如何在Perl替换运算符的替换端使用变量? [英] How to use a variable in the replacement side of the Perl substitution operator?

问题描述

我想执行以下操作:

$find="start (.*) end";
$replace="foo \1 bar";

$var = "start middle end";
$var =~ s/$find/$replace/;

我希望$ var包含"foo middle bar"，但是它不起作用.都没有:

I would expect $var to contain "foo middle bar", but it does not work. Neither does:

$replace='foo \1 bar';

以某种方式，我在逃逸方面缺少了一些东西.

Somehow I am missing something regarding the escaping.

我修复了丢失的's'

推荐答案

在替换方面，您必须使用$ 1，而不是\ 1.

On the replacement side, you must use $1, not \1.

并且您只能通过替换一个给出您想要的结果的evalable表达式并使用/ee修饰符告诉s///来对其进行评估，来做您想做的事情，如下所示:

And you can only do what you want by making replace an evalable expression that gives the result you want and telling s/// to eval it with the /ee modifier like so:

$find="start (.*) end";
$replace='"foo $1 bar"';

$var = "start middle end";
$var =~ s/$find/$replace/ee;

print "var: $var\n";

要了解为什么需要使用"和double/e，请在此处查看double eval的效果:

To see why the "" and double /e are needed, see the effect of the double eval here:

$ perl
$foo = "middle";
$replace='"foo $foo bar"';
print eval('$replace'), "\n";
print eval(eval('$replace')), "\n";
__END__
"foo $foo bar"
foo middle bar

(尽管正如池上所指出的那样，单个/e或双e的第一个/e并不是真正的eval()；而是，它告诉编译器该替换是要编译的代码，而不是字符串.尽管如此. ，eval(eval(...))仍然说明了为什么需要做所需的工作才能使/ee正常工作.)

(Though as ikegami notes, a single /e or the first /e of a double e isn't really an eval(); rather, it tells the compiler that the substitution is code to compile, not a string. Nonetheless, eval(eval(...)) still demonstrates why you need to do what you need to do to get /ee to work as desired.)

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