我试图在数据库的两个日期之间计算天数。 Date_from和Date_to。而且我收到了一个错误。任何帮助,将不胜感激。代码:
<?php
require('config/conn.php');
// Select all from table 'reguests'
$query = 'SELECT * FROM requests';
//Results from table
$result = mysqli_query($conn, $query);
//Fetch data
$requests = mysqli_fetch_all($result, MYSQLI_ASSOC);
//Free result from fetch
mysqli_free_result($result);
//Get Days Count Between Dates From And To
$dateFrom = $request['date_from'];
$dateTo = $requests['date_to'];
$daysDiff = floor(abs(strtotime($dateTo) - strtotime($dateFrom)) / (60*60*24));
//Close conn
mysqli_close($conn);
?>
输出:
<td><?php echo $daysDiff ?>
你的问题标题说要排除周末,但你的代码似乎没有尝试?
为了解释更复杂的逻辑,计算一个时期并使用它来迭代几天可能是谨慎的。
像这样的东西:
$dateFrom = new DateTime();
$dateTo = new DateTime( '+1 month +1 second' ); // Add 1s so period includes last day.
$period = new DatePeriod( $dateFrom, new DateInterval( 'P1D' ), $dateTo );
$days = 0;
foreach ( $period as $date ) {
$day = $date->format( 'l' );
if ( 'Saturday' !== $day && 'Sunday' !== $day ) {
$days ++;
}
}
echo $days; // 23
请检查您的标题,这与您的问题不明确。但要查看日期之间的差异,请查看链接上的Php手册:http://php.net/manual/en/datetime.diff.php
如果你真的想检查一天是否是周末,请检查:Checking if date is weekend PHP