我有一个1:n的关系:
Rule 1:n Examples
为简化起见,您可以假设Rule和Examples仅包含一个字符串属性。我正在寻找一个sql语句,我得到一个包含以下列的表:
rule; example_1; example_2; example_3
我不关心4和更高的例子......
答案 0 :(得分:0)
我觉得有一个更好的方法可以做到这一点,我忽略了,但这似乎工作得很好。我自动加入了有序的examples
表,其中每个连续的示例值必须大于前一个值(因此产生前三个中的每一个)。
SELECT
r.rule,
e1.example AS example_1,
e2.example AS example_2,
e3.example AS example_3
FROM rules r
LEFT JOIN (SELECT * FROM examples e ORDER BY example) e1
ON e1.ruleID = r.ruleID
LEFT JOIN (SELECT * FROM examples e ORDER BY example) e2
ON e2.ruleID = r.ruleID
AND e2.example > e1.example
LEFT JOIN (SELECT * FROM examples e ORDER BY example) e3
ON e3.ruleID = r.ruleID
AND e3.example > e2.example
GROUP BY r.ruleID
结果:
| RULE | EXAMPLE_1 | EXAMPLE_2 | EXAMPLE_3 |
|-------|-----------|-----------|-----------|
| rule1 | ex1 | ex2 | ex3 |
| rule2 | ex5 | ex6 | ex7 |
| rule3 | ex8 | (null) | (null) |
| rule4 | (null) | (null) | (null) |
示例数据:
/* rules table */
| RULEID | RULE |
|--------|-------|
| 1 | rule1 |
| 2 | rule2 |
| 3 | rule3 |
| 4 | rule4 |
/* examples table */
| EXAMPLE | RULEID |
|---------|--------|
| ex1 | 1 |
| ex2 | 1 |
| ex3 | 1 |
| ex4 | 1 |
| ex5 | 2 |
| ex6 | 2 |
| ex7 | 2 |
| ex8 | 3 |
这是一个有效的SQL Fiddle example。