对于这种特定的转换，我们还可以使用格式字符串。

$new = vsprintf('%3$s-%2$s-%1$s', explode('-', $old));

显然，这不适用于许多其他日期格式转换，但由于我们在本例中只是重新排列子字符串，因此这是另一种可能的方法。

Use:

implode('-', array_reverse(explode('-', $date)));

如果没有日期转换开销，我不确定它会有多大影响。

Note: Because this post's answer sometimes gets upvoted, I came back here to kindly ask people not to upvote it anymore. My answer is ancient, not technically correct, and there are several better approaches right here. I'm only keeping it here for historical purposes. Although the documentation poorly describes the strtotime function, @rjmunro correctly addressed the issue in his comment: it's in ISO format date "YYYY-MM-DD". Also, even though my Date_Converter function might still work, I'd like to warn that there may be imprecise statements below, so please do disregard them.

投票最多的答案其实是错误的!

PHP strtotime手册在这里声明“该函数期望得到一个包含英文日期格式的字符串”。它实际上的意思是，它期望美国日期格式，如“m-d-Y”或“m/d/Y”。

这意味着以“Y-m-d”形式提供的日期可能会被strtotime错误解释。您应该以预期的格式提供日期。

我写了一个小函数，以几种格式返回日期。随意使用和修改。如果有人真的把它变成了一个类，我很高兴它能被分享。

function Date_Converter($date, $locale = "br") {

    # Exception
    if (is_null($date))
        $date = date("m/d/Y H:i:s");

    # Let's go ahead and get a string date in case we've
    # been given a Unix Time Stamp
    if ($locale == "unix")
        $date = date("m/d/Y H:i:s", $date);

    # Separate Date from Time
    $date = explode(" ", $date);

    if ($locale == "br") {
        # Separate d/m/Y from Date
        $date[0] = explode("/", $date[0]);
        # Rearrange Date into m/d/Y
        $date[0] = $date[0][1] . "/" . $date[0][0] . "/" . $date[0][2];
    }

    # Return date in all formats
        # US
        $Return["datetime"]["us"] = implode(" ", $date);
        $Return["date"]["us"]     = $date[0];

        # Universal
        $Return["time"]           = $date[1];
        $Return["unix_datetime"]  = strtotime($Return["datetime"]["us"]);
        $Return["unix_date"]      = strtotime($Return["date"]["us"]);
        $Return["getdate"]        = getdate($Return["unix_datetime"]);

        # BR
        $Return["datetime"]["br"] = date("d/m/Y H:i:s", $Return["unix_datetime"]);
        $Return["date"]["br"]     = date("d/m/Y", $Return["unix_date"]);

    # Return
    return $Return;

} # End Function

您可以尝试strftime()函数。简单示例:strftime($time， '%d %m %Y');

有两种实现方法:

1.

    $date = strtotime(date);
    $new_date = date('d-m-Y', $date);

2.

    $cls_date = new DateTime($date);
    echo $cls_date->format('d-m-Y');

function dateFormat($date)
{
    $m = preg_replace('/[^0-9]/', '', $date);
    if (preg_match_all('/\d{2}+/', $m, $r)) {
        $r = reset($r);
        if (count($r) == 4) {
            if ($r[2] <= 12 && $r[3] <= 31) return "$r[0]$r[1]-$r[2]-$r[3]"; // Y-m-d
            if ($r[0] <= 31 && $r[1] != 0 && $r[1] <= 12) return "$r[2]$r[3]-$r[1]-$r[0]"; // d-m-Y
            if ($r[0] <= 12 && $r[1] <= 31) return "$r[2]$r[3]-$r[0]-$r[1]"; // m-d-Y
            if ($r[2] <= 31 && $r[3] <= 12) return "$r[0]$r[1]-$r[3]-$r[2]"; //Y-m-d
        }

        $y = $r[2] >= 0 && $r[2] <= date('y') ? date('y') . $r[2] : (date('y') - 1) . $r[2];
        if ($r[0] <= 31 && $r[1] != 0 && $r[1] <= 12) return "$y-$r[1]-$r[0]"; // d-m-y
    }
}

var_dump(dateFormat('31/01/00')); // return 2000-01-31
var_dump(dateFormat('31/01/2000')); // return 2000-01-31
var_dump(dateFormat('01-31-2000')); // return 2000-01-31
var_dump(dateFormat('2000-31-01')); // return 2000-01-31
var_dump(dateFormat('20003101')); // return 2000-01-31