Question

如何在codeigniter中编写以下MySQL查询。我测试了查询，它按预期工作。 select p.post_title, count(d._id) from posts p INNER JOIN discussions d ON p._id = d.post_id INNER JOIN accounts a ON a._id = p.account_id where a._id = '1494900911hRs5kjPXV9591a60afa434f' group by p._id

Answer 1

当您看到文档here时，您可以使用查询构建器类来构建查询。所以你的查询构建器将如下所示：

$this->db->select('p.post_title, count(d._id)');
$this->db->from('post p');
$this->db->join('discussions d', 'p._id = d.post_id', 'inner');
$this->db->join('accounts a', 'a._id = p.account_id', 'inner');
$this->db->where('a._id', '1494900911hRs5kjPXV9591a60afa434f');
$this->db->group_by('p._id');
$query = $this->db->get();

然后您可以看到文档here到生成查询结果

// get result.
$rows = $query->result();

Answer 2

$this->db->select('p.post_title, count(d._id) as count')->from('post p')->join('discussions d', 'p._id = d.post_id', 'inner')->join('accounts a', 'a._id = p.account_id', 'inner')->where('a._id', '1494900911hRs5kjPXV9591a60afa434f')->group_by('p._id');
$query = $this->db->get();

获取结果数据

$rows = $query->result(); // object format
$rows = $query->result('array'); // Array Format

如何在Codeigniter查询中编写连接选择查询

2 个答案: