仅更新MySQL中重复条目的第一条记录 [英] Update only first record from duplicate entries in MySQL

问题描述

问题已解决！

更新:

不太正确，我需要在具有ID，NAME，COVER字段的简单表上进行示例

Not quite right what I need, lets do example on simple table with fields ID,NAME,COVER

我有100个条目，其中包含100个名称，其中一些名称是重复的，但是我只想从重复项中更新第一个.

I have got 100 entries with 100 names, some of the names are duplicated, but I want only update first one from duplicates.

试图从数据库中所有重复项中更新所有第一行，真的很难做到，我知道怎么做吗?下面是我要重建的代码，但是对于所有重复项，此代码将每个第一个替换为最后一个.

Trying to update all the 1st rows from all the duplicates in database, really hard to do it, any idea how I can make it? Below is the code I am trying to rebuild, but this code replace every 1st one with the last one for all the duplicates.

模式，我希望它如何在下面工作

Schema, how I want it work below

ID NAME COVER
1  Max   1
2  Max   0
3  Andy  1
4  Andy  0
5  Andy  0

UPDATE table t
  JOIN (
    SELECT MinID, b.Name LatestName
    FROM table b
    JOIN (
      SELECT MIN(ID) MinID, MAX(ID) MaxID
      FROM table
      GROUP BY tag
      HAVING COUNT(*) > 1
    ) g ON b.ID = g.MaxID
  ) rs ON t.ID = rs.MinID
SET t.Name = LatestName;

推荐答案

目前尚不清楚您想要什么.也许是这样:

It's not clear at all what you want. Perhaps this:

UPDATE table AS t
  JOIN 
    ( SELECT MIN(ID) MinID
      FROM table
      GROUP BY Name
      HAVING COUNT(*) > 1
    ) AS m 
    ON t.ID = m.MinID
SET t.Cover = 1 ;

---

对于这个(以及将来的)问题，在编写问题时请记住:

---

For this (and future) question, keep in mind, when you write a question:

1. a description of your problem, as clear as possible    --- you have that
2. data you have now (a few rows of the tables)           --- ok, nice
3. the code you have tried                 --- yeah, but better use same names
                                           --- as the data and description above
4. the error you get (if you get an error)      --- doesn't apply here
5. the result you want (the rows after the update in your case)
                                       --- so we know what you mean in case we 
                                       --- haven't understood from all the rest 

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