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{$2}/g; print; }'
這個 perl one-liner 將令牌交換為轉義序列。
它按預期作業，但前提是輸入被換行符包圍。
IE
echo "\x0a${MSG_OUT}\x0a" | perl -ne '.... etc.
從標準輸入讀取時如何避免這個問題？
uj5u.com熱心網友回復：
-n將您的代碼包裝在while (<>) { ... }* (cf perldoc perlrun ) 中。因此，您的單線相當于：
perl -e '
while(<>) {
%ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }
}
'
[為可讀性添加了換行符。如果您愿意，可以將它們移除。]
看到雙了while (<>) { ... }嗎？那是你的問題：第一個while（由 增加的那個-n）讀取一行，然后第二個while（你寫的那個）讀取第二行，做你的s///（在第二行），并列印第二行更新。因此，在要處理的實際行之前需要一個空行。
要解決此問題，請洗掉內部while(<>)，或洗掉-n標志。例如：
perl -e '
%ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
while (<>) { s/(<([BINORSU])>)/\e$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{$2}/g; print;
'
請注意，您可以使用，而不是使用-nand print，-p這-n與print末尾額外的**相同：
perl -pe '
BEGIN { %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m") };
s/(<([BINORSU])>)/\e$ES{$2}/g;
'
*為完整起見，請注意在回圈 ( )之前-n添加標簽，盡管這在您的情況下無關緊要。LINEwhileLINE: while(<>) { ... }
**的print增加通過-p實際上是在continue之后的塊while，盡管再次，這件事情并沒有你的情況。
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標籤：perl
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