在 PHP 中为日期添加前导零 - adding leading zero to dates in php

adding leading zero to dates in php

本文关键字：添加日期 PHP | 更新日期: 2023-09-27

我正在尝试在日期 1 到 9 的前面添加一个 0，因为日期列在下拉菜单中。这是我的代码，我以为使用 d 会添加前导零，但似乎不起作用。我没有太多的 php 经验，所以这是一个远景......提前谢谢你！

<?PHP 
FUNCTION DateSelector($inName, $useDate=0) 
{ 
    /* create array so we can name months */ 
    $monthName = ARRAY(1=> "January", "February", "March", 
        "April", "May", "June", "July", "August", 
        "September", "October", "November", "December"); 
    /* if date invalid or not supplied, use current time */ 
    IF($useDate == 0) 
    { 
        $useDate = TIME(); 
    } 
    /* make month selector */ 
    ECHO "<SELECT NAME=" . $inName . "month>'n"; 
    FOR($currentMonth = 1; $currentMonth <= 12; $currentMonth++) 
    { 
        ECHO "<OPTION VALUE='""; 
        ECHO INTVAL($currentMonth); 
        ECHO "'""; 
        IF(INTVAL(DATE( "m", $useDate))==$currentMonth) 
        { 
            ECHO " SELECTED"; 
        } 
        ECHO ">" . $monthName[$currentMonth] . "'n"; 
    } 
    ECHO "</SELECT>"; 
    /* make day selector */ 
    ECHO "<SELECT NAME=" . $inName . "day>'n"; 
    FOR($currentDay=1; $currentDay <= 31; $currentDay++) 
    { 
        ECHO "<OPTION VALUE='"$currentDay'""; 
        IF(INTVAL(DATE( "d", $useDate))==$currentDay) 
        { 
            ECHO " SELECTED"; 
        } 
        ECHO ">$currentDay'n"; 
    } 
    ECHO "</SELECT>"; 
    /* make year selector */ 
    ECHO "<SELECT NAME=" . $inName . "year>'n"; 
    $startYear = DATE( "Y", $useDate); 
    FOR($currentYear = $startYear - 0; $currentYear <= $startYear+2;$currentYear++) 
    { 
        ECHO "<OPTION VALUE='"$currentYear'""; 
        IF(DATE( "Y", $useDate)==$currentYear) 
        { 
            ECHO " SELECTED"; 
        } 
        ECHO ">$currentYear'n"; 
    } 
    ECHO "</SELECT>"; 
} 
?>

如果我理解正确，你需要这个：

$day = 1;    
echo str_pad($day, 2, 0, STR_PAD_LEFT);

您可以替换：

ECHO INTVAL($currentMonth);

跟：

printf("%02s", $currentMonth);

$day_with_leading_zeroes = sprintf("%02d", $day);

而不是：

ECHO ">$currentDay'n";

您可以键入：

echo ">".($currentDay<10 ? "0" : "").$currentDay."'n";

$date =4
$month = 6
$year = 2013

如果要以这种格式显示在上面。 2013/6/4

printf('%02d/%02d/%04d', $date, $month, $year);
$date =14
$month = 12
$year = 2013

如果要以这种格式显示在上面。 2013/12/14

printf('%02d/%02d/%04d', $date, $month, $year);