Question

我的查询工作正常，但我希望显示今天的日期数量+提前2天添加，并在今天的日期之后订购数据，然后是明天和之后的日期？

因为记录太多所以无法以其他格式更改 varchar 行meeting_date

这是MySQL查询

日期存储为月/日/年，如2016年1月1日。

$today = date("m/d/Y");

$query = "SELECT COUNT(*) as num FROM records WHERE meeting = 'Yes' AND (meeting_date LIKE '%".$today."%') ORDER BY id DESC";
$total_pages = mysql_fetch_array(mysql_query($query));

感谢您的时间和帮助。

Answer 1

尝试

$today = date("m/d/Y");//today's date
$after2days = date('m/d/Y', strtotime("+2 days"));//2 days from now
$query = "SELECT COUNT(*) as num FROM records WHERE meeting = 'Yes' AND meeting_date BETWEEN '".$today."' AND '".$after2days."' ORDER BY meeting_date";

我建议您不要使用mysql_ *函数，因为它们已被弃用。请使用mysqli_ *或PDO函数。

Answer 2

你应该在获取日期时在php中执行此操作

$today = date("m/d/Y");
$tommarow= date('m/d/Y', strtotime("+1 days"));
$twoday= date('m/d/Y', strtotime("+2 days"));
$query = "SELECT COUNT(*) as num FROM records WHERE meeting = 'Yes' AND 
(meeting_date = '".$today."' or  meeting_date = '".$tommarow."' or 
meeting_date = '".$twoday."') group by meeting_date ORDER BY meeting_date";

Answer 3

你需要这样做：

// get 3 days used in the query
$today = date("m/d/Y");
$day1 =  date("m/d/Y", time() + 86400 ); 
$day2 =  date("m/d/Y", time() + 86400*2 ); 

// updated query without LIKE
 $query = "SELECT COUNT(*) as num FROM records WHERE meeting = 'Yes' AND (meeting_date = '$today' OR meeting_date = '$day1' meeting_date = '$day2') ORDER BY id DESC";

需要使用php排序数组后。

字符串格式的日期提前显示两条记录

3 个答案: