如何在刚刚选择的值中插入值
本文关键字:插入 选择 | 更新日期: 2023-09-27 18:00:38
sb.Append("SELECT u.Reputation");
sb.Append(" FROM Users AS u");
sb.Append(" INNER JOIN Comments AS c ON c.UsersID=u.UsersID");
sb.Append(" WHERE c.CommentsID=@CommentsID");
我想选择那个值。。从用户表中删除信誉,然后在其中插入数字5。我该怎么做。。?我希望用户的信誉在数据库中的值为5。
UPDATE Users u
SET u.Reputation = (u.Reputation + 5)
INNER JOIN Comments AS c ON c.UsersID=u.UsersID
WHERE c.CommentsID=@CommentsID