SQL查询占用时间超过1小时

Question

我们的一些SQL查询需要3个多小时才能执行，考虑到我们的数据库目前大约有5000个条目，这确实很长。

这里两个最长声明：

/* Alte Jobs anzeigen */ 

update vwr_synch.listings set vwr_synch.listings.Alt="alt" 
where vwr_synch.listings.JobKey not in (select vwr_synch.jobs.JobKey from vwr_synch.jobs) 
; 

update vwr_synch.listings set vwr_synch.listings.Alt="alt" where vwr_synch.listings.VWRStatus="NoJobs" or vwr_synch.listings.VWRStatus="Problem" 
; 


update vwr_synch.listings set vwr_synch.listings.Alt=NULL 
where vwr_synch.listings.VWRStatus="Active" and vwr_synch.listings.VWRRetry!="0" and vwr_synch.listings.Alt="alt" 
; 


/* Neue Jobs anzeigen */ 

update vwr_synch.jobs set vwr_synch.jobs.NeuAlt="Neu" 
where vwr_synch.jobs.JobKey not in (select vwr_synch.listings.JobKey from vwr_synch.listings) and (vwr_synch.jobs.`Status`="Active" and vwr_synch.jobs.Retry="0") 
; 

Answer 1

示例代码中有多个语句，所以我就重点放在了第一位。

我更喜欢not in，因为使用NULL的语义，虽然有证据表明not in可能更有效（请参阅评论中的链接）。这是第一个查询：

update vwr_synch.listings 
    set vwr_synch.listings.Alt = 'alt' 
    where vwr_synch.listings.JobKey not in (select vwr_synch.jobs.JobKey from vwr_synch.jobs); 

我将其更改为：

update vwr_synch.listings l 
    set l.Alt = 'alt' 
    where not exists (select 1 from vwr_synch.jobs.JobKey jk where jk.JobKey = l.JobKey); 

然后，这有效地工作，你需要vwr_synch.jobs.JobKey(JobKey)的索引。

接下来的两个语句是：

update vwr_synch.listings l 
    set l.Alt = 'alt' 
    where l.VWRStatus in ('NoJobs', 'Problem'); 

update vwr_synch.listings l 
    set l.Alt = NULL 
    where l.VWRStatus = 'Active' and l.VWRRetry <> '0' and l.Alt = 'alt'; 

对于这些，你想对vwr_synch.listings(VWRStatus, Alt, VWRRetry)的索引。