とみたさん、ＭＬ皆さんおせわになります。

select article.id,name1,sum(number) as sum1
from article left outer join stock on article.id=stock.article_id
group by article.id;
で
+-----------+-------+-----+
|article_id | name1 | sum1|
+-----------+-------+-----+
|         1 | aaaa  | 100 |
+-----------+-------+-----+
|         2 | bbbb  | 200 |
+-----------+-------+-----+
|         3 | cccc  | NULL|
+-----------+-------+-----+
|         4 | dddd  | 300 |
+-----------+-------+-----+
願ってた結果を得ることができました。

そこでさらに、テーブルstockにWHEREを適用したくて

SELECT article.id, article.name1,SUM(number) AS sum1
FROM article
LEFT OUTER JOIN (SELECT * FROM stock WHERE number >150) AS stock
ON ( article.id = stock.article_id ) 
GROUP BY article_id
ORDER BY article.id;
と書きましたが、結果は
+-----------+-------+-----+
|article_id | name1 | sum1|
+-----------+-------+-----+
|         1 | aaaa  | NULL|
+-----------+-------+-----+
|         2 | bbbb  | 200 |
+-----------+-------+-----+
|         4 | dddd  | 200 |
+-----------+-------+-----+
こうなりました、

+-----------+-------+-----+
|article_id | name1 | sum1|
+-----------+-------+-----+
|         1 | aaaa  | NULL|
+-----------+-------+-----+
|         2 | bbbb  | 200 |
+-----------+-------+-----+
|         3 | cccc  | NULL|
+-----------+-------+-----+
|         4 | dddd  | 200 |
+-----------+-------+-----+
このような結果を得るにはどのようなＳＱＬを書けばいいのでしょうか
よろしくお願いします。
奮闘中http://hdkk.com/notobe/CustomView1list.php
-------
平野