在 Perl 中读取 stdin 需要在输入周围换行。如何避免？

Question

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 单行代码将标记替换为转义序列。
它按预期工作，但前提是输入被换行符包围。

IE

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

从 stdin 读取时如何避免此问题？

Answer 1

-n将您的代码包装在while (<>) { ... }* 中（参见perldoc perlrun）。因此，你的一句话相当于：

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

[添加换行符以提高可读性。如果您愿意，可以将它们删除。]

看到双while (<>) { ... }吗？这就是你的问题：第一个while（由 所添加的-n）读取一行，然后第二个while（你编写的）读取第二行，执行你的s///（在第二行上），并打印更新的第二行。因此，您需要在要处理的实际行之前有一个空行。

要解决此问题，请删除 insidewhile(<>)或删除该-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 ，这与末尾额外的**相同：print-p-nprint

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

*为了完整性，请注意在循环 ( )之前-n添加标签，尽管这在您的情况下并不重要。LINEwhileLINE: while(<>) { ... }

**添加print的-p实际上是在continue后面的块中while，尽管这对于您的情况来说并不重要。