perl 正则表达式替换中的条件

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如果匹配一组，我正在尝试使用 perl regex one-liner 返回不同的替换结果.到目前为止，我已经有了这个:

I'm trying to return different replacement results with a perl regex one-liner if it matches a group. So far I've got this:

echo abcd | perl -pe "s/(ab)(cd)?/defined($2)?\1\2:''/e"

但我明白

Backslash found where operator expected at -e line 1, near "1\" (Missing operator before \?) syntax error at -e line 1, near "1\" Execution of -e aborted due to compilation errors.

如果输入是 abcd 我想得到 abcd ，如果它是 ab 我想得到一个空字符串.我哪里出错了?

If the input is abcd I want to get abcd out, if it's ab I want to get an empty string. Where am I going wrong here?

推荐答案

您使用了正则表达式原子 \1 和 \2(匹配第一个或第二个捕获捕获的内容)在正则表达式模式之外.您打算使用 $1 和 $2(就像您在另一个地方所做的那样).

You used regex atoms \1 and \2 (match what the first or second capture captured) outside of a regex pattern. You meant to use $1 and $2 (as you did in another spot).

此外，双引号字符串中的美元符号对您的外壳有意义.最好在程序周围使用单引号[1].

Further more, dollar signs inside double-quoted strings have meaning to your shell. It's best to use single quotes around your program[1].

echo abcd | perl -pe's/(ab)(cd)?/defined($2)?$1.$2:""/e'

更简单:

echo abcd | perl -pe's/(ab(cd)?)/defined($2)?$1:""/e'

更简单:

echo abcd | perl -pe's/ab(?!cd)//'

避免在程序中使用单引号[2]，或者使用 '\'' 来转义"它们.

您通常可以使用 q{} 而不是单引号.您也可以切换到使用双引号.在双引号内，您可以使用 \x27 作为撇号.

Either avoid single-quotes in your program[2], or use '\'' to "escape" them.

You can usually use q{} instead of single-quotes. You can also switch to using double-quotes. Inside of double-quotes, you can use \x27 for an apostrophe.