循环全局替换 [英] Loop through global substitution

问题描述

$num = 6;
$str = "1 2 3 4";
while ($str =~ s/\d/$num/g)
{
    print $str, "\n";
    $num++;
}

是否可以在 perl 中执行上述操作?我希望循环只运行 4 次，并以 $str 为 6 7 8 9 结束.

Is it possible to do something like the above in perl? I would like the loop to run only 4 times and to finish with $str being 6 7 8 9.

推荐答案

您不需要循环:/g 修饰符会导致替换重复进行多次匹配.您想要的是/e 修饰符来计算替换.假设你得到的效果是每个数字加5，示例代码如下.

You don't need the loop: the /g modifier causes the substitution to be repeated as many times as it matches. What you want is the /e modifier to compute the substitution. Assuming the effect you were after is to add 5 to each number, the example code is as follows.

$str = "1 2 3 4";
$str =~ s/(\d)/$1+5/eg;
print "$str\n";

如果您真的想替换以 6 开头的数字，那么这行得通.

If you really wanted to substitute numbers starting with 6, then this works.

$num = 6;
$str = "1 2 3 4";
$str =~ s/\d/$num++/eg;
print "$str\n";

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