SQL查詢佔用時間超過1小時

Question

我們的一些SQL查詢需要3個多小時才能執行，考慮到我們的數據庫目前大約有5000個條目，這確實很長。

這裏兩個最長聲明：

/* 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

示例代碼中有多個語句，所以我就重點放在了第一位。

我更喜歡not in，因爲使用NULL的語義，雖然有證據表明not in可能更有效（請參閱評論中的鏈接）。這是第一個查詢：

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); 

我將其更改爲：

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); 

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

接下來的兩個語句是：

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'; 

對於這些，你想對vwr_synch.listings(VWRStatus, Alt, VWRRetry)的索引。