问题描述
我有一个文件 a.out，其中包含多行.每行只有一个字符，Unicode 字符 U+2013 或小写字母 a-z.
I have a file, a.out, which contains a number of lines. Each line is one character only, either the unicode character U+2013 or a lower case letter a-z.
对 a.out 执行文件命令会得到 UTF-8 Unicode 文本.
Doing a file command on a.out elicits the result UTF-8 Unicode text.
locale 命令报告:
The locale command reports:
LANG=en_US.UTF-8
LC_CTYPE="en_US.UTF-8"
LC_NUMERIC="en_US.UTF-8"
LC_TIME="en_US.UTF-8"
LC_COLLATE="en_US.UTF-8"
LC_MONETARY="en_US.UTF-8"
LC_MESSAGES="en_US.UTF-8"
LC_PAPER="en_US.UTF-8"
LC_NAME="en_US.UTF-8"
LC_ADDRESS="en_US.UTF-8"
LC_TELEPHONE="en_US.UTF-8"
LC_MEASUREMENT="en_US.UTF-8"
LC_IDENTIFICATION="en_US.UTF-8"
LC_ALL=
如果我发出命令 grep -P -n "[^x00-xFF]" a.out 我希望只有包含 U+2013 的行被退回.如果我在 cygwin 下进行测试就是这种情况.然而，问题环境是 Oracle Linux Server 6.5 版，问题是 grep 命令不返回任何行.如果我发出 grep -P -n "[x00-xFF]" a.out 然后返回所有行.
If I issue the command grep -P -n "[^x00-xFF]" a.out I would expect only the lines containing U+2013 to be returned. And this is the case if I carry out the test under cygwin. The problem environment however is Oracle Linux Server release 6.5 and the issue is that the grep command returns no lines. If I issue grep -P -n "[x00-xFF]" a.out then all lines are returned.
我意识到[grep -P]...是高度实验性的，grep -P 可能会警告未实现的功能."但没有发出警告.
I realise that "[grep -P]...is highly experimental and grep -P may warn of unimplemented features." but no warnings are issued.
我错过了什么吗?
推荐答案
我建议避免狡猾的 grep -P 实现并使用真实的东西.这有效:
I recommend avoiding dodgy grep -P implementations and use the real thing. This works:
perl -CSD -nle 'print "$.: $_" if /P{ASCII}/' utfile1 utfile2 utfile3 ...
地点:
-CSD选项表示 stdio 三重奏(stdin、stdout、stderr)和磁盘文件都应被视为 UTF-8 编码.
The
-CSDoptions says that both the stdio trio (stdin, stdout, stderr) and disk files should be treated as UTF-8 encoded.
$.代表当前记录(行)号.
The $. represents the current record (line) number.
$_ 代表当前行.
P{ASCII} 匹配任何非 ASCII 的代码点.
The P{ASCII} matches any code point that is not ASCII.
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