将 group_concat 的结果集限制为实际结果

Question

我有以下查询，它大部分都有效，但是 returns 在 group_concat() 中的结果太多，而其中一个加入的 table 返回了不同数量的结果：

select 
    a.sku, a.ek, a.mwst, 
    concat('[[', group_concat('{"offer": ', b.offer, ', "minQuantity": ', b.minQuantity, '}') , ']]') offersA,
    concat('[[', group_concat('{"offer": ', c.offer, ', "minQuantity": ', c.minQuantity, '}') , ']]') offersB,
    concat('[[', group_concat('{"offer": ', d.offer, ', "minQuantity": ', d.minQuantity, '}') , ']]') offersC
from all_prices a 
    left join all_prices_a b on a.sku = b.sku 
    left join all_prices_b c on a.sku = c.sku 
    left join all_prices_c d on a.sku = d.sku
where a.sku in (123,456) 
group by a.sku

我得到的结果是（请运行看片断table）或者看fiddle

<table border=1>
<tr>
<td bgcolor=silver class='medium'>sku</td>
<td bgcolor=silver class='medium'>ek</td>
<td bgcolor=silver class='medium'>mwst</td>
<td bgcolor=silver class='medium'>offersA</td>
<td bgcolor=silver class='medium'>offersB</td>
<td bgcolor=silver class='medium'>offersC</td>
</tr>

<tr>
<td class='normal' valign='top'>123</td>
<td class='normal' valign='top'>154.32</td>
<td class='normal' valign='top'>19</td>
<td class='normal' valign='top'>[[{&quot;offer&quot;: 9.65, &quot;minQuantity&quot;: 3},{&quot;offer&quot;: 9.86, &quot;minQuantity&quot;: 1}]]</td>
<td class='normal' valign='top'>[[{&quot;offer&quot;: 9.66, &quot;minQuantity&quot;: 1},{&quot;offer&quot;: 9.66, &quot;minQuantity&quot;: 1}]]</td>
<td class='normal' valign='top'>[[{&quot;offer&quot;: 9.65, &quot;minQuantity&quot;: 1},{&quot;offer&quot;: 9.65, &quot;minQuantity&quot;: 1}]]</td>
</tr>

<tr>
<td class='normal' valign='top'>456</td>
<td class='normal' valign='top'>48.48</td>
<td class='normal' valign='top'>19</td>
<td class='normal' valign='top'>[[{&quot;offer&quot;: 13.30, &quot;minQuantity&quot;: 1},{&quot;offer&quot;: 13.30, &quot;minQuantity&quot;: 1}]]</td>
<td class='normal' valign='top'>[[{&quot;offer&quot;: 13.30, &quot;minQuantity&quot;: 1},{&quot;offer&quot;: 122.00, &quot;minQuantity&quot;: 3}]]</td>
<td class='normal' valign='top'>NULL</td>
</tr>
</table>

如您所见，例如 offersB 有两个结果

[[{"offer": 9.66, "minQuantity": 1},{"offer": 9.66, "minQuantity": 1}]]

两者相等，给定 sku 123 在数据库中只有一个条目，但 offersA 有两个不同数量的不同报价：

[[{"offer": 9.65, "minQuantity": 3},{"offer": 9.86, "minQuantity": 1}]]

我正在使用 JavaScript 稍后处理结果，所以我可以删除重复的结果 - 但我想知道是否有

a) 一种更聪明的数据查询方式

b) 一种删除查询本身中重复项的方法

Answer 1

试试这个：

SELECT a.sku, a.ek, a.mwst, 
      CONCAT('[[', b.offersA , ']]') offersA,
      CONCAT('[[', c.offersB , ']]') offersB,
      CONCAT('[[', d.offersC , ']]') offersC
FROM all_prices a 
LEFT JOIN ( SELECT b.sku, GROUP_CONCAT('{"offer": ', b.offer, ', "minQuantity": ', b.minQuantity, '}') AS offersA
            FROM all_prices_a b 
            GROUP BY b.sku
          ) AS b ON a.sku = b.sku 
LEFT JOIN ( SELECT c.sku, GROUP_CONCAT('{"offer": ', c.offer, ', "minQuantity": ', c.minQuantity, '}') AS offersB
            FROM all_prices_b c 
            GROUP BY c.sku
          ) AS c ON a.sku = c.sku 
LEFT JOIN ( SELECT d.sku, GROUP_CONCAT('{"offer": ', d.offer, ', "minQuantity": ', d.minQuantity, '}') AS offersC
            FROM all_prices_c d 
            GROUP BY d.sku
          ) AS d ON a.sku = d.sku 
WHERE a.sku IN (123, 456);

勾选这个SQL FIDDLE DEMO

::输出::

| sku |    ek | mwst |                                                                   offersA |                                offersB |                                offersC |
|-----|-------|------|---------------------------------------------------------------------------|----------------------------------------|----------------------------------------|
| 123 | 12.48 |   19 | [[{"offer": 12.28, "minQuantity": 1},{"offer": 11.24, "minQuantity": 3}]] | [[{"offer": 12.28, "minQuantity": 1}]] | [[{"offer": 12.28, "minQuantity": 1}]] |
| 456 | 13.24 |   19 |  [[{"offer": 10.00, "minQuantity": 1},{"offer": 9.00, "minQuantity": 3}]] |  [[{"offer": 9.00, "minQuantity": 3}]] |  [[{"offer": 9.00, "minQuantity": 3}]] |

Answer 2

这里有一个稍微简单的方法来完成您需要的事情。使用 GROUP_CONCAT(DISTINCT...).

select 
  a.sku, a.ek, a.mwst, 
  concat('[[', group_concat(DISTINCT '{"offer": ', b.offer, ', "minQuantity": ', b.minQuantity, '}') , ']]') offersA,
...

我称它为 sleazy，因为它会不正确地消除数据中确实存在的重复项。 http://sqlfiddle.com/#!9/2481e/6/0

将 group_concat 的结果集限制为实际结果

limit group_concat's resultset to actual results

mysql

sql

select

join

group-concat