我有一个脚本提供某些IP地址的信息。
我想从文本中提取国家。
在以下文字的国家/地区中为"Country: US"
我想只显示US
案文是:
[Querying whois.arin.net]
[whois.arin.net]
#
# Query terms are ambiguous. The query is assumed to be:
# "n 173.194.74.100"
#
# Use "?" to get help.
#
#
# The following results may also be obtained via:
# http://whois.arin.net/rest/nets;q=173.194.74.100?showDetails=true&showARIN=false&ext=netref2
#
NetRange: 173.194.0.0 - 173.194.255.255
CIDR: 173.194.0.0/16
OriginAS: AS15169
NetName: GOOGLE
NetHandle: NET-173-194-0-0-1
Parent: NET-173-0-0-0-0
NetType: Direct Allocation
RegDate: 2009-08-17
Updated: 2012-02-24
Ref: http://whois.arin.net/rest/net/NET-173-194-0-0-1
OrgName: Google Inc.
OrgId: GOGL
Address: 1600 Amphitheatre Parkway
City: Mountain View
StateProv: CA
PostalCode: 94043
Country: US
RegDate: 2000-03-30
Updated: 2011-09-24
Ref: http://whois.arin.net/rest/org/GOGL
OrgTechHandle: ZG39-ARIN
OrgTechName: Google Inc
OrgTechPhone: +1-650-253-0000
OrgTechEmail: arin-contact@google.com
OrgTechRef: http://whois.arin.net/rest/poc/ZG39-ARIN
OrgAbuseHandle: ZG39-ARIN
OrgAbuseName: Google Inc
OrgAbusePhone: +1-650-253-0000
OrgAbuseEmail: arin-contact@google.com
OrgAbuseRef: http://whois.arin.net/rest/poc/ZG39-ARIN
#
# ARIN WHOIS data and services are subject to the Terms of Use
# available at: https://www.arin.net/whois_tou.html
#
答案 0 :(得分:2)
如果它只是你需要的正则表达式 - 试试这个 - 国家ID将在第一组
Country:\s*([A-Z]{2})
Country:
- 匹配文字\s*
- 匹配任意数量的空格,标签等。([A-Z]{2})
- 匹配并捕获任意字母(大写)两次如果您需要出现此模式,请使用preg_match_all
答案 1 :(得分:2)
使用preg_match,您可以执行以下操作:
if (preg_match('/^Country:\s*([A-Z]{2,3)$/m', $str, $match)) {
echo $match[1];
}
答案 2 :(得分:1)
有一个用于处理whois数据的phpwhois库。它会以阵列的形式为您提供响应。
答案 3 :(得分:0)
使用preg_match
进行提取preg_match("/Country:(.*)\"/siU", $str, $match);
echo trim($match[1]);
答案 4 :(得分:0)
$regex = "/country:[\ \t\r\n\f][A-Z]+\s/";
$txt = "descr: NCC#200X44704917
country: FR
admin-c: ACPSA223-RIPE
tech-c: TCWQQP8-RIPE";
preg_match($regex, $txt, $result);
print_r($result);
------------------------------------
数组([0] =>国家:FR)