Question

是否有人知道如何编写查询以使所需结果与此数据库中的工具/升降计划/ cranimax相似

resource_category_id    category_name   parent_category     
    1                            Tools                  0    
    2                    product literature             0   
    3                   Terms and Conitions             0   
    4                        crane library              1   
    5                         geniune                   1   
    6                       lift planning               1   
    8                         cranimax                  6

为此，我写了一个这样的查询，但是没有完成，例如，如果我从cranimax中选择它有一个parent_category 6，那么这就是我的想法，所以它应该与那行有resource_category_id，那行有parent_category 1，它应该重复，直到父类别0出现。

$data['breadcrumbs'] = $this->Manito_model->get_breadcrumbs_details($resource_id);

public function get_breadcrumbs_details($resource_id)
{
    $query=$this->db->query("select * from resource_category as m WHERE m.resource_category_id = $resource_id union (select * from resource_category where parent_category != 0) as m2 on m.parent_category = m2.resource_parent_category"); 


    return $query->result();


}

我希望我的结果像这样

array (size=3)
 0 => 
   object(stdClass)[35]
    public 'resource_category_id' => string '8' (length=1)
    public 'category_name' => string 'cranimax' (length=8)
    public 'parent_category' => string '6' (length=1)
    public 'created_at' => string '2017-11-04 13:59:39' (length=19)
 1 => 
   object(stdClass)[35]
    public 'resource_category_id' => string '6' (length=1)
    public 'category_name' => string 'lift planning' (length=8)
    public 'parent_category' => string '1' (length=1)
    public 'created_at' => string '2017-11-04 13:59:39' (length=19)

 2 => 
   object(stdClass)[35]
    public 'resource_category_id' => string '1' (length=1)
    public 'category_name' => string 'Tools' (length=8)
    public 'parent_category' => string '0' (length=1)
    public 'created_at' => string '2017-11-04 13:59:39' (length=19)

Answer 1

UNION用于将多个SELECT语句的结果合并到一个结果集中（即使它变得非常简单：你需要做两个选择，但你只需要一个结果，然后你就建立一个联合）在你的情况下，你不需要UNION，而是一个简单的JOIN，用其他数据“完成”你的结果：

select m.*, m2.category_name AS parent_name FROM resource_category AS m LEFT JOIN resource_category m2 ON m2.resource_category_id = m.resource_parent_category WHERE m.resource_category_id = $resource_id

查询联合条件

1 个答案: