我很多时候发现自己作为某个select max(superstar)
(SELECT superstars.name,COUNT(superstar01) as superstar FROM smackdown_teams JOIN superstars on smackdown_teams.superstar01 = superstars.id GROUP by superstar01 Order by COUNT(superstar01) desc
UNION
SELECT superstars.name,COUNT(superstar02) as superstar FROM smackdown_teams JOIN superstars on smackdown_teams.superstar02 = superstars.id GROUP by superstar02 Order by COUNT(superstar02) desc)
类型的arg传入但是每个键都是可选的，只需要至少一个。
例如：
Shape是否有实用程序功能可以从type Shape = {
+isFetching: boolean,
+errorFetching: null | string
}
type ShapeOpt = {
isFetching?: boolean,
errorFetching?: boolean
}
function set(data: ShapeOpt) {
for (const key in data) {
global[key] = data[key];
}
}
转换为Shape？
答案 0 :(得分：1)
有一个$Shape<Type>助手用于生成对象类型，其中每个键都是可选的。但是我不知道如何说自动需要至少一个项目。