我有一个函数,它返回数组中两个日期之间的所有日期,但我需要排除该数组中的星期日。
public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
$dates[] = date($format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
排除星期日后,我有一个表,我将在其中存储一些日期,我也需要从数组中排除这些日期。
比如,如果我输入日期范围为 01-05-2012(日-月-年)到 10-05-2012,06-05-2012 将是星期日 &日期 01-05-2012 和 08-05-2012 将在我上面提到的表格中,最后的输出应该是这样的,
02-05-2012
03-05-2012
04-05-2012
05-05-2012
07-05-2012
09-05-2012
10-05-2012
如何在PHP中执行此操作?我尝试了一些,但找不到正确的方法。
对于星期日部分:
public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
if (date("D", $current) != "Sun")
$dates[] = date($format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
对于假期部分:
首先,您需要将日期加载到某种数组中,然后遍历每个日期的数组并检查它们是否匹配。
我找到了问题的答案,感谢帮助过我的人。
public function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
$sql = "SELECT * FROM ost_holidays where holiday_date='".date('Y-m-d', $current)."' LIMIT 1";
$sql = db_query($sql);
$sql = db_fetch_array($sql);
if($sql['holiday_date'] != date('Y-m-d',$current))
if (date('w', $current) != 0)
$dates[] = date($format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
上面的代码用于删除给定范围内的假期和星期日。
我在 Jquery 中做了同样的上述方法
//Convert dates into desired formatt
function convertDates(str) {
var date = new Date(str),
mnth = ("0" + (date.getMonth() + 1)).slice(-2),
day = ("0" + date.getDate()).slice(-2);
return [date.getFullYear(), mnth, day].join("-");
}
// Returns an array of dates between the two dates
var getDates = function(startDate, endDate, holidays) {
var dates = [],
currentDate = startDate,
addDays = function(days) {
var date = new Date(this.valueOf());
date.setDate(date.getDate() + days);
return date;
};
while (currentDate <= endDate) {
dates.push(currentDate);
currentDate = addDays.call(currentDate, 1);
}
return dates;
};
//Indise Some Function
var datesTemp = [];
var dates = getDates(new Date(prodDet.details.date1), new Date(prodDet.details.date2));
dates.forEach(function(date) {
if (date.getDay() != 0) {
datesTemp.push(convertDates(date));
}
});
datesTemp.forEach(function(date) {
for (var j = 0; j < prodDet.holidays.length; j++) {
if ((prodDet.holidays[j] != date)) {
ideal.idates.push(date);
}
}
});
console.log(ideal.idates);
//Function Ends Here