Mysql 如何返回外键存在信息选择结果?
我有一个问题:Mysql 如何返回外键存在信息选择结果?,mysql,select,exists,Mysql,Select,Exists,我有一个问题: SELECT `Name`, `ID_dir`, 999 as `children` FROM `dir` dir WHERE dir.`fid_parent` IS NULL AND ( EXISTS ( SELECT 1 FROM `file` f WHERE dir.ID_dir = f.fid_parent ) OR ( S
SELECT `Name`, `ID_dir`, 999 as `children`
FROM `dir` dir WHERE dir.`fid_parent` IS NULL
AND (
EXISTS (
SELECT 1
FROM `file` f
WHERE dir.ID_dir = f.fid_parent
)
OR (
SELECT 1
FROM `dir` d2
WHERE dir.ID_dir = d2.fid_parent
)
)
我检查目录是否有外键。
如何在“选择…999作为子项”
中移动该信息以代替999?
我想在该位置返回(0或1)异或布尔值作为
子项
将存在
子查询放入选择
列表中
SELECT Name, ID_Dir, (
EXISTS (
SELECT 1
FROM `file` f
WHERE dir.ID_dir = f.fid_parent
)
OR (
SELECT 1
FROM `dir` d2
WHERE dir.ID_dir = d2.fid_parent
)
) AS children
FROM dir WHERE fid_parent IS NULL
谢谢,看起来我以前试过这个,但有一些错误,所以没用。