如何使用Perl Glob读取远程位置? [英] How to use Perl glob to read remote location?

问题描述

我的测试脚本:

my $loc = "\\\\ant\\d1_sp\\test__18716093";
####$loc = "c:\\temp"; #this works good if I un-comment.
system("dir $loc\\*.log"); #added system command just for debugging.
my @test = glob qq("$loc\\*.log");
print "\narray=@test\n";

我想将 $ loc 中的文件名保存到数组中以进行进一步处理，但是这样做没有，我缺少什么?输出是:

I want to save the file names in $loc into array for further processing, but its not doing so, what am I missing? output is:

C:\>perl c:\temp\foo.pl

Directory of \\ant\d1_sp\test__18716093

03/14/2016  01:09 PM               959 build_1.8980.log
03/14/2016  01:20 PM           102,402 build_2.98981.log
           2 File(s)        103,361 bytes
           0 Dir(s)  1,589,522,239,488 bytes free

array=
C:\>

推荐答案

您要在以下共享中列出.log文件:

You want to list the .log files in the following share:

\\ant\d1_sp\test__18716093

为此，您需要使用以下glob模式:

To do so, you need to use the following glob pattern:

\\\\ant\\d1_sp\\test__18716093\\*.log

以下是产生该字符串的字符串文字:

The following is a string literal that produces that string:

"\\\\\\\\ant\\\\d1_sp\\\\test__18716093\\\\*.log"

解决方案是

glob("\\\\\\\\ant\\\\d1_sp\\\\test__18716093\\\\*.log")

最好只使用/而不是\，因为您无需以全局模式或字符串文字对其进行转义.

Best to just use / instead of \ since you don't need to escape it in either glob patterns or string literals.

glob("//ant/d1_sp/test__18716093/*.log")

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