如何在MySQL中創建嵌套的SQL查詢？

當寫下面的查詢時，它會抱怨「so.id」列是未知的！下面的查詢在MS SQL中正常工作，但不在MySQL中？如何在MySQL中編寫嵌套的select語句？如何在MySQL中創建嵌套的SQL查詢？

這是什麼等價物？

select so.* 
from smart_objects as so 
    inner join (
     SELECT st.objectid, st.issueid 
     FROM smart_targets as st    
     WHERE st.issueid != 0 
      AND 
      ( (st.objectid = so.id) 
       OR 
       (    
       st.objectid in (
        SELECT sor.parent 
        FROM smart_objectrelations as sor  
        WHERE sor.child = so.id) 
       ) 
      )) as soi 
    on soi.objectid = so.id 
where so.id < 100;

我需要內部連接嵌套選擇SQL基本上。我的select語句比上面更復雜，上面只是一個簡化版本來解釋嵌套表。

我也嘗試過其他的建議：

select so.* 
    from smart_objects as so 
     inner join (
      SELECT st.objectid, st.issueid 
      FROM smart_targets as st, smart_objects as so   
      WHERE st.issueid != 0 
       AND 
       ( (st.objectid = so.id) 
        OR 
        (    
        st.objectid in (
         SELECT sor.parent 
         FROM smart_objectrelations as sor  
         WHERE sor.child = so.id) 
        ) 
       )) as soi 
     on soi.objectid = so.id 
    where so.id < 100;

上面的代碼進入一個永無止境的循環，而我只是想嵌套的SQL來爲母公司的每一行運行。

已更新與答案：

您是如何簡化下面的查詢？

select so.* 
from smart_objects as so 
    inner join (
     SELECT st.objectid, st.issueid 
     FROM smart_targets as st, smart_objects as so   
     WHERE st.objectid = so.id 
      and st.issueid != 0 
      AND 
      ( (st.objectid = so.id) 
       OR 
       (    
       st.objectid in (
        SELECT sor.parent 
        FROM smart_objectrelations as sor, smart_objects as so  
        WHERE sor.child = so.id) 
       ) 
      ) 
     ) as soi 
    on soi.objectid = so.id

來源

2013-02-12 The Light

難道真的在MS SQL工作？這個小提琴在MySQL和MS SQL 2008或者2012中都不起作用：http：//www.sqlfiddle.com/#!3/f32ce/1 – 2013-02-12 08:48:19

對不起，但是如何解決這個問題，so.id是未知的，而且查詢觸發無限循環？ – 2013-02-12 08:52:47

已更新。你怎麼在MySQL中創建嵌套的select語句？你如何用父母的每條記錄運行另一條select語句？ – 2013-02-12 08:56:07

試試這個：我想這應該工作

select so.* 
    from smart_objects as so 
    inner join (SELECT st.objectid, st.issueid FROM smart_targets as st,smart_objects as so   
     WHERE st.objectid = so.id) as soi 
    on soi.objectid = so.id

來源

2013-02-12 08:31:08 user2001117

但它會有不同的含義，但？我想要嵌套的Select與每個父行一起運行。 – 2013-02-12 08:46:50

如何在MySQL中創建嵌套的SQL查詢？

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