[英]Date range array excluding the Sunday & the holiday in PHP
我有一個函數,該函數返回數組中兩個日期之間的所有日期,但是我需要在該數組中排除星期日。
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 $dates;
}
在排除星期日之后,我有了一個要存儲一些日期的表,我也需要從數組中排除這些日期。
例如,如果我輸入日期范圍為01-05-2012(DD-MM-YYYY)到10-05-2012,則06-05-2012將為星期日,日期為01-05-2012和08-05-我上面提到的表格是2012年,最終的結果應該是這樣,
02-05-2012
03-05-2012
04-05-2012
05-05-2012
07-05-2012
09-05-2012
10-05-2012
如何在PHP中做到這一點? 我嘗試了一些,但找不到正確的方法。
對於星期日部分:
public function dateRange($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 $dates;
}
對於假期部分:
首先,您需要將日期加載到某種數組中,然后為每個日期循環遍歷數組,並檢查它們是否匹配。
我找到了問題的答案,謝謝幫助我的人們。
public function dateRange($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;
}
上面的代碼用於刪除給定范圍內的假日和星期日。
我在jQuery中執行了上述相同的方法
//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
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