Question

我有一个单行的Perl脚本（Perl v5.10.1），其行为方式我无法理解或解释：

$ cat packer-build.log |
>   perl -MPOSIX -pe \
>     's/^(\d+)/my $x = strftime("%Y-%m-%d %H:%M", localtime($1)); print "$x "/e' | tail -5
2017-06-21 11:11 1,,error-count,1
2017-06-21 11:11 1,,ui,error,\n==> Some builds didn't complete successfully and had errors:
2017-06-21 11:11 1,amazon-ebs,error,unexpected EOF
2017-06-21 11:11 1,,ui,error,--> amazon-ebs: unexpected EOF
2017-06-21 11:11 1,,ui,say,\n==> Builds finished but no artifacts were created.

未过滤的输出：

$ tail -5 packer-build.log 
1498007467,,error-count,1
1498007467,,ui,error,\n==> Some builds didn't complete successfully and had errors:
1498007467,amazon-ebs,error,unexpected EOF
1498007467,,ui,error,--> amazon-ebs: unexpected EOF
1498007467,,ui,say,\n==> Builds finished but no artifacts were created.

请注意，未经过滤的输出中不存在1之后的2017-06-21 11:11。

导致打印的原因是什么？如何修改代码以便我将字段1中的纪元时间替换为人类可读的输出？

Answer 1

在替换中使用/e时，匹配将替换为替换代码评估的内容。您在代码中调用print()，并且print()在打印成功时返回真值（1）。这就是你在替换中所看到的。

但你不需要print()。用以下代码替换您的代码：

s/^(\d+)/my $x = strftime("%Y-%m-%d %H:%M", localtime($1)); $x/e

或者，甚至更简单：

s/^(\d+)/strftime("%Y-%m-%d %H:%M", localtime($1))/e

Perl / e regexp添加意外字符

1 个答案: