MSG_OUT="<B><I>Skipping<N> all libraries and fonts...<N>"

perl -ne '%ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m"); while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }'

这个 perl one-liner 将令牌交换为转义序列。
它按预期作业，但前提是输入被换行符包围。

IE

echo "\x0a${MSG_OUT}\x0a" | perl -ne '.... etc.

从标准输入读取时如何避免这个问题？

uj5u.com热心网友回复：

-n将您的代码包装在while (<>) { ... }* (cf perldoc perlrun ) 中。因此，您的单线相当于：

perl -e '
   while(<>) {
      %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
      while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }
   }
'

[为可读性添加了换行符。如果您愿意，可以将它们移除。]

看到双了while (<>) { ... }吗？那是你的问题：第一个while（由 增加的那个-n）读取一行，然后第二个while（你写的那个）读取第二行，做你的s///（在第二行），并列印第二行更新。因此，在要处理的实际行之前需要一个空行。

要解决此问题，请洗掉内部while(<>)，或洗掉-n标志。例如：

perl -e '
   %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
   while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }
' 

或者，

perl -ne '
   BEGIN { %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m") };
   s/(<([BINORSU])>)/\e$ES{$2}/g; print;
'

请注意，您可以使用，而不是使用-nand print，-p这-n与print末尾额外的**相同：

perl -pe '
    BEGIN { %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m") };
    s/(<([BINORSU])>)/\e$ES{$2}/g;
'

*为完整起见，请注意在回圈 ( )之前-n添加标签，尽管这在您的情况下无关紧要。LINEwhileLINE: while(<>) { ... }

**的print增加通过-p实际上是在continue之后的块while，尽管再次，这件事情并没有你的情况。