Question

我有以下表格

,--------------------------------------------,
| contacts                                   |
|------------,---------------,---------------|
| contact_id | contact_score | contact_email |
|------------|---------------|---------------|
| 1          | 53            | email1@xx.com |
| 2          | 53            | email2@xx.com |
| 3          | 4             | email3@xx.com |
'------------'---------------'---------------'

,------------------------,
| reputation_email_score |
|---------------,--------,
| email         | score  |
|---------------|--------|
| email1@xx.com | 3      |
| email3@xx.com | 4      |
'---------------'--------'

,----------------------,
| x_message_batch_4_pq |
|----------------------|
| prequeue_contact_id  |
|----------------------|
| 1                    |
| 2                    |
'----------------------'

我想更新分数为53但仅在prequeue表中的联系人的所有分数。如果它们不存在于reputation表中，则必须更新为2分。

所以在这个例子中：

contact_score上的

email1@xx.com将更新为3而contact_score上的email2@xx.com将更新为2。

我该怎么做？我尝试编写CASE语句，但我正在努力从声誉表中获取值。

编辑1

这是我尝试过的，但我收到了Error in query (1093): You can't specify target table 'x_message_batch_4_pq' for update in FROM clause错误

UPDATE x_message_batch_4_pq 
LEFT JOIN contacts ON prequeue_contact_id = contact_id
LEFT JOIN reputation_email_score ON contact_email = email
SET contact_contact_score =
    (CASE WHEN 
        (SELECT COUNT(contact_id) 
        FROM x_message_batch_4_pq 
        JOIN contacts ON prequeue_contact_id = contact_id 
        JOIN reputation_email_score ON contact_email = email 
        WHERE contact_contact_score = '53') = 0
    THEN '2' 
    ELSE (SELECT score FROM reputation_email_score WHERE email = contact_email)
    END)
WHERE contact_contact_score = '53';

Answer 1

您可以尝试使用JOIN这样的UPDATE语句，像这样使用LEFT JOIN

UPDATE contacts a 
       JOIN x_message_batch_4_pq b 
         ON a.contact_id = b.prequeue_contact_id 
       LEFT JOIN reputation_email_score c 
              ON a.contact_email = c.email 
SET    a.contact_score = CASE 
                           WHEN c.score IS NULL THEN 2 
                           ELSE c.score 
                         END 
WHERE  a.contact_score = 53

我创建了一个演示here，请检查

根据条件值更新行

编辑1

1 个答案: