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如果我找回场馆与他们的关闭时间从另一个表像这样的列表:即使连接表为空也显示行吗?
$fullDate = date('Y-m-d H:i:s');
$query="SELECT
venues.ID,
venues.VENUE_NAME,
locations.LOCATION,
hrs.opening_day,
hrs.opening_time,
hrs.closing_time
FROM venues
INNER JOIN venue_types
ON venues.VENUE_TYPE = venue_types.ID
INNER JOIN locations
ON venues.VENUE_LOCATION = locations.ID
INNER JOIN hours_of_operation hrs
on hrs.venue_id = venues.id
and ((hrs.opening_day=date_format('$fullDate' - INTERVAL (closing_time<opening_time and time('$fullDate')<closing_time) DAY, '%a')))
GROUP BY venues.VENUE_NAME";
我得到的,我已经指定了打开/关闭时间某一天在hours_of_operation
场馆的结果。无论当天hours_of_operation
行是否为空,该如何显示所有场所?
谢谢! :)
您在此查询没有聚集功能,使GROUP BY是不合适的。我怀疑你的意思是使用SELECT DISTINCT ... – Strawberry