Question

我有以下表格：

projects: id, name, language
          1 |test | php
          2 |test | java
          3 |hello| php


attrs:  id, name,      value,   tb1_id
        1  | status    | finish | 1
        2  | reference | 2      | 1

我需要一个查询来选择所有项目，这些项目具有另一个项目的“引用”属性并且“状态”完成。

示例输出：

id, name      
1 |test

你能帮帮我吗？

Answer 1

SELECT P.id, P.name
FROM projects P inner join attrs A on P.id = a.tb1_id
WHERE (A.name = 'status' AND value = 'finish')
UNION
SELECT P.id, P.name
FROM attrs A INNER JOIN projects P ON A.value = P.id
WHERE A.name='reference'

查看this fiddle

中的结果

Answer 2

你可以：

对连接的表进行分组，在HAVING子句中使用合适的聚合函数过滤这些组：

SELECT   projects.id, projects.name
FROM     projects
    JOIN attrs ON attrs.tb1_id = projects.id
GROUP BY projects.id
HAVING   SUM(attrs.name='reference')
     AND SUM(attrs.name='status' AND attrs.value='finish')

或多次加入attrs：

SELECT   p.id, p.name
FROM     projects AS p
    JOIN attrs    AS r ON  r.tb1_id = p.id
                       AND r.name   = 'reference'
    JOIN attrs    AS s ON  s.tb1_id = p.id
                       AND s.name   = 'status'
WHERE    s.value = 'finish'

Answer 3

试试这个：

SELECT p.id, p.name
FROM projects p 
INNER JOIN (SELECT tb1_id, MAX(IF(a.name = 'status', a.value, '')) attrStatus, MAX(IF(a.name = 'reference', a.value, '')) attrReference
                FROM attrs a GROUP BY tb1_id) A ON p.id = A.tb1_id AND A.attrStatus = 'finish' 
WHERE EXISTS (SELECT 1 FROM projects p1 WHERE A.attrReference = p1.id)

检查SQL FIDDLE DEMO

<强>输出

| ID | NAME |
|----|------|
|  1 | test |

Answer 4

您需要加入attrs两次，每个属性一次;

SELECT p.*, ref_attr.value reference FROM projects p
JOIN attrs ref_attr  ON ref_attr.tb1_id = p.id
                    AND ref_attr.name = 'reference'
JOIN attrs stat_attr ON stat_attr.tb1_id = p.id
                    AND stat_attr.name = 'status'
                    AND stat_attr.value = 'finish'

An SQLfiddle to test with

SQL Query选择多行

4 个答案: