我想SELECT基于ID值的字段。
Products
PRODUCT_ID Name
19 Chair
20 Table
Product_fields
ID PRODUCT_ID TYPE DESCRIPTION
1 19 C White
2 19 S Modern
3 20 C Black
4 20 S Classic
我需要一个结果：
Product Type_C Type_S
Chair White Modern
Table Black Classic
我能够在product_fields表上使用两个LEFT JOINs生成此内容，但这会使查询速度过慢。还有更好的方法吗？
答案 0 :(得分：0)
减慢查询的速度是多少？什么是可以接受的？
如果您真的不想使用联接（您必须有一个联接），请使用视图或嵌套查询。但我不认为它们会更快，不过你可以尝试一下。
上的观看次数select p.PRODUCT_ID, p.Name, f.CDescription, f.SDescription
from Products p
join(
SELECT PRODUCT_ID, Max( CDescription ) as CDescription,
Max( SDescription ) as SDescription
FROM(
select PRODUCT_ID,
case Type when 'C' then Description end as CDescription,
case Type when 'S' then Description end as SDescription
from Fields
) x
group by PRODUCT_ID
) f
on f.PRODUCT_ID = p.PRODUCT_ID;
答案 1 :(得分：0)
完整的陈述是：
SELECT
NL.product_name,
PRD.product_sku AS product_sku,
CF.virtuemart_product_id AS virtuemart_product_id,
GROUP_CONCAT(distinct CFA.customsforall_value_name
ORDER BY CFA.customsforall_value_name ASC
separator ' | ' ) AS Name_exp_3,
ROUND((((prices.product_price * CALC.calc_value) / 100) + prices.product_price),
2) AS Prijs,
VMCF_L.custom_value AS latijn,
VMCF_T.custom_value AS THT
VMCF_B.custom_value AS Batch
from j25_virtuemart_products AS PRD
LEFT join j25_virtuemart_product_custom_plg_customsforall AS CF ON CF.virtuemart_product_id = PRD.virtuemart_product_id
join j25_virtuemart_product_prices AS prices ON PRD.virtuemart_product_id = prices.virtuemart_product_id
join j25_virtuemart_calcs AS CALC ON prices.product_tax_id = CALC.virtuemart_calc_id
join j25_virtuemart_products_nl_nl AS NL ON NL.virtuemart_product_id = PRD.virtuemart_product_id
LEFT join j25_virtuemart_product_customfields AS VMCF ON VMCF.virtuemart_product_id = PRD.virtuemart_product_id
LEFT join j25_virtuemart_custom_plg_customsforall_values AS CFA ON CFA.customsforall_value_id = CF.customsforall_value_id
LEFT JOIN j25_virtuemart_product_customfields AS VMCF_L ON VMCF.virtuemart_product_id = VMCF_L.virtuemart_product_id AND VMCF_L.virtuemart_custom_id = 16
LEFT JOIN j25_virtuemart_product_customfields AS VMCF_T ON VMCF.virtuemart_product_id = VMCF_T.virtuemart_product_id AND VMCF_T.virtuemart_custom_id = 3
LEFT JOIN j25_virtuemart_product_customfields AS VMCF_B ON VMCF.virtuemart_product_id = VMCF_B.virtuemart_product_id AND VMCF_B.virtuemart_custom_id = 18
WHERE
PRD.product_sku like '02.%'
group by PRD.virtuemart_product_id
order by NL.product_name;
其中名为'Latijn'，'THT'和'Batch'的三个SELECT结果是我之前比较为黑/白和经典/现代值的结果。 希望这有任何意义。 正如您所看到的，这涉及到Virtuemart的安装，因此我无法对模式进行多少讨论。
当我排除底部的3个JOINS并且存在相关的FIELDS时，查询大约需要0.5秒。包含JOINS和FIELDS后，查询大约需要19秒。
我从这个完整的查询中创建了一个视图，我从我的标签应用程序中查询。
答案 2 :(得分：0)
谢谢大家！根据您的输入我创建：
select
NL .这里给AS这里给,
PRD . product_sku AS product_sku ,
CF . virtuemart_product_id {{ 1}} virtuemart_product_id AS CFA ,
group_concat(distinct customsforall_value_name . CFA
order by customsforall_value_name . Name_exp_3 ASC
separator ' | ') AS价格,
round(((( PRODUCT_PRICE {{1} } CALC . calc_value *价格. PRODUCT_PRICE ) / 100) + Prijs . F到),
2) AS Latijn ,
Latijn . F到AS THT ,
THT . F到AS批次,
批次. j25_virtuemart_products AS PRD
from
((((((( j25_virtuemart_product_custom_plg_customsforall {{ 1}} CF CF
left join virtuemart_product_id PRD ON (( virtuemart_product_id . j25_virtuemart_product_prices =价格. PRD {{1} } virtuemart_product_id )))
join价格 virtuemart_product_id ON (( j25_virtuemart_calcs . CALC =价格. product_tax_id )))
join CALC virtuemart_calc_id ON (( j25_virtuemart_products_nl_nl . NL = NL . virtuemart_product_id )))
join PRD virtuemart_product_id ON (( j25_virtuemart_product_customfields . VMCF {{ 1}} {VMCF {1}} {virtuemart_product_id {1}} {PRD {1}} virtuemart_product_id {{1 }} j25_virtuemart_custom_plg_customsforall_values = CFA . CFA )))
left join customsforall_value_id CF ON (( customsforall_value_id . vw_batch_Latijn_THT_grouped = F到. ˚F)))
left join virtuemart_product_id PRD ON (( virtuemart_product_id . PRD = product_sku . PRD )))
left join virtuemart_product_id NL { {1}} product_name``
执行需要1.4秒，比我开始的19秒快一点。