SQL：如果不存在則先插入然后更新

Question

我知道有很多關於此主題的帖子，但這略有不同。 我有2個表variant_detail (vad)和variant_external_analysis (vaea) 。 vaea主要是空的，我需要為vad所有行更新它。 如果我插入vaea並且該行已經存在，它將復制該行，從而破壞數據庫。

在1行測試中，我解決了以下問題：

IF NOT EXISTS (SELECT vaea_vad_id FROM variant_external_analysis 
            JOIN variant_detail ON vad_id = vaea_vad_id 
            WHERE vad_variant_code = 16469)
BEGIN
    INSERT variant_external_analysis (vaea_vad_id) 
    SELECT vad_id
    FROM variant_detail
END

UPDATE variant_external_analysis

SET vaea_last_amended_on = GETDATE(), vaea_last_amended_by = 13, vaea_n_1 = 1

FROM variant_external_analysis
JOIN variant_detail ON vad_id = vaea_vad_id

WHERE vad_variant_code = 16469

直到我刪除WHERE語句，它才可以正常工作，這樣它將影響整個表。 下面的代碼可正確更新，但現在不再插入

IF NOT EXISTS (SELECT vaea_vad_id FROM variant_external_analysis

JOIN variant_detail ON vad_id = vaea_vad_id 

            WHERE vad_id = vaea_vad_id)
BEGIN
    INSERT variant_external_analysis (vaea_vad_id) 
    SELECT vad_id
    FROM variant_detail
END

UPDATE variant_external_analysis

SET vaea_last_amended_on = GETDATE(), vaea_last_amended_by = 13, vaea_n_1 = 1

FROM variant_external_analysis
JOIN variant_detail ON vad_id = vaea_vad_id

我可能缺少一些簡單的東西。

Answer 1

該代碼看起來與T-SQL非常相似，因此這里是一個示例，說明如何在MS SQL Server中使用MERGE進行操作：

merge variant_external_analysis vea
--using here left join (along with where condition) we'll get only rows 
--that are in variand_details but aren't in variant_external_analysis
using(select vd.vad_id 
  from variant_detail vd
  left join variant_external_analysis vaea on vd.vad_id = vaea.vaea_vad_id 
  where vaea.id is null) t --or any another column of vaea
on (vea.vaea_vad_id = t.vad_id)
when not matched then 
    insert (vaea_vad_id, vaea_last_amended_on, vaea_last_amended_by, vaea_n_1)
    values(t.vad_id, getdate(), 13, 1)
;

UPD要用單行測試子句：

1. 在using子句中添加code作為返回列
2. 用code條件擴展when子句

---

merge variant_external_analysis vea
    --using here left join (along with where condition) we'll get only rows 
    --that are in variand_details but aren't in variant_external_analysis
    using(select vd.vad_id, vd.vad_variant_code 
      from variant_detail vd
      left join variant_external_analysis vaea on vd.vad_id = vaea.vaea_vad_id 
      where vaea.id is null) t --or any another column of vaea
    on (vea.vaea_vad_id = t.vad_id)
    when not matched and t.vad_variant_code = 16469 then 
        insert (vaea_vad_id, vaea_last_amended_on, vaea_last_amended_by, vaea_n_1)
        values(t.vad_id, getdate(), 13, 1)
    ;

Answer 2

您的where子句中有“ vad_variant_code = vad_id = vaea_vad_id”。