Question

这里我正在开发一个逻辑，其中我想根据要求给出小时条件，示例：考虑，

案例1：如果小时在17到20之间，那么它将在日期执行操作为date-1

case2：如果不是，它将显示日期

以下查询我有设计，但它不起作用.. :( 为我提供更好的解决方案......谢谢！

<?php
$mysqli = mysqli_connect('localhost','root','','company');
if (mysqli_connect_errno($mysqli)){
    echo "Failed to connect to MySQL:".mysqli_connect_error();
  }
$arr=array();

$query=mysqli_query($mysqli, "SELECT substr(date,1,2) as DAY,substr(date,4,2) as MONTH,substr(date,7,4) as YEAR 
from tbl_users
where substr(time,1,2)=17 OR substr(time,1,2)=18 OR substr(time,1,2)=19 OR substr(time,1,2)=20 AND id=3");

while($d=mysqli_fetch_array($query, MYSQL_ASSOC))
{
     $arr[]=$d['DAY'];
     $arr[]=$d['MONTH'];
     $arr[]=$d['YEAR'];
     $arr[]=$d['HOUR'];
     $arr[]=$d['MINUTE'];

}
$time=$arr[3]-1;
echo $time;
?>

Answer 1

根据我的评论，您必须检查PHP级别的条件而不是查询。从表中检索日期和时间，然后检查检索到的值的条件

$query=mysqli_query($mysqli, "SELECT date, time from tbl_users where id=3");

while($d = mysqli_fetch_array($query, MYSQL_ASSOC))
{
   $day = date('D', strtotime($d['date']));
   //or you may use substr here instead of writing them in queries
}

如何根据要求在当天给予条件

1 个答案: