MySQL查询不适用于我Codeigniter Php

Question

我正在使用带有Codeigniter的mysql，这是我的表结构：

表A：

id(doctorid)            name    
-----------------------------
1                       abc
2                       xyz
3                       ahd
4                       djd

表B：

id          doctor_id       type
-------------------------------------
1           1               Nuter
2           3               Spay

现在，我想获取所有具有类型计数的医生记录，我想要以下结果：

id          name        Nuter     Spay
---------------------------------------
1           abc         1         0
2           xyz         0         Spay

我尝试使用以下代码，但对我不起作用，该怎么办？

$this->db->select('A.*');
$this->db->from('A');
$this->db->join('B', 'B.doctor_id = (SELECT COUNT(Type) FROM B )');
$query = $this->db->get();

Answer 1

试试这个代码：

$this->db->select('A.*,B.type');
$this->db->from('A');
$this->db->join('B', 'B.doctor_id = A.id');
$query = $this->db->get();

Answer 2

似乎您正在寻找条件聚合。

在纯SQL中，查询应类似于：

SELECT
    A.id,
    A.name,
    SUM(CASE WHEN B.type = 'Nuter' THEN 1 ELSE 0 END) as Nuter,
    SUM(CASE WHEN B.type = 'Spay' THEN 1 ELSE 0 END) as Spay
FROM A
INNER JOIN B ON B.doctor_id = A.id
GROUP BY A.id, A.name

如果表A和B之间的关系是1-1，而不是1-N（如示例数据所示），则可以简化为：

SELECT
    A.id,
    A.name,
    CASE WHEN B.type = 'Nuter' THEN 1 ELSE 0 END as Nuter,
    CASE WHEN B.type = 'Spay' THEN 1 ELSE 0 END as Spay
FROM A
INNER JOIN B ON B.doctor_id = A.id

尝试：

$this->db->select(
    "A.id,
    A.name,
    CASE WHEN B.type = 'Nuter' THEN 1 ELSE 0 END as Nuter,
    CASE WHEN B.type = 'Spay' THEN 1 ELSE 0 END as Spay"
);
$this->db->from('A');
$this->db->join('B', 'B.doctor_id = A.id');
$query = $this->db->get();