请检查查询。
SELECT DATE_FORMAT( CAST( CONCAT( table.date, " ", table.time ) ) , '%p' )
FROM table AS table
Error:
[hash]1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '), '%p') FROM exp_visitorsday as vd LIMIT 0, 30' at line 2
我在以下情况下尝试了此查询。
SELECT DATE_FORMAT( CAST( CONCAT( table.date, " ", table.time ) as datetimecol ) as datetimecol, '%p' )
FROM table AS table
SELECT DATE_FORMAT( CAST( CONCAT( table.date, " ", table.time ) as datetimecol ), '%p' )
FROM table AS table
示例数据:
table.date => 01-08-2013
table.time => 10:00 AM
如何查询?这个查询有什么问题?
答案 0 :(得分:1)
你可以:
SELECT *
FROM (
SELECT DATE_FORMAT(
CONCAT( SUBSTRING_INDEX( "01-08-2013", "-", -1), "-",
SUBSTRING_INDEX(SUBSTRING_INDEX( "01-08-2013", "-", 2), "-", -1), "-",
SUBSTRING_INDEX( "01-08-2013", "-", 1), " ",
IF( SUBSTRING_INDEX( "10:00 PM", " ", -1) = "AM",
SUBSTRING_INDEX( "10:00 AM", " ", 1),
CONCAT( SUBSTRING_INDEX(SUBSTRING_INDEX( "10:00 AM", " ", 1), ":", 1) + 12, ":", SUBSTRING_INDEX(SUBSTRING_INDEX( "10:00 AM", " ", 1), ":", -1)) )) , '%Y-%m-%d %h:%i') AS datformat ) AS tmp
ORDER BY tmp.datformat;
说明:
SELECT SUBSTRING_INDEX( "01-08-2013", "-", 1); #Day
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX( "01-08-2013", "-", 2), "-", -1);#Month
SELECT SUBSTRING_INDEX( "01-08-2013", "-", -1); #Year