Question

我正在寻找查询结果，我只能看到表2中没有的表1数据;这是我的表定义和数据;

表1

id  name    father name age
1      a    a father    60
2      b    b father    70
3      c    c father    60
4      d    d father    50
5      e    e father    20
6      f    f father    32
7      g    g father    40

表2

id  account_amount
1   42
3   90
5   80
7   49

现在我希望表1中的所有记录在表2中不可用或表2中的对应account_amount小于50.这里将是所需的输出

id  name    father name age
1      a    a father    60
2      b    b father    70
4      d    d father    50
6      f    f father    32
7      g    g father    40

提前感谢codeigniter中的解决方案查询

Answer 1

使用反连接：

SELECT t1.id, t1.name, t1.father_name, t1.age
FROM table1 t1
LEFT JOIN table2 t2
    ON t1.id = t2.id AND t2.account_amount >= 50
WHERE t2.id IS NULL

不存在：

SELECT t1.id, t1.name, t1.father_name, t1.age
FROM table1 t1
WHERE NOT EXISTS
    (SELECT 1 FROM table2 t2 WHERE t2.id = t1.id AND t2.account_amount >= 50)

由于您使用MySQL标记了问题，因此第一个选项会提高性能。

在模型中

$this->db->query("SELECT t1.id, t1.name, t1.father_name, t1.age
       FROM table1 t1
       LEFT JOIN table2 t2
       ON t1.id = t2.id AND t2.account_amount >= 50
       WHERE t2.id IS NULL");

Imo Active Record对此类查询不方便

Answer 2

你在这里：

$this->db->select('id');
$excludedData = $this->db->get('table2')->result_array();
foreach($excludedData as $key => $record){
    $excludedData[$key] = $record['id'];
}
$this->db->where_not_in('id',$excludedData);
$desiredData = $this->db->get('table1')->result_array();

拉取另一个表codeigniter中没有的记录

2 个答案: