sql 取最新一條記錄

1. 選出某個條件最新的一條記錄

// 選出最新狀態下的每乙個單號對應的子單資料

select
distinct
a.receiving_code,a.product_barcode,a.rd_putaway_qty
from
odoo_ykd_oversea_shipping_information a 
inner
join odoo_ykd_receiving_order_manage b on a.receiving_code=b.receiving_code and a.data_version=
b.data_version 
inner
join odoo_sync_control_scrapy c on b.data_version=
c.data_version 
inner
join
(select
max(a.odoo_create_time) as odoo_create_time, a.receiving_code 
from odoo_ykd_receiving_order_manage a 
inner
join odoo_sync_control_scrapy b on b.data_version=a.data_version 
where a.odoo_create_time <
'2020-11-19
'group
bya.receiving_code) 
as d 
on d.receiving_code=b.receiving_code and d.odoo_create_time=
b.odoo_create_time
where c.task_type=
1and c.task_status=
2and c.result_type=
1and c.record_status=
1and c.data_date <=
'2020-11-18
'and c.data_date >
'2020-08-10
'// 或者用這種方式，沒寫完整，自己補充
with res as(
select id,consignment_no, row_number()over(partition by b.consignment_no  order by b.odoo_create_time desc) as row_n  from odoo_dsf_inbound_order_info)

2. 對上面sql的簡化解釋

1. 

select a.receiving_code,a.product_code 
from
table_a 
inner join 
--找出receiving_code維度的最大的create_time（如果可以確定id大的就是最大的話，用id更好），
( select max(c.create_time),c.receiving_code 
from table_c where 條件1 group by receiving_code) b on a.receiving_code=b.receiving_code and a.create_time=b.create_time
where 條件
2. 分組排序，去除排序之後的序號為1的（mysql不支援分組排序函式）

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