如何获取所选年份的选定月份的日期和日期,并在表格中显示。 例如: 我到目前为止已经尝试过了。
<?php
$num_of_days = cal_days_in_month(CAL_GREGORIAN, 9, 2003);
for( $i=1; $i<= $num_of_days; $i++)
$dates[]= str_pad($i,2,'0', STR_PAD_LEFT);
/*echo "<pre>";
print_r($dates);
echo "</pre>";*/
?>
<table>
<tr>
<?php
foreach($dates as $date){
echo"<td>".$date."</td>";
}
?>
</tr>
</table>
这为我执行了这段代码。
<table>
<tbody><tr>
<td>01</td><td>02</td><td>03</td><td>04</td><td>05</td><td>06</td><td>07</td><td>08</td><td>09</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td><td>29</td><td>30</td> </tr>
</tbody></table>
但我希望在日期行下方另行。 哪个应显示与该日期相关的日期?
白天我的意思是:星期一,星期二,星期三等 按日期我的意思是:1,2,3,4等
所以就像
<tr><td>1</td><td>2</td><td>3</td><td>4</td>
<tr><td>Mon</td><td>Tues</td><td>Wed</td><td>Thursday</td>
我希望我能解释一下自己......
答案 0 :(得分:10)
您可以使用date('l')
获取相应的日期名称:
<?php
$date = '2003-09-01';
$end = '2003-09-' . date('t', strtotime($date)); //get end date of month
?>
<table>
<tr>
<?php while(strtotime($date) <= strtotime($end)) {
$day_num = date('d', strtotime($date));
$day_name = date('l', strtotime($date));
$date = date("Y-m-d", strtotime("+1 day", strtotime($date)));
echo "<td>$day_num <br/> $day_name</td>";
}
?>
</tr>
</table>
答案 1 :(得分:9)
另一种方法可能是使用DateTime对象:
$aDates = array();
$oStart = new DateTime('2014-12-01');
$oEnd = clone $oStart;
$oEnd->add(new DateInterval("P1M"));
while ($oStart->getTimestamp() < $oEnd->getTimestamp()) {
$aDates[] = $oStart->format('D d');
$oStart->add(new DateInterval("P1D"));
}
然后打印:
foreach ($aDates as $day) {
echo $day;
}
有关格式参数的详细信息,请参阅:http://php.net/manual/en/function.date.php
答案 2 :(得分:4)
您可以尝试这样的事情
$myYearMonth = '2003-09';
$start = new DateTime(date('Y-m-01', strtotime($myYearMonth)));
$end = new DateTime(date('Y-m-t', strtotime($myYearMonth)));
$diff = DateInterval::createFromDateString('1 day');
$periodStart = new DatePeriod($start, $diff, $end);
foreach ( $periodStart as $dayDate ){
echo '<td>'.$dayDate->format( "d\n" ).'</td><td>'.$dayDate->format( "l\n" ).'</td>';
}
答案 3 :(得分:3)
您不应该使用date()
函数,因为根据PHP文档,日期范围限制在1970-2038:
时间戳的有效范围通常是从格林尼治标准时间1901年12月13日20:45:54到格林威治标准时间2038年1月19日星期二03:14:07。 (这些是与32位有符号整数的最小值和最大值对应的日期)。但是,在PHP 5.1.0之前,在某些系统(例如Windows)上,此范围仅限于01-01-1970至19-01-2038。
使用DateTime()
课程,您可以使用以下功能检索给定月份所需的信息。这使用DateTime::format()
获取英文日期名称(无本地化)。
function getMonth($year, $month) {
// this calculates the last day of the given month
$last=cal_days_in_month(CAL_GREGORIAN, $month, $year);
$date=new DateTime();
$res=Array();
// iterates through days
for ($day=1;$day<=$last;$day++) {
$date->setDate($year, $month, $day);
$res[$day]=$date->format("l");
}
return $res;
}
这将返回一个关联的数组,如下所示:
$res=getMonth(2015, 2);
print_r($res);
Array
(
[1] => Sunday
[2] => Monday
[3] => Tuesday
[4] => Wednesday
[5] => Thursday
[...]
)
要在两行表中输出数据,可以使用以下代码:
<?php
echo '<table><tr><td>'.implode('</td><td>', array_keys($res)).'</td></tr>';
echo '<tr><td>'.implode('</td><td>', $res).'</td></tr></table>';
由于Datetime::format()
函数不支持翻译的语言环境,您可以使用关联数组来获取另一种语言的翻译。
答案 4 :(得分:3)
以下完整解决方案。
我确定给定月/年组合中的天数。然后我循环过去几天同时创建两个必需的行。
完成后,这两行将包装在一个表中并返回给调用者。
<?php
echo buildDate(12, 2017);
function buildDate($month, $year)
{
// start with empty results
$resultDate = "";
$resultDays = "";
// determine the number of days in the month
$daysInMonth = cal_days_in_month(CAL_GREGORIAN, $month, $year);
for ($i = 1; $i <= $daysInMonth; $i++)
{
// create a cell for the day and for the date
$resultDate .= "<td>".sprintf('%02d', $i)."</td>";
$resultDays .= "<td>".date("l", mktime(0, 0, 0, $month, $i, $year))."</td>";
}
// return the result wrapped in a table
return "<table>".PHP_EOL.
"<tr>".$resultDate."</tr>".PHP_EOL.
"<tr>".$resultDays."</tr>".PHP_EOL.
"</table>";
}
?>
phpfiddle链接:http://phpfiddle.org/main/code/ffjm-hqsu
答案 5 :(得分:2)
试试这个脚本:
// Day of month, e.g. 2014-12-14 if you need the table for december 2014
$date = time();
// Array containing the dates and weekdays
$days = array();
// loop to populate the array
for(
$day = strtotime('midnight', strtotime(date('1 F Y', $date))); /* first day of month */
$day < strtotime(date('1 F Y', strtotime('next month', $date))); /* first day of next month */
$day = strtotime('next day', $day)
){
// insert current day into the array
$days[date('d', $day)] = date('l', $day);
}
// print the row containing all day numbers
echo '<tr><td>'.implode('</td><td>', array_keys($days)).'</td></tr>';
// print the row containing all weekday names
echo '<tr><td>'.implode('</td><td>', $days).'</td></tr>';
答案 6 :(得分:1)
在您的情况下date('D', strtotime($date))
应该有效,但您需要格式为yyyy-mm-dd
的日期
我做了一些测试,结果如下:
for( $i=1; $i<= $num_of_days; $i++){
$dates[]= str_pad($i,2,'0', STR_PAD_LEFT);
$d = "2003-09-".$i;
$days[] = date('D', strtotime($d));
}
添加了另外tr
天:
<tr>
<?php
foreach($days as $day){
echo"<td>".$day."</td>";
}
?>
</tr>
答案 7 :(得分:1)
构建日期关联数组的DateTime方法 - &gt;一天。
2015年12月的结果如下:
array(31) {
[1] = string(7) "Tuesday"
[2] = string(9) "Wednesday"
[3] = string(8) "Thursday"
[4] = string(6) "Friday"
[5] = string(8) "Saturday"
[6] = string(6) "Sunday"
[7] = string(6) "Monday"
[8] = string(7) "Tuesday"
[9] = string(9) "Wednesday"
[10] = string(8) "Thursday"
[11] = string(6) "Friday"
[12] = string(8) "Saturday"
[13] = string(6) "Sunday"
[14] = string(6) "Monday"
[15] = string(7) "Tuesday"
[16] = string(9) "Wednesday"
[17] = string(8) "Thursday"
[18] = string(6) "Friday"
[19] = string(8) "Saturday"
[20] = string(6) "Sunday"
[21] = string(6) "Monday"
[22] = string(7) "Tuesday"
[23] = string(9) "Wednesday"
[24] = string(8) "Thursday"
[25] = string(6) "Friday"
[26] = string(8) "Saturday"
[27] = string(6) "Sunday"
[28] = string(6) "Monday"
[29] = string(7) "Tuesday"
[30] = string(9) "Wednesday"
[31] = string(8) "Thursday"
}
获取所需表格的完整代码:
<?php
// Get an array of days
$arrayDays = getDays(12, 2015);
// Function to get an array of days
function getDays($month, $year){
// Start of Month
$start = new DateTime("{$year}-{$month}-01");
$month = $start->format('F');
// Prepare results array
$results = array();
// While same month
while($start->format('F') == $month){
// Add to array
$day = $start->format('l');
$date = $start->format('j');
$results[$date] = $day;
// Next Day
$start->add(new DateInterval("P1D"));
}
// Return results
return $results;
}
?>
<!-- Output the Table -->
<table>
<tr>
<?php foreach (array_keys($arrayDays) as $someDate): ?>
<td><?= $someDate; ?></td>
<?php endforeach; ?>
</tr>
<tr>
<?php foreach (array_values($arrayDays) as $someDate): ?>
<td><?= $someDate; ?></td>
<?php endforeach; ?>
</tr>
</table>