Question

我有以下几部PHP代码,用于配对一年:

$pattern = "/20dd/";

$exp = "BORN ON:        Friday, June 11, 2001 ";

$match = preg_match($pattern,$exp, $matches);

echo $matches[0];

This matches "2001"...which is what I am trying to match--the birth year. But this is clearly a hack of a solution and can run into problems if there are other years before $exp for some reason. How can I include the "BORN ON" information to pick out just the year?

Answer 1

还搜索“BORN ON”并利用一个捕获组。

$pattern = "/BORN ON.*(20dd)s$/";
$exp = "BORN ON:        Friday, June 11, 2001 ";
$match = preg_match($pattern,$exp, $matches);
echo $matches[1];

由于现在的方括号,我把国际年归入一个捕获组,你将这个组带上<代码>$matches >> (而不是0)。此外,我把年限到座尾(代号为:>>,与$。

Answer 2

为此:

 /BORN ON:s+w+, w+ d+, (20dd)/

Answer 3

首先,所有投入数据都实现正常化。如果你重新划分为固定宽度或采用具体、简单的格式,那么就会注意到,并开发出一种能够更好地适应投入数据的可操作性。如您需要快速参考,请查询。

Answer 4

为此:

$pattern = "BORN ON.*/20dd/";

有时,从一个使用

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