[英]How to match a fixed number of digits with regex in PHP?
I want to retrieve the consecutive 8 digits out of a string. 我想从字符串中检索连续的8位数。
"hello world,12345678anything else"
should return 12345678
as result(the space in between is optional). 应该返回
12345678
(两者之间的空格是可选的)。
But this should not return anything: 但这不应该返回任何东西:
"hello world,123456789anything else"
Because it has 9 digits,I only need 8 digits unit. 因为它有9位数,所以我只需要8位数单位。
尝试
'/(?<!\d)\d{8}(?!\d)/'
$var = "hello world,12345678798anything else";
preg_match('/[0-9]{8}/',$var,$match);
echo $match[0];
You need to match the stuff on either side of the 8 digits. 你需要匹配8位数字两边的东西。 You can do this with zero-width look-around assertions, as exemplified by @S Mark, or you can take the simpler route of just creating a backreference for the 8 digits:
您可以使用零宽度环视断言来实现此操作,例如@S Mark,或者您可以采用更简单的方法来创建8位数的反向引用:
preg_match('/\D(\d{8})\D/', $string, $matches)
$eight_digits = $matches[1];
But this won't match when the digits start or end a line or string; 但是当数字开始或结束一条线或字符串时,这将不匹配; for that you need to elaborate it a bit:
为此你需要详细说明:
preg_match('/(?:\D|^)(\d{8})(?:\D|$)/', $string, $matches)
$eight_digits = $matches[1];
The (?:...)
in this one allows you to specify a subset of alternates, using |
此中的
(?:...)
允许您使用|
指定替换子集 , without counting the match as a back-reference (ie adding it to the elements in the array $matches
). ,不将匹配计为反向引用(即将其添加到数组
$matches
的元素)。
For many more gory details of the rich and subtle language that is Perl-Compatible Regular Expression syntax, see http://ca3.php.net/manual/en/reference.pcre.pattern.syntax.php 有关Perl兼容正则表达式语法的丰富和微妙语言的更多详细信息,请参阅http://ca3.php.net/manual/en/reference.pcre.pattern.syntax.php
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