Question

我使用下面的代码来回显当前月份。如何增强它，以便显示所有月份的名称，日期和日期..

代码：

 <?php
$today    = getdate();
$firstDay = getdate(mktime(0,0,0,$today['mon'],1,$today['year']));
$lastDay  = getdate(mktime(0,0,0,$today['mon']+1,0,$today['year']));


?>

<?php

echo '<table>';
echo '  <tr><th colspan="7">'.$today['month']." - ".$today['year']."</th></tr>";
echo '<tr class="days">';
echo '  <td>Mo</td><td>Tu</td><td>We</td><td>Th</td>';
echo '  <td>Fr</td><td>Sa</td><td>Su</td></tr>';
?> 

<?php
echo '<tr>';
for($i=1;$i<$firstDay['wday'];$i++){
    echo '<td>&nbsp;</td>';
}
$actday = 0;
for($i=$firstDay['wday'];$i<=7;$i++){
    $actday++;
    echo "<td>$actday</td>";
}
echo '</tr>';
?> 

<?php
$fullWeeks = floor(($lastDay['mday']-$actday)/7);

for ($i=0;$i<$fullWeeks;$i++){
    echo '<tr>';
    for ($j=0;$j<7;$j++){
        $actday++;
        echo "<td>$actday</td>";
    }
    echo '</tr>';
    }
    ?> 

    <?php
    if ($actday < $lastDay['mday']){
    echo '<tr>';

    for ($i=0; $i<7;$i++){
        $actday++;
        if ($actday <= $lastDay['mday']){
            echo "<td>$actday</td>";
        }
        else {
            echo '<td>&nbsp;</td>';
        }
    }

    echo '</tr>';
}
?>

Answer 1

试试这个：

function getDates($year)
{
    $dates = array();

    date("L", mktime(0,0,0, 7,7, $year)) ? $days = 366 : $days = 365;
    for($i = 1; $i <= $days; $i++){
        $month = date('m', mktime(0,0,0,1,$i,$year));
        $wk = date('W', mktime(0,0,0,1,$i,$year));
        $wkDay = date('D', mktime(0,0,0,1,$i,$year));
        $day = date('d', mktime(0,0,0,1,$i,$year));

        $dates[$month][$wk][$wkDay] = $day;
    } 

    return $dates;   
}

它将返回传递给函数的一年中 - 月 - >周 - >日 - >工作日的数组。希望通过数组遍历可以很容易地打印出所有内容。我确定你可以做很多调整，但这是一个开始。

我也会尝试远离使用echo打印html，例如代替;

echo '<tr>';
for($i=1;$i<$firstDay['wday'];$i++){
    echo '<td>&nbsp;</td>';
}

做;

<tr>;
<?php for($i=1;$i<$firstDay['wday'];$i++){ ?>
    <td><?php echo $var; ?></td>
<?php } ?>

我认为它使代码更具可读性。

编辑：我想我应该包括一个用例示例，如下所示：

<?php $dates = getDates(2011); 

$weekdays = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'); ?>
<?php foreach($dates as $month => $weeks) { ?>
<table>
    <tr>
        <th><?php echo implode('</th><th>', $weekdays); ?></th>
    </tr>
    <?php foreach($weeks as $week => $days){ ?>
    <tr>
        <?php foreach($weekdays as $day){ ?>
        <td>
            <?php echo isset($days[$day]) ? $days[$day] : '&nbsp'; ?>
        </td>               
        <?php } ?>
    </tr>
    <?php } ?>
</table>
<?php } ?>

它为您提供输出：

enter image description here

Answer 2

您可以使用此功能将整年转换为数组

  function year2array($year) {
    $res = $year >= 1970;
    if ($res) {
      // this line gets and sets same timezone, don't ask why :)
      date_default_timezone_set(date_default_timezone_get());

      $dt = strtotime("-1 day", strtotime("$year-01-01 00:00:00"));
      $res = array();
      $week = array_fill(1, 7, false);
      $last_month = 1;
      $w = 1;
      do {
        $dt = strtotime('+1 day', $dt);
        $dta = getdate($dt);
        $wday = $dta['wday'] == 0 ? 7 : $dta['wday'];
        if (($dta['mon'] != $last_month) || ($wday == 1)) {
          if ($week[1] || $week[7]) $res[$last_month][] = $week;
          $week = array_fill(1, 7, false);
          $last_month = $dta['mon'];
          }
        $week[$wday] = $dta['mday'];
        }
      while ($dta['year'] == $year);
      }
    return $res;
    }

称之为

print_r(year2array(2011));

您会在来源（月 - >周 - >天）中看到这一点：

所以，现在很容易使用这样的东西为每个月创建月表

  function month2table($month, $calendar_array) {
    $ca = 'align="center"';
    $res = "<table cellpadding=\"2\" cellspacing=\"1\" style=\"border:solid 1px #000000;font-family:tahoma;font-size:12px;background-color:#ababab\"><tr><td $ca>Mo</td><td $ca>Tu</td><td $ca>We</td><td $ca>Th</td><td $ca>Fr</td><td $ca>Sa</td><td $ca>Su</td></tr>";
    foreach ($calendar_array[$month] as $month=>$week) {
      $res .= '<tr>';
      foreach ($week as $day) {
        $res .= '<td align="right" width="20" bgcolor="#ffffff">' . ($day ? $day : '&nbsp;') . '</td>';
        }
      $res .= '</tr>';
      }
    $res .= '</table>';
    return $res;
    }

使用这些功能，如

  $calarr = year2array(2011);
  echo month2table(1, $calarr); // January
  echo month2table(2, $calarr); // February
  ...
  echo month2table(12, $calarr); // December

..或在for循环中放置数月。

所以...例如对于 2011年1月，您将在浏览器中看到此

enter image description here

希望这有帮助。

Answer 3

$daysArr = array('Mon','Tue','Wed','Thu','Fri','Sat','Sun');
$monthtotdays= cal_days_in_month(CAL_GREGORIAN,date('m'),date("Y"));
$currdays=jddayofweek (cal_to_jd(CAL_GREGORIAN, date('m'),1, date("Y")) , 2 );
$currdaysval = 0;
echo "<table border=1px>";
echo "<tr>";
for($d=0;$d<=6;$d++){
echo "<td>". $daysArr[$d]."</td>";
if($daysArr[$d]==$currdays)  $currdaysval = $d;
}

echo "</tr>";
echo "<tr>";
if($currdaysval > 0 ){
echo '<td colspan="'.$currdaysval.'">&nbsp;</td>';
}
for($i=1;$i<=$monthtotdays;$i++){
echo "<td>".$i."</td>";
if(($i+$currdaysval )%7 <= 0 ){
echo "</tr><tr>";
}
}
echo "</tr></table>"

Answer 4

最佳答案有不准确！如果年份是一个周期之前的闰年，您应该添加一个条件，否则您将遇到第一个月的第一天see screenshot的输出问题。

必要条件：date("L", mktime(0,0,0, 7,7, $year)) ? $days = 366 : $days = 365; （如果年份是闰年，那么周期的计数器= 366，否则365）

Answer 5

将当前代码包装在函数中，然后使用所需日期将参数传递给它。

使用日历回显所有月份

5 个答案: