如何从值列表更新值的列一组记录

Question

我按行做一个表行的更新：

UPDATE table 
SET col = $value 
WHERE id = $id 

现在，如果我更新例如每个记录10000条记录获得$value，但哪个$id获得哪个$value并不重要。我唯一的要求是，我正在更新的所有记录都以$value结尾。
所以我怎么会这个更新转换为类似

UPDATE table 
SET col ?????? what here from a $value_list??? 
WHERE id IN ($id_list) 

即通过列表ID和不知何故的价值观和ids的范围得到A值

Answer 1

让我们假设你有两个逗号分隔的列表你的ID和你的值与相同的项目数。然后，你可以做你的更新像那些语句：

-- the list of the ids 
SET @ids = '2,4,5,6'; 
-- the list of the values 
SET @vals = '17, 73,55, 12'; 

UPDATE yourtable 
INNER JOIN (
    SELECT 
     SUBSTRING_INDEX(SUBSTRING_INDEX(t.ids, ',', n.n), ',', -1) id, 
     SUBSTRING_INDEX(SUBSTRING_INDEX(t.vals, ',', n.n), ',', -1) val 
    FROM (SELECT @ids as ids, @vals as vals) t 
    CROSS JOIN (
     -- build for up to 1000 separated values 
     SELECT 
      1 + a.N + b.N * 10 + c.N * 100 AS n 
     FROM 
      (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a 
      ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b 
      ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) c 
     ORDER BY n 
    ) n 
    WHERE n <= (1 + LENGTH(t.ids) - LENGTH(REPLACE(t.ids, ',', ''))) 
) t1 
ON 
    yourtable.id = t1.id 
SET 
    yourtable.val = t1.val; 

说明

内一系列工会建立从1到1000的数字表，您应该能够扩大这一机制您的需求：

-- build for up to 1000 separated values 
SELECT 
    a.N + b.N * 10 + c.N * 100 + 1 AS n 
FROM 
    (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a 
    ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b 
    ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) c 
ORDER BY n 

我们用这个数字来获取项目从我们的名单与嵌套调用SUBSTRING_INDEX

SUBSTRING_INDEX(SUBSTRING_INDEX(t.ids, ',', n.n), ',', -1) id, 
SUBSTRING_INDEX(SUBSTRING_INDEX(t.vals, ',', n.n), ',', -1) val 

WHERE子句获得的项目数量（只确定了两个中的一个）列表：

WHERE n <= (1 + LENGTH(t.ids) - LENGTH(REPLACE(t.ids, ',', ''))) 

因为我们已经得到了分离器的一个occurence少，我们加1的区别带有分隔符的列表的长度和没有出现分隔符的所有列表的长度。

然后我们使用JOIN操作对外部UPDATE语句中的id值执行UPDATE。

看来它的工作this fiddle。

相信我：这比痛苦的逐行更新要快得多。