Perl搜索并替换变量和捕获组

As the question says, I am trying to do a search replace using a variable and a capture group. That is, the replace string contains $1. I followed the answers here and here, but they did not working for me; $1 comes through in the replace. Can you help me spot my problem?

正如问题所说，我正在尝试使用变量和捕获组进行搜索替换。也就是说，替换字符串包含$ 1。我按照这里和这里的答案，但他们没有为我工作;更换中需要1美元。你能帮我发现我的问题吗？

I am reading my regular expressions from a file like so:

我正在从这样的文件中读取我的正则表达式：

while( my $line = <$file>) {
    my @findRep = split(/:/, $line);
    my $find = $findRep[0];
    my $replace = '"$findRep[2]"'; # This contains the $1
    $allTxt =~ s/$find/$replace/ee;
}

If I manually set my $replace = '"$1 stuff"' the replace works as expected. I have played around with every single/double quoting and /e combination I can think of.

如果我手动设置我的$ replace ='“$ 1 stuff”，则替换按预期工作。我已经玩过我能想到的每一个/双引号和/或组合。

2 个解决方案

#1

You're using single quotes so $findRep[2] isn't interpolated. Try this instead:

你使用的是单引号，所以$ findRep [2]没有插值。试试这个：

my $replace = qq{"$findRep[2]"};

#2

Why regex replacement when you already have your values in @findRep

当你已经在@findRep中拥有值时，为什么要更换正则表达式

while( my $line = <$file>) {
    my @findRep = split(/:/, $line);
    $findRep[0] = $findRep[2];
    my $allTxt = join(":", @findRep);
}

#1