Question

我有2个桌子有趣（主）和blk（奴隶）。

sqlite> select * from fun;
id          mod_id      name         freq      
----------  ----------  -----------  ----------
1           1           adpcm_coder  99108     
2           1           adpcm_decod  0

我想知道有多少blk是有趣的，所以我使用“group by”：

sqlite> SELECT fun.*, count(blk.id) as no_blk FROM fun, blk WHERE fun.id=blk.fun_id GROUP BY (blk.fun_id);
id          mod_id      name         freq        no_blk     
----------  ----------  -----------  ----------  ----------
1           1           adpcm_coder  99108       12

第二行被拒绝，因为它的blks不存在。 如何得到这样的结果？

id          mod_id      name         freq        no_blk     
----------  ----------  -----------  ----------  ----------
1           1           adpcm_coder  99108       12        
2           1           adpcm_decod  0           0

Answer 1

你想要一个OUTER JOIN 此外，您应该GROUP BY SELECT子句中的所有非汇总列。{/ p>

尝试这样的事情：

  SELECT fun.id, fun.mod_id, fun.name, fun.freq, count(blk.id) as no_blk 
    FROM fun
         LEFT OUTER JOIN blk
         ON blk.fun_id = fun.id
GROUP BY fun.id, fun.mod_id, fun.name, fun.freq;

Answer 2

亚当是正确的;另一种方法：

SELECT fun.*, count(blk.id) as no_blk FROM fun, blk 
WHERE fun.id=blk.fun_id GROUP BY (blk.fun_id) 
UNION 
SELECT fun.*, 0 as no_blk FROM fun, blk 
WHERE fun.id not in (SELECT fun.id from fun, blk WHERE fun.id=blk.fun_id);

sqlite3分组依据

2 个答案: