我想从“日期”类型列中选择值。我有像29.04.2019这样的值,但是当我选择它并在视图中打印时,我得到29.04.2019 00:00:00。
我的查询:
SELECT Id, project_name, start_date, end_date, Team_Id FROM Projects;
答案 0 :(得分:1)
使用Convert
函数将日期时间转换为日期
SELECT Id, project_name, convert(date,start_date) start_date, convert(date, end_date)end_date, Team_Id FROM Projects;
答案 1 :(得分:1)
如果要使用DD.MM.YYYY
格式,请使用CONVERT(varchar(10), Getdate(), 104)
SELECT Id, project_name, CONVERT(varchar(10), start_date, 104) as start_date,
CONVERT(varchar(10), end_date, 104) as end_date, Team_Id FROM Projects;
所有sql格式check the link here
答案 2 :(得分:0)
SELECT Id, project_name,
convert(date,start_date) as start_date,
convert(date,end_date) as end_date,
Team_Id FROM Projects