lemon

这是一种正确的迭代方式,就像你正在尝试的那样:

a = [[[0, 0], [0, 0]], [[1, 0], [0, 0]], [[2, 0], [0, 0]] ]
for i in range(len(a)):
    for z in range(len(a[i])):
        print((a[i])[z])

如果对您来说更简单,请尝试迭代列表中的元素,而不是使用索引来迭代列表的长度:

a = [[[0, 0], [0, 0]], [[1, 0], [0, 0]], [[2, 0], [0, 0]] ]
for a1 in a:
    for a2 in a1:
        print(a2)

尝试使用以下方法:

l = re.findall('\d{3}_S_\d{4}', ''.join(all_files))

你可以先计算 COUNT 分别购买和销售,然后离开 LEFT JOIN 作为使查询更高效的最后一个操作。

SELECT 
    ProductEAN.EAN                          AS EAN,
    COALESCE(Purchases.num_purchases, 0)    AS num_purchases,
    COALESCE(Sales.num_sales, 0)            AS num_sales
FROM 
    ProductEAN
LEFT JOIN 
    (
    SELECT 
        EAN,
        COUNT(EAN) AS num_purchases
    FROM 
        Purchase
    GROUP BY 
        EAN
    ) Purchases
ON 
    ProductEAN.EAN = Purchases.EAN
LEFT JOIN 
    (
    SELECT 
        EAN,
        COUNT(EAN) AS num_sales
    FROM 
        Sale
    GROUP BY 
        EAN
    ) Sales
ON 
    ProductEAN.EAN = Sales.EAN

因此,您可以将此代码存储在视图中。

请参阅此小提琴进行测试: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=b2d777f367960cf90d6539980c2ad935 .