带替换和选项的正则表达式

Question

我正在尝试写一个正则表达式来修改电话号码。如果它是一个国际号码(非美国)，我希望保留+符号(%2B后被URL编码)。如果它是一个国内号码，%2B应该被剥离，并更改为11位格式与一个1在前面。

这4个用例是：

• %2B2125551000变成0112125551000(这应该被看作是一个国际数字，因为它以+[2-9]开头--用011代替+ )
• %2B12125551000变成12125551000 (因为这是+1，它是一个国内号码，去掉+)
• 2125551000变成12125551000 (没有+的国内号码)
• 12125551000变成12125551000 (没有+的国内号码)

我一直试图在Linux上使用sed测试它：

进行匹配的表达式是：

((%2B)|)?((1)|)?([0-9]{10})

然而，我不一定总是需要所有的5个论点。我只需要在字符串为%2B的情况下保留%2B[2-9]。

$ for line in %2B2125551000 %2B12125551000 12125551000 2125551000;do echo $line | sed -r 's/^((%2B|))?((1)|)?([0-9]{10})/one:\1   two:\2   three:\3   four:\4   five:\5/';done
one:%2B   two:%2B   three:   four:   five:2125551000
one:%2B   two:%2B   three:1   four:1   five:2125551000
one:   two:   three:1   four:1   five:2125551000
one:   two:   three:   four:   five:2125551000

Answer 1

好吧，让我看看，你想要什么：

s/^%2B(?=[2-9])/011/ || # this will do first rule and change %2B to 011  
s/^%2B(?=1)// ||        # this will do second rule and strip off %2B  
s/^(?:[2-9])/1$&/ ;     # this will do third rule and add 1 to the beginning of the number  

# Perl code 
my @all = <DATA>;

for (@all){
    s/^%2B(?=[2-9])/011/ or 
    s/^%2B(?=1)// or
    s/^(?:[2-9])/1$&/ ; 
    print "line is $_ \n";  
}

__DATA__
%2B2125551000
%2B1125551000
225551000
125551000