Question

我有以下查询主要工作，但当其中一个连接表返回的结果数不同时，在group_concat()内返回的结果太多：

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（请运行代码段以查看表格）或查看 fiddle

＆＃13;

<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>

＆＃13;

如您所见，例如offersB包含两个结果

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

两者相同，数据库中只有一个条目用于给定的sku 123，但offersA对此sku的不同数量有两个不同的要约：

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

我之后使用JavaScript来处理结果，所以我可以删除重复的结果 - 但我想知道是否有

a）一种更聪明的查询数据的方法

b）在查询本身中删除这些重复项的方法

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);

选中此SQL FIDDLE DEMO

<强> :: OUTPUT ::

| 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

这是一种稍微轻松的方式来做你需要的。使用GROUP_CONCAT(DISTINCT...)。

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

我称之为肮脏，因为它会不正确地消除数据中真正存在的重复项。 http://sqlfiddle.com/#!9/2481e/6/0

将group_concat的结果集限制为实际结果

2 个答案: