我正在尝试编写一个实用程序,它将通过一个看起来像这样的文件：
# Directory | file name | action | # of days without modification to the file for the command to take action /work/test/|a*|delete|1 /work/test/|b*|compress|0 /work/test/|c*|compress|1
我的脚本将通过文件决定是否,例如,/ work / test /下的文件以’a’开头,在过去1天内没有被修改过,如果有,它会删除它们.
为此,我使用find命令.例：
my $command = "find " . $values[0] . $values[1] . " -mtime +" . $values[3] . " -delete ;\n"; system ($command);
但是,我被要求检索每个步骤的返回码,以验证每个步骤是否正常.
解决方法
运行反引号后,返回代码在
$?中可用.
$?The status returned by the last pipe close,backtick (“) command,successful call to wait() or waitpid(),or from the system() operator. This is just the 16-bit status word returned by the traditional Unix wait() system call (or else is made up to look like it).
$output = `$some_command`; print "Output of $some_command was '$output'.\n"; print "Exit code of $some_command was $?\n";