Question

我有以下代码：

function calendarDay ($month, $year)
    {
        $num = cal_days_in_month(CAL_GREGORIAN, $month, $year);
        $today = strtotime(date("Y-m-d"));
        $result = array();
        $str = "";
        $strMonth = "";
        for ($i = 0; $i < $num; $i++) {
            $date = strtotime($year . "-" . $month . "-" . ($i + 1));
            $day=strftime("%a", $date);
            $month = strftime("%b", $date);
            if ($today == $date) {
                if ($day == "Sat" || $day == "Sun") {
                    $str .= "<th style='background-color:mediumseagreen'>" . $day. "</th>";
                    $strMonth = $strMonth . "<th style='background-color:mediumseagreen'>".($i + 1) . "-" . $month." ". "</th>";
                }
                else {
                    $str .= "<th style='background-color:#888888'>" . $day. "</th>";
                    $strMonth = $strMonth . "<th style='background-color:#888888'>".($i + 1) . "-" . $month." ". "</th>";
                }
            }
            else if ($today != $date) {
                if ($day == "Sat" || $day == "Sun") {
                    $str .= "<th style='background-color:mediumseagreen'>" . $day. "</th>";
                    $strMonth = $strMonth . "<th style='background-color:mediumseagreen'>".($i + 1) . "-" . $month." ". "</th>";
                }
                else {
                    $str .= "<th>" . $day. "</th>";
                    $strMonth = $strMonth . "<th>".($i + 1) . "-" . $month." ". "</th>";
                }

            }
            $result = array_merge($result, array("Month" => $strMonth));
            $result = array_merge($result, array("Day" => $str));
        }
        return $result;
    }

当我删除使用strftime（“％b”，$ date）将数字$ month从参数转换为字符串的行时，它表现出良好的行为。当我添加这行代码时，var $ day开始在每月的第一天重复9次...直到星期二，并且无法获得解决方案，这对我来说是个错误...

Answer 1

您要替换在以下位置使用的变量：

$date = strtotime($year . "-" . $month . "-" . ($i + 1));

因此，在每月的第二天，不是解析2019-1-2，而是解析2019-Jan-1。我不确定为什么，但是当您将此格式与一位数字的日期结合使用时，它总是像2019-01-01一样进行解析。

解决方案是使用其他变量。

function calendarDay ($month, $year)
{
    $num = cal_days_in_month(CAL_GREGORIAN, $month, $year);
    $today = strtotime(date("Y-m-d"));
    $result = array();
    $str = "";
    $strMonth = "";
    for ($i = 0; $i < $num; $i++) {
        $date = strtotime($year . "-" . $month . "-" . ($i + 1));
        $day=strftime("%a", $date);
        $monthName = strftime("%b", $date);
        if ($today == $date) {
            if ($day == "Sat" || $day == "Sun") {
                $str .= "<th style='background-color:mediumseagreen'>" . $day. "</th>";
                $strMonth = $strMonth . "<th style='background-color:mediumseagreen'>".($i + 1) . "-" . $monthName." ". "</th>";
            }
            else {
                $str .= "<th style='background-color:#888888'>" . $day. "</th>";
                $strMonth = $strMonth . "<th style='background-color:#888888'>".($i + 1) . "-" . $monthName." ". "</th>";
            }
        }
        else if ($today != $date) {
            if ($day == "Sat" || $day == "Sun") {
                $str .= "<th style='background-color:mediumseagreen'>" . $day. "</th>";
                $strMonth = $strMonth . "<th style='background-color:mediumseagreen'>".($i + 1) . "-" . $monthName." ". "</th>";
            }
            else {
                $str .= "<th>" . $day. "</th>";
                $strMonth = $strMonth . "<th>".($i + 1) . "-" . $monthName." ". "</th>";
            }

        }
        $result = array_merge($result, array("Month" => $strMonth));
        $result = array_merge($result, array("Day" => $str));
    }
    return $result;
}

Answer 2

替换代码

$date = strtotime($year . "-" . $month . "-" . ($i + 1));

进入

$date = strtotime($year . "-" . $month . "-" .str_pad(($i + 1), 2, '0', STR_PAD_LEFT) );

Answer 3

哦，我走了，谢谢大家！！！很多天以来，我一直在搜索自己在做错什么...

一个变量名与我的数字$ month和带strftime的字符串month冲突。所以我将字符串的名称替换为一个。

严重错误！ ^^

strftime令人难以置信的行为

3 个答案: