SQL：列出用户及其电话号码和电子邮件地址

Question

我想在PostgreSQL中查询，列出我的用户及其电子邮件地址和电话号码（用逗号分隔），如下所示：

| user1 | email1@mail.com, email2@mail.com | +3612123123, +3623234234 |

这些表是：

user - stores the user's data

user_email - stores the user's email addresses

user_phone - stores the user's phone numbers

我尝试了一下：

SELECT user.id, user.name
(
  SELECT array_agg(user_email.email)
  FROM user_email
  WHERE user_email.user_id = user.id
) AS EmailAddresses,
(
  SELECT array_agg(user_phone.phone)
  FROM user_phone
  WHERE user_phone.user_id = user.id
) AS PhoneNumbers
FROM user
ORDER BY user.id

但这会导致荒谬的查询时间（34秒）。 比我尝试过的：

SELECT user.id, user.name, array_agg(user_email.email), array_agg(user_phone.phone)
FROM user
LEFT JOIN user_email ON user_email.user_id = user.id
LEFT JOIN user_phone ON user_phone.user_id = user.id
GROUP BY user.id
ORDER BY user.id

这将导致非常好的查询时间（大约100毫秒）。 但是这样可以列出电话号码和电子邮件地址的所有组合。 因此，如果我有3个电子邮件地址和3个电话号码，那么它将列出9个电子邮件地址（每个地址三倍）和9个电话号码。

有一种有效的方式来做我想要的吗？

Answer 1

DISTINCT也可以在聚合表达式中使用：

SELECT "user".id, name, array_agg(DISTINCT email) emails, array_agg(DISTINCT phone) phones
FROM "user"
LEFT JOIN user_email ON user_email.user_id = "user".id
LEFT JOIN user_phone ON user_phone.user_id = "user".id
GROUP BY "user".id
ORDER BY "user".id;

注意：如果只需要逗号分隔的列表，则可能要使用string_agg()而不是array_agg() 。

SQLFiddle

Answer 2

要么DISTINCT

SELECT DISTINCT user.id, user.name, array_agg(user_email.email), array_agg(user_phone.phone)
FROM user
LEFT JOIN user_email ON user_email.user_id = user.id
LEFT JOIN user_phone ON user_phone.user_id = user.id
GROUP BY user.id
ORDER BY user.id

或INNER加盟

SELECT user.id, user.name, array_agg(user_email.email), array_agg(user_phone.phone)
FROM user
INNER JOIN user_email ON user_email.user_id = user.id
INNER JOIN user_phone ON user_phone.user_id = user.id
GROUP BY user.id
ORDER BY user.id

或两者

SELECT DISTINCT user.id, user.name, array_agg(user_email.email), array_agg(user_phone.phone)
FROM user
INNER JOIN user_email ON user_email.user_id = user.id
INNER JOIN user_phone ON user_phone.user_id = user.id
GROUP BY user.id
ORDER BY user.id