用-p将参数传递给perl

Question

我有一个简短的Perl脚本，用于对文件运行正则表达式替换： perl -pi -e 's/x/y/' <file>我想将参数传递给脚本，因此它将用命令行参数替换y （例如，类似perl -pi -e 's/x/$argv[1]/' <file>但是当使用-pi参数时，我使用$ argv [1]不起作用。

Answer 1

I have never understood the preoccupation with one-line Perl programs, and I don't understand why you can't write a Perl script file to do this.

However, you can remove an item from @ARGV and save it in a variable before the -p loop starts. Like this

perl -p -i -e 'BEGIN{ $r = pop } s/x/$r/' <file> <replacement>

Answer 2

Solution 1: Roll out your own loop.

perl -i -e'$r = shift; while (<>) { s/x/$r/; print }' "$replacement" "$file"

Solution 2: Grab the var at compile time.

perl -i -pe'BEGIN { $r = shift; } s/x/$r/' "$replacement" "$file"

Solution 3: Use an env var instead of an argument

R="$replacement" perl -i -pe's/x/$ENV{R}/' "$file"

Answer 3

perl -pi -e 's/x/y/' <file> is equivalent to this:

while (<>) {
   s/x/y/;
} 
continue {
   print or die "-p failed: $!\n";
}

This will pass every line of <file> through the expression s/x/y/; and print the result. (Note the -i flag will modify <file> inplace). If you really need to pass in a replacement, why don't you list it directly in the substitution. Either way it is still listed only once.

perl -pi -e 's/x/<replacement>/;' <file>