Question

替换多个字符串并在单个列中返回替换操作的结果的最佳方法是什么。

以下查询完美无缺，但它显然生成了5列。

SELECT 
REPLACE(l.status, 'New Onboarding Request', '1. New Lead'),
REPLACE(l.status, 'Appointment Accepted', '2. Appointment Accepted'),
REPLACE(l.status, 'Accepted', '3. Property Eligible'),
REPLACE(l.status, 'Owner Accepted', '4. Terms Accepted'),
REPLACE(l.status, 'Onboarding', '5. Onboarding'), 
l.id, o.name as partner, v.lat as latitude, v.lng as longitude, v.postcode, 
l.status, v.bedrooms, v.bathrooms, "v"."houseOrFlat", to_char("v"."created_at", 'YYYYMMDD') as valuation_date, to_char("l"."created_at", 'YYYYMMDD') as referral_date, to_char("l"."updated_at", 'YYYYMMDD') as updated,  v.estimated_nightly_rate, v.furnished   FROM valuations v
LEFT JOIN agent_onboarding_requests l on v.id = l.valuation_id
INNER JOIN organisations o on o.id = l.organisation_id
WHERE l.organisation_id != 1
ORDER BY referral_date ASC

Answer 1

最后，这让我觉得它比嵌套的REPLACE

更优雅

SELECT 

    CASE 
        WHEN(l.status = 'New Onboarding Request') THEN '1. New Lead'
        WHEN(l.status = 'Appointment Accepted') THEN '2. Appointment Accepted'
        WHEN(l.status = 'Accepted') THEN '3. Property Eligible'
        WHEN(l.status = 'Owner Accepted') THEN '4. Terms Accepted'
        WHEN(l.status = 'Onboarding') THEN '5. Onboarding'
    END AS status_modified,

Answer 2

您还可以使用VALUES创建内联表，并将其加入现有表：

SELECT
    s.status old_status,
    n.new_status
FROM
  statuses s
JOIN
  (VALUES ('New Onboarding Request', '1. New Lead'),
          ('Appointment Accepted','2. Appointment Accepted'),
          ('Accepted', '3. Property Eligible'),
          ('Owner Accepted', '4. Terms Accepted'),
          ('Onboarding', '5. Onboarding')) n(old_status, new_status)
  ON s.status = n.old_status

或使用CTE（WITH）：

WITH n(old_status, new_status) AS
(VALUES ('New Onboarding Request', '1. New Lead'),
        ('Appointment Accepted','2. Appointment Accepted'),
        ('Accepted', '3. Property Eligible'),
        ('Owner Accepted', '4. Terms Accepted'),
        ('Onboarding', '5. Onboarding'))
SELECT
    s.status old_status,
    n.new_status
FROM
  statuses s
JOIN n ON s.status = n.old_status

PostgreSQL SELECT和REPLACE多个字符串成一个列

2 个答案: