Php preg_replace-don';如果$4为空,则不包括字符串
我有这样的表达:Php preg_replace-don';如果$4为空,则不包括字符串,php,regex,preg-replace,Php,Regex,Preg Replace,我有这样的表达: $regex_phone = '/^(?:1(?:[. -])?)?(?:\((?=\d{3}\)))?([2-9]\d{2})' .'(?:(?<=\(\d{3})\))? ?(?:(?<=\d{3})[.-])?([2-9]\d{2})' .'[. -]?(\d{4})(?: (?i:ext)\.? ?(\d{1,5}))?$/'; if(!preg_match($regex_phone, $data[
$regex_phone = '/^(?:1(?:[. -])?)?(?:\((?=\d{3}\)))?([2-9]\d{2})'
.'(?:(?<=\(\d{3})\))? ?(?:(?<=\d{3})[.-])?([2-9]\d{2})'
.'[. -]?(\d{4})(?: (?i:ext)\.? ?(\d{1,5}))?$/';
if(!preg_match($regex_phone, $data['phone'])){
$error[] = "Please enter a valid phone number.";
}else{
$data['phone'] = preg_replace($regex_phone, '($1) $2-$3 ext.$4', $data['phone']);
}
$regex\u phone='/^(?:1(?:[.-])?(?:\((?=\d{3}\))?([2-9]\d{2})
.(?:(?我不确定这个花哨的正则表达式是否真的必要
但是让我们看看
啊,我修改了大约4美元的区域,以便
考虑到缺少分机号码。你能发现吗
区别是什么
...
$regex_phone =
'/^
(?:
1(?:[. -])?
)?
(?:
\( (?=\d{3} \) ) # possible mistake? (probably correct)
)?
([2-9]\d{2}) # $1
(?:
(?<=\(\d{3})\)
)?
\s?
(?:
(?<=\d{3})
[.-]
)?
([2-9]\d{2}) # $2
[.\s-]?
(\d{4}) # $3
(?:
\s? ext [.\s]?
(\d{1,5})? # $4
)?
$/ix';
if( preg_match( $regex_phone, $data['phone'], $m) ) {
$data['phone'] = "$m[1] $m[2]-$m[3]" . (isset($m[4]) ? " ext.$m[4]" : '');
}
else {
$error[] = "Please enter a valid phone number.<br />";
}
...
。。。
$regex_电话=
'/^
(?:
1(?:[. -])?
)?
(?:
\((?=\d{3}\)#可能的错误?(可能正确)
)?
([2-9]\d{2})#$1
(?:
(?甚至没有考虑过获取preg_match变量。工作非常出色。我做了一些修改以适应我的编码风格,但isset()和$m是我所需要的。谢谢rbo