Question

我试图加入3个表格,有些是等级内的人加入,并从第3个表中获取数据。我的出发点是,从该条表中删除第1条(156118)。这里是工作说明和表格结构,但必须有一个办法,以正确的方式把所有这些内容结合起来?

//  Get the parent task from the article
select task_parent
from article a, tasks t
where a.task_id = t.task_id
and a.article_number = 156118

// Get the task id for the  Blog  task
select task_id 
from tasks 
where task_parent = 26093 
and task_name like  %blog% 

// Get ALL the blog record
select * 
from blogs
where task_id = 26091

---------Tables------------

* article table *
id | article_number | task_id
1  | 156118         | 26089


* tasks table *
id    | task_name | task_parent
26089 | article   | 26093 
26091 | blogs     | 26093
26093 | Main Task | 26093


* blog table *
id | task_id | content
1  | 102     | blah
2  | 102     | blah 
3  | 102     | blah

-------------

<><>>> 我如何获得所有博客数据,用1份SQl声明只用第_条编号?

提前感谢!

Answer 1

www.un.org/Depts/DGACM/index_spanish.htm 重读后的回答。你需要两人加入任务表。一是让该条的母子承担任务,二是同第条任务相同的母子承担博客任务。

select b.id, b.task_id, b.content
    from article a
        inner join tasks t1
            on a.task_id = t1.task_id
        inner join tasks t2
            on t1.task_parent = t2.task_parent
                and t2.task_name like  %blog% 
        inner join blogs b
            on t2.task_id = b.task_id
    where a.article_number = 156118

Answer 2

Looks like you re wanting to tie them all together and just use the article number as the parameter... Try:


select b.* 
from blogs b, tasks t, tasks tp, article a  
where b.task_id = t.task_id 
and t.task_parent = tp.task_id 
and tp.task_id = a.task_id 
and a.article_number = 156118

Answer 3

Here you go.

SELECT * FROM a
INNER JOIN tasks USING (a.article_number)
INNER JOIN blogs USING (a.article_number)
WHERE a.article_number = 156118;

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