SELECT *,(SELECT COUNT(*) FROM item WHERE category = category.id) as item_count FROM category
还有别的更优化的方法吗？
还有别的更优化的方法吗？
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参考文档
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MySQL 相关项目
› MariaDB
› Drizzle
参考文档
› http://mysql-python.sourceforge.net/MySQLdb.html
This topic created in 4869 days ago, the information mentioned may be changed or developed.
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fangzhzhJul 9, 2013
1
可不可以不要select *, 表结构一变,你这代码就随时可能爆炸啊
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fangzhzhJul 9, 2013
3
除了select*不能接受, 你这写法我能接受
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sutarJul 9, 2013
4
JOIN
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6mN361AJx04in471Jul 9, 2013
6
@iloveyou 好多框架带的orm，new model不都是*吗？这是错的，可以自定义筛选字段。
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lichaoJul 9, 2013
7
select category.field1, item.field2 from category, item where item.categoryID = category.categoryID
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10
select category.categoryName, count(item.itemID) as YOUR_COUNT from category, item where item.categoryID = category.categoryID group by category.categoryID
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11
select category.categoryID, count(item.itemID) as YOUR_COUNT from category, item where item.categoryID = category.categoryID group by category.categoryID
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12
select category, count(*) from item group by category，将此临时表与category表做联合查询。
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msg7086Jul 9, 2013
13
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masterJul 9, 2013
14
话说为什么不用JOIN ？
比如像这样： SELECT category.*, COUNT(item.id) as item_count FROM category LEFT JOIN item ON item.category = category.id |
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15
抱歉，漏了 GROUP
SELECT category.*, COUNT(item.id) as item_count FROM category LEFT JOIN item ON item.category = category.id GROUP BY category.id |
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