first blog post!
        [slug] => my-first-blog-post
        [content] => This is my first blog post. Don't worry, it's just a test, my real blog won't be this boring, hopefully!
        [author] => Joni
        [date] => 2012-01-23
        [time] => 00:00:00
        [tags] => Testing
        [status] => 
        [categories] => Testing-More Tests-Test
    )

[1] => Array
    (
        [id] => 2
        [title] => This is another test-post
        [slug] => this-is-another-test-post
        [content] => Well you guessed it. another boring test post, enjoy!
        [author] => Joni
        [date] => 2012-01-23
        [time] => 00:00:00
        [tags] => Sexy
        [status] => 
        [categories] => Test
    )

)

现在，当我修改查询以实现 cmets 的第三个连接时，如下所示：

            $this->db->select('posts.id,
                            posts.title,
                            posts.slug,
                            posts.content,
                            posts.author,
                            posts.date,
                            posts.time,
                            posts.tags,
                            posts.status,
                            GROUP_CONCAT(categories.name SEPARATOR \'-\') AS categories,
                            count(comments.id) as total_comments
                            ');
        $this->db->group_by(array('posts.id'));
        $this->db->from('posts');
        $this->db->join('posts_categories', 'posts_categories.blog_entry_id = posts.id');
        $this->db->join('categories', 'posts_categories.blog_category_id = categories.category_id');
        $this->db->join('comments', 'comments.post_id = posts.id');
        $query = $this->db->get();
        return $query->result_array();

我最终得到了这个

(
[0] => Array
    (
        [id] => 1
        [title] => My first blog post!
        [slug] => my-first-blog-post
        [content] => This is my first blog post. Don't worry, it's just a test, my real blog won't be this boring, hopefully!
        [author] => Joni
        [date] => 2012-01-23
        [time] => 00:00:00
        [tags] => Testing
        [status] => 
        [categories] => Testing-More Tests-Test
        [total_comments] => 3
    )

)

如果你已经做到了这一步，对不起，拖了这么久，只是想提前说声谢谢！

干杯乔尼

Answer 1

您需要使用 LEFT OUTER JOIN 否则您只会收到包含 cmets 的帖子。当您执行 INNER JOIN（默认）时，它将要求左侧的任何内容在联接的右侧都有匹配的元素。如果在右侧找不到匹配项，则将其省略。 LEFT OUTER JOIN 将保留连接左侧的所有元素，无论右侧是否存在匹配项。

改变这个：

$this->db->join('comments', 'comments.post_id = posts.id'); 

到

$this->db->join('comments', 'comments.post_id = posts.id', 'left outer' );

Answer 2

->join 是做什么的？如果它正在进行内部连接，您的问题可能是它会排除没有任何 cmets 的帖子。您需要在那里使用左外连接来确保包含没有 cmets 的帖子。

Answer 3

通过快速查看您的查询，我假设您的第二个帖子没有任何 cmets，并且当您尝试在 cmets 上进行连接时，它不会收到第二个帖子。