Question

$month = ['January', 'February', 'March', 'April', 'May', 'June', 'July',  'August', 'September', 'October', 'November', 'December'];
$str = "The fly went away in 1990-11-20";
$pattern = array('/(\d{4})-(\d{1,2})-(\d{1,2})/');
$replacement = array('${3}.'.'$month[\2 -1]<-- I don't know'.' ${1}');
$res = preg_replace($pattern, $replacement, $str);
echo $res;

结果应该是“飞行于1990年11月20日消失”。（我知道这种模式可能是毛坯发动机。但现在还可以。）什么是“$ month [\ 2 -1]”的替换？

Answer 1

您无法直接使用preg_replace。试试这个

function doReplace($matches) {
    $month = ['January', 'February', 'March', 'April', 'May', 'June', 'July',  'August', 'September', 'October', 'November', 'December'];
    return $matches[3] . ' ' . $month[$matches[2] - 1] . ' ' . $matches[1];
}

$str = "The fly went away in 1990-11-20";
$pattern = array('/(\d{4})-(\d{1,2})-(\d{1,2})/');
$res = preg_replace_callback($pattern, 'doReplace', $str);
echo $res;

使用preg_replace_callback - 有关详细信息，请参阅手册。

Answer 2

使用preg_replace_callback()：

$str = preg_replace_callback(
           $pattern,
           function (array $matches) use ($month) {
              return $matches[3].'.'.$month[$matches[2]-1].' '.$matches[1];
           },
           $str
       );

Answer 3

试试这个工作示例，它会为您提供所需的输出

    //echo date('d.F Y',strtotime('1990-11-20 00:00:00'));
    $month = array('','January','February','March', 'April', 'May', 'June', 'July',  'August', 'September', 'October', 'November', 'December');
    $str = "The fly went away in 1990-11-20";
    $pattern = array('/(\d{4})-(\d{1,2})-(\d{1,2})/');
    $replacement = array('${3}.'.$month[12-1].' ${1}');
    $res = preg_replace($pattern, $replacement, $str);
    echo $res;

数组中的Preg_replace索引替换为模式$ arr [{...}]

3 个答案: