mysql if 函数 - 子查询作为条件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13840181/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
mysql if function - subquery as a condition
提问by shealtiel
I believe the following query is self explanatory:
我相信以下查询是不言自明的：
SELECT IF (SELECT COUNT(*) FROM mytable > 0, 'yes', 'no');
Why doesn't it work? And how should I correct it?
为什么不起作用？我应该如何纠正它？
回答by MichaelRushton
Enclose the subquery in parentheses:
将子查询括在括号中：
SELECT IF ((SELECT COUNT(*) FROM mytable), 'yes', 'no');
回答by Devart
It this what you want?
这是你想要的吗？
SELECT IF(COUNT(*) > 0, 'yes', 'no') FROM mytable;
1:
1：
SELECT
t1.*,
(SELECT IF(COUNT(*) > 0, 'yes', 'no') FROM mytable) AS col1
FROM
table t1;
2:
2：
SELECT
t1.*,
t2.*
FROM
table t1,
(SELECT IF(COUNT(*) > 0, 'yes', 'no') AS col1 FROM mytable) t2
回答by Luca Rainone
If your query is more complex and this is only a reduced problem, I think this is the better solution for you
如果您的查询更复杂并且这只是一个简化的问题，我认为这对您来说是更好的解决方案
SELECT IF ( (SELECT COUNT(*) AS counter FROM myTable HAVING counter>0) , 'yes', 'no')
so you can do more complex check (i.e. counter > N or multiple conditions)
所以你可以做更复杂的检查（即计数器> N 或多个条件）
回答by Saharsh Shah
Try this:
尝试这个：
SELECT IF((SELECT COUNT(*) FROM mytable) > 0, 'yes', 'no');
