Question

我正在为我的某个项目使用日历渲染库（simplecalender / donatj），但我无法弄清楚如何仅显示当月的剩余天数，而不是整月。

据我了解，这是构建日历的部分：

public function show( $echo = true ) {
    if( $this->wday_names ) {
        $wdays = $this->wday_names;
    } else {
        $today = (86400 * (date("N")));
        $wdays = array();
        for( $i = 0; $i < 7; $i++ ) {
            $wdays[] = strftime('%a', time() - $today + ($i * 86400));
        }
    }

    $this->arrayRotate($wdays, $this->offset);
    $wday    = date('N', mktime(0, 0, 1, $this->now['mon'], 1, $this->now['year'])) - $this->offset;
    $no_days = cal_days_in_month(CAL_GREGORIAN, $this->now['mon'], $this->now['year']);

如果我将no_days更改为25，则日历中仅显示1-25天。所以几乎是我想要的。

知道如何才能显示25到31（7月剩下的天数）？

感谢。

以下是代码的其余部分：

        $out = '<table cellpadding="0" cellspacing="0" class="SimpleCalendar"><thead><tr>';

    for( $i = 0; $i < 7; $i++ ) {
        $out .= '<th>' . $wdays[$i] . '</th>';
    }

    $out .= "</tr></thead>\n<tbody>\n<tr>";

    $wday = ($wday + 7) % 7;

    if( $wday == 7 ) {
        $wday = 0;
    } else {
        $out .= str_repeat('<td class="SCprefix">&nbsp;</td>', $wday);
    }

    $count = $wday + 1;
    for( $i = 1; $i <= $no_days; $i++ ) {
        $out .= '<td' . ($i == $this->now['mday'] && $this->now['mon'] == date('n') && $this->now['year'] == date('Y') ? ' class="today"' : '') . '>';

        $datetime = mktime(0, 0, 1, $this->now['mon'], $i, $this->now['year']);

        $out .= '<time datetime="' . date('Y-m-d', $datetime) . '">' . $i . '</time>';

        $dHtml_arr = false;
        if( isset($this->dailyHtml[$this->now['year']][$this->now['mon']][$i]) ) {
            $dHtml_arr = $this->dailyHtml[$this->now['year']][$this->now['mon']][$i];
        }

        if( is_array($dHtml_arr) ) {
            foreach( $dHtml_arr as $dHtml ) {
                $out .= '<div class="event">' . $dHtml . '</div>';
            }
        }

        $out .= "</td>";

        if( $count > 6 ) {
            $out .= "</tr>\n" . ($i != $count ? '<tr>' : '');
            $count = 0;
        }
        $count++;
    }
    $out .= ($count != 1 ? str_repeat('<td class="SCsuffix">&nbsp;</td>', 8 - $count) : '') . "</tr>\n</tbody></table>\n";
    if( $echo ) {
        echo $out;
    }

    return $out;
}

private function arrayRotate( &$data, $steps ) {
    $count = count($data);
    if( $steps < 0 ) {
        $steps = $count + $steps;
    }
    $steps = $steps % $count;
    for( $i = 0; $i < $steps; $i++ ) {
        array_push($data, array_shift($data));
    }
}

PHP日历天只剩下当月

0 个答案: