SQL查询最近7天基于日期的前3个售罄产品

Question

我希望上周销售量排名前三的产品。

这是我的SQL查询。

select   ProductId,sum(Quantity) as quantity,createdOn from

(SELECT  inv.Id,invd.ProductId, invd.Quantity  ,cast (inv.CreatedOn as date) as createdOn FROM Invoice as inv 
INNER JOIN 
InvoiceDetail invd
ON
invd.InvoiceId = inv.id
WHERE inv.CreatedOn  >= DATEADD(day,-11117, GETDATE()) ) as tbl 
group by createdOn , ProductId 
ORDER BY createdOn DESC 

但是我没有获得每个日期的前三名产品。如果我使用TOP 3，那么它只会给出前3种产品，而我希望上周每天都可以获得前3种产品。

这是我的输出。 但是我每天只需要3条记录。

期望的输出：

Answer 1

如果我理解正确，您可以将<div class="grid__item medium-up--one-fifth"> <span class="filter-label">Brands:</span> {% assign tags = settings.brands_filter | escape | split: ',' %} <select class="coll-filter"> <option value="" class="default-tag">All</option> {% for tag in tags %} {% if current_tags contains tag %} <option value="{{ tag | handle }}" selected>{{ tag }}</option> {% elsif collection.all_tags contains tag %} <option value="{{ tag | handle }}">{{ tag }}</option> {% endif %} {% endfor %} </select> </div> <script> var allFilters = jQuery('.coll-filter, .coll-picker'); allFilters.change(function() { var newTags = []; jQuery('.coll-filter').each(function() { if (jQuery(this).val()) { newTags.push(jQuery(this).val()); } }); if (newTags.length) { var query = newTags.join('+'); if (jQuery('.coll-picker').length) { window.location.href = '/collections/' + jQuery('.coll-picker').val() + '/' + query; } else { window.location.href = jQuery('{{ '' | link_to_tag: '' }}').attr('href') + query; } } else { if (jQuery('.coll-picker').length) { window.location.href = '/collections/' + jQuery('.coll-picker').val(); } else { {% if collection.handle %} window.location.href = '/collections/{{ collection.handle }}'; {% elsif collection.products.first.type == collection.title %} window.location.href = '{{ collection.title | url_for_type }}'; {% elsif collection.products.first.vendor == collection.title %} window.location.href = '{{ collection.title | url_for_vendor }}'; {% endif %} } } }); </script> 与windows function结合使用，以每天获得前三名Row_number。

根据quantity列从高到低的顺序，每天按createdon行号。

quantity

sqlfiddle

[结果] ：

   ;WITH CTE AS(
    SELECT productid,quantity,createdon,Row_number() over(partition by createdon ORDER BY quantity DESC,productid DESC) as RN
    FROM 
    ( 
        SELECT    invd.productid, 
                  sum(invd.quantity) as quantity, 
                  cast(inv.createdon AS date) AS createdon
        FROM      invoice  AS inv INNER JOIN invoicedetail invd
        ON         invd.invoiceid = inv.id 
        WHERE      inv.createdon >= dateadd(day,-11117, getdate()) 
        GROUP BY cast(inv.createdon AS date), invd.productid
    ) AS tbl 
)
SELECT * 
FROM CTE
WHERE RN <= 3