使用ARGV for perl获取目录(get directories using ARGV for perl)

使用ARGV for perl获取目录(get directories using ARGV for perl) $args = @ARGV; foreach $ARGV (@ARGV) { $guitestsDir = "v:\\stuf\\comment\\logs\\$ARGV\\"; ############################open logs folder######################### if ( -d $guitestsDir ) { opendir( DIR, $guitestsDir ); while ( defined( $pathofdir = readdir(DIR) ) ) { next if ( $pathofdir eq "." ); next if ( $pathofdir eq ".." ); next unless (-d "$guitestsDir/$pathofdir"); push (@pathallSuites, $pathofdir); } closedir(DIR); } print "@pathallSuites\n";

嗨，我会给目录1和2作为参数。 在dir 1中，将dir a和b在dir 2 c和d中，当我运行它时print "@pathallSuites\n"; 输出将是：

a b a b c d

应该是

a b c d

我究竟做错了什么？ 谢谢

$args = @ARGV; foreach $ARGV (@ARGV) { $guitestsDir = "v:\\stuf\\comment\\logs\\$ARGV\\"; ############################open logs folder######################### if ( -d $guitestsDir ) { opendir( DIR, $guitestsDir ); while ( defined( $pathofdir = readdir(DIR) ) ) { next if ( $pathofdir eq "." ); next if ( $pathofdir eq ".." ); next unless (-d "$guitestsDir/$pathofdir"); push (@pathallSuites, $pathofdir); } closedir(DIR); } print "@pathallSuites\n";

hi, I will give directory 1 and 2 as argument. in dir 1 will be dir a and b in dir 2 c and d when I run it for print "@pathallSuites\n"; the output will be:

a b a b c d

and should be

a b c d

what am I doing wrong? thanks

最满意答案

您永远不会重置@pathAllSuites变量。 在你的foreach循环开始时，你需要自己做这件事。

foreach $ARGV (@ARGV) { $#pathAllSuites = -1; # Set the 'last index' to -1, effectively clearing it. $guitestsDir = "v:\\stuf\\comment\\logs\\$ARGV\\";

You never reset the @pathAllSuites variable. At the beginning of your foreach loop, you need to do this yourself.

foreach $ARGV (@ARGV) { $#pathAllSuites = -1; # Set the 'last index' to -1, effectively clearing it. $guitestsDir = "v:\\stuf\\comment\\logs\\$ARGV\\";