Question

我正在尝试重命名文件夹中的所有tif文件，并将它们从1编号为x。防爆。初始文件名“image-2.tif”和“image-3.tif”将重命名为“file1.tif”和“file2.tif”。

这是我的代码：

my $dirname = "../folder";
opendir (DIR, $dirname) or die "cannot open directory $dirname";
my @files = grep /.tif/, readdir DIR;
closedir (DIR);

my $basename = "file";
my $count = 1;
my $new;
foreach (@files) {
$new = "${basename}${count}.tif";
print "rename $_ ${basename}${count}.tif\n";
rename $_, $new;
$count++;
}

虽然所有文件都已正确读取，但它们不会重命名。

Answer 1

当您从另一个文件中获取文件时，您需要为rename()函数使用一个好的路径。模块File::Spec可以帮助实现它：

use File::Spec;

my $dirname = "../folder";
my $abs_path = File::Spec->rel2abs($dirname);

和

foreach (@files) {
    $new = "${basename}${count}.tif";
    print "rename $_ ${basename}${count}.tif\n";
    rename File::Spec->catfile($abs_path, $_), File::Spec->catfile($abs_path, $new);
    $count++;
}

Answer 2

我建议改变你的行：

rename $_, $new;

...更像是：

rename($_, $new) or warn "rename: $_: $new: $!\n";

...你应该能够看到它无法正常工作 - 因为rename失败时会返回false（$!会告诉你失败的原因）。

此外，rename操作的目标（在您的情况下：$new）也需要包含目录组件（否则，您尝试将文件移动到进程的当前工作目录中（可能不是你想要的））。

最后，我建议，不要将$dirname的值硬编码到相对路径，而应将其作为命令行参数接受。这允许您从任何$PWD运行脚本。请参阅@ARGV联机帮助页和/或内置perlvar命令中的shift。

因此，像这样：

my $dirname = shift; # Accept dirname as commandline argument
opendir (DIR, $dirname) or die "cannot open directory $dirname";
my @files = grep /\.tif$/, readdir DIR; # Escape regex meta-char (.) and anchor to end of string ($).
closedir (DIR);

my $basename = "file";
my $count = 1;
my $new;
foreach (@files) {
    $new = "${dirname}/${basename}${count}.tif"; # Include dirname in target path
    print "rename $_ $new\n";
    rename($_, $new) or warn "rename: $_: $new: $!\n"; #warn when rename fails
    $count++;
}

Answer 3

我终于找到了我正在寻找的东西。在Tim Peoples的回应的帮助下。最简单的回答是：

my $dirname = "../folder";
opendir (DIR, $dirname) or die "cannot open directory $dirname";
my @files = grep /[.]tif\z/, readdir DIR; #corrected pattern based on Sinan Ünür's comment
closedir (DIR);

my $basename = "file";
my $count = 1;
my $new;
foreach (@files) {
$new = "${basename}${count}.tif";
print "rename $_ $new\n";
rename ("$dirname/$_", "$dirname/$new") or warn "rename: $_: $new: $!\n"; #directory name     added to both the original file name and new file name. Thanks to Tim for helping me found this error using warn.
$count++;
}

重命名和编号目录中的文件

3 个答案: