Question

SELECT  distinct referrals.listing_id,submissions.listing_name
FROM submissions
INNER JOIN referrals ON referrals.listing_id=submissions.listing_id;

如果我这样做

SELECT  distinct referrals.listing_id,submissions.listing_name,referrals.timestamp
FROM submissions
INNER JOIN referrals ON referrals.listing_id=submissions.listing_id;

它为我提供了许多其他没有distinct listing_id

的记录

我希望distinct listing_id有他们的时间戳（这些时间戳不同，但是根据distinct listing_id）

Answer 1

如果您希望每个列表包含一行，请使用聚合。例如：

GROUP BY
    Treatment.Treatment_Date, 
    Salutation.Value,
    Customer.Firstname,
    Customer.Lastname,
    Address.Value, 
    Postal.Postal, 
    Postal.City, 
    Patient.Name,
    TreatmentType.Type,
    Treatment.Price,
    Tax.Value

SELECT r.listing_id, s.listing_name, max(r.timestamp) as most_recent_timestamp FROM submissions s INNER JOIN referrals r ON r.listing_id = s.listing_id GROUP BY r.listing_id, s.listing_name;适用于SELECT DISTINCT。

中的所有表达式

我希望在连接mysql时有明显和非独特的记录

1 个答案: