如何使用perl获取值的索引或变量长度(类似于java的动态列表长度)？

Question

我有2个输出: -

7: ib1: <BROADCAST,MULTICAST,SLAVE,UP,LOWER_UP>    
9: bondib0: <BROADCAST,MULTICAST,MASTER,UP,LOWER_UP>

使用perl我想获得substr ib1和bondib0.目前,如果我设置

substr($line,3,3); this will return ib1
substr($line,3,7); this will return bondib0

我想在上面有一个substr,怎么做？

像substr($ line,3,index($ line,":"));

请告诉我如何获得3和7个以上索引的动态值,因为值的长度不同.

Answer 1

你的使用方法substr,index应该工作.但是,还有另外两种方法可以做到这一点.

更容易的是使用正则表达式并进行模式匹配.

while (my $line = <DATA>) {
    if ($line =~ m/: ([^:]+):/) {
        print $1, "\n";
    }
}

__DATA__
7: ib1: <BROADCAST,MULTICAST,SLAVE,UP,LOWER_UP>    
9: bondib0: <BROADCAST,MULTICAST,MASTER,UP,LOWER_UP>

这将匹配冒号:和空格上的字符串, and then capture everything that is not a colon, until it encounters another colon. See regex101.com for an explanation, and take a look at perlretut for a gentle introduction to regex.

An alternative would be to use split在冒号和空间上使用.那里只有两个.

while (my $line = <DATA>) {
    ( undef, my $interface ) = split /: /, $line;
    print $interface;
}

__DATA__
7: ib1: <BROADCAST,MULTICAST,SLAVE,UP,LOWER_UP>    
9: bondib0: <BROADCAST,MULTICAST,MASTER,UP,LOWER_UP>

它会将弦打成碎片": ",丢弃第一部分,保存第二部分并丢弃其余部分.

然而,正则表达式解决方案更好.