preg_replace 输出中未评估时间

Question

我写了一个使用 preg_replace 转换日期格式的代码。代码如下：

$pattern=array(
    "#Y#",//full year
    "#y#",//short year

    "#M#",//month short name
    "#F#",//month full name
    "#m#",//month number 0 lead
    "#n#",//month number
    "#t#",//days in month

    "#l#",//full week day
    "#D#",//short week day

    "#d#",//day number of month
    "#j#",//day number of month

    "#a#",//AM/PM short view
    "#A#",//AM/PM full view
            );
$replace=array(
    $d->ENnum2FA($converted[0]),//year 13xx
    $d->ENnum2FA(substr($converted[0],2),true),//year xx lead zero

    $d->shmonths[$converted[1]],//month name
    $d->months[$converted[1]],//month name
    $d->ENnum2FA($converted[1],true), //month number
    $d->ENnum2FA($converted[1]), //month number
    //$converted[1],
    $d->j_days_in_month[$converted[1]],

    $d->days[strtolower(gmdate("D",$stamp))],//week day {full view}
    $d->ldays[strtolower(gmdate("D",$stamp))],//week day ‍‍{short view}

    $d->ENnum2FA($converted[2],true),//day of month
    $d->ENnum2FA($converted[2],true),//day of month

    $d->pmam[gmdate('a',$stamp)],
    $d->pmam[gmdate('A',$stamp)],

    );
// $format = "Y/m/d"; example
$date= preg_replace($pattern,$replace,$format);

它完美地改变了日期格式，但问题是它将时间输出为 H:i:s 而不是时间值！例如输出是 1398/5/21, H:i:s 而不是 1398/5/21, 22:15:36。所以，我添加了以下代码：

$time_f = preg_replace_callback(
    "#([His])#",
    function ($matches) {
        return(gmdate($matches[1],$stamp));
    },
    $date
);

问题解决了。但现在它总是显示时间为：00:00:00 例如：1398/5/21，00:00:00

我该如何解决这个问题？

Answer 1

在您的 preg_replace_callback() 函数中，您使用了未定义的局部变量 $stamp。如果需要从外部作用域继承这个变量，需要使用use()选项。

$time_f = preg_replace_callback(
    "#([His])#",
    function ($matches) use ($stamp) {
        return(gmdate($matches[1],$stamp));
    },
    $date
);

preg_replace 输出中未评估时间

Time is not evaluated in preg_replace output

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