如何使用 Perl 在目录中搜索特定文件? [英] How do I search for a particular file in a directory using Perl?

问题描述

我在目录中有一个文件，我想一次从目录中选择一个特定的文件.

I have a file in a directory and i want to pick one particular file from the directory at a time.

代码如下:

$xml_file_name = <STDIN>;
chomp ($xml_file_name);

@o = file_search($xml_file_name);
print "file::@o\n";

sub file_search
{
    opendir (DIR, "/home/sait11/Desktop/test/Test cases") or die "Failed to open directory\n";
    @dirs_found = grep { /$xml_file_name/ } readdir DIR;
    closedir (DIR);
#   print "dir ::@dirs_found\n";
    return @dirs_found;
}

我正在输入要作为 sample.xml 返回的文件名.但是在 @dirs_found 变量中，所有以 's' 开头的文件名都会被存储.

I am inputting the file name to be returned as sample.xml. But in the @dirs_found variable, all the file names that starts with 's' are getting stored.

如何一次找出确切的一个文件?

How to find out the exact one file at a time?

推荐答案

要查找特定文件，只需打开该文件，或对其运行文件测试:

To find a specific file, simply open the file, or run a file test on it:

my $file = "/home/sait11/Desktop/test/Test cases/$xml_file_name";

print "$file found\n" if -f $file;

运行您的代码，它似乎确实有效.如果您输入一个模式，那么它会从目标目录中正确选择与您键入的 Perl 正则表达式匹配的文件名.您要查找什么名称?

Running your code, it certainly seems to work. If you enter a pattern, then it correctly picks up the file names from the target directory that match that Perl regex you type in. What name were you looking for?

如果将grep修改为:

my @dirs_found = grep { /^$xml_file_name$/ } readdir DIR;

然后它将排除正则表达式与整个名称不匹配的值.另一方面，当你这样做时，你会放弃一些灵活性.

then it will exclude values where the regex doesn't match the entire name. On the other hand, you give up some flexibility when you do that.

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