不可思议的行为与 strftime

Question

我有这个代码：

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;
    }

当我删除将我的数字$month从参数转换为带有 strftime（"%b"，$date）的字符串的行时，它给出了良好的行为。当我添加这一行时，var $day在每月的第一天开始重复 9 次......这是星期二，无法得到解决方案，这对我来说是一个错误......

Answer 1

您要替换在以下位置中使用的变量：

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

因此，在每月的第二天，您不是尝试解析2019-1-2而是解析2019-Jan-1。我不知道为什么，但是当您将这种格式与个位数的日期一起使用时，它总是解析得好像它是2019-01-01.

解决方案是使用不同的变量。

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));

到

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

Answer 3

哦，

我的去!!非常感谢你们，几天来我一直在寻找我做错了什么......

只是一个变量名称与我的数字$month冲突，字符串月份与 strftime 冲突。所以我替换了字符串的名称。

大错特错！ ^^