我试图找出如何在if / else块中使用相同的变量。例如,
$var1
$var2
$var3
if(condition) {
...
}
else {
...
}
我需要在块中访问$ vars1,2和3,但我不能。我想我有一个范围问题。我尝试将它们设为全局,但我看到很多错误并且程序运行不正常。我错过了什么?
这是我的代码。我的问题是服务器需要底部的变量(@varList等),因为这是一个也使用CGI的apache服务器上的脚本。它吐出这样的错误:
[error] Global symbol "$questionslist" requires explicit package name at /home/megaoff/www/vi
ewquestions.dhtml line 46.\nGlobal symbol "$site" requires explicit package name at /home/megaoff/www/viewquestions.dhtm
l line 46.\nGlobal symbol "$xs" requires explicit package name at /home/megaoff/www/viewquestions.dhtml line 46.\nGlobal
symbol "$username" requires explicit package name at /home/megaoff/www/viewquestions.dhtml line 46.\n
#!/usr/bin/perl
use strict;
use CGI;
use BarryP;
my $pagev = BarryP::makeP("noextracook", 1);
my $bvga = $pagev->{'vga'};
my %vga = %$bvga;
my $cgi = CGI->new;
my $usePage = "answerquestions.html";
my $anslist = "/home/megaoff/www/limages/anslist.txt";
my $unanslist = "/home/megaoff/www/limages/unanslist.txt";
my $action = $vga{"action"};
if($action eq 'adminmode') {
my @a_list = $pagev->listFile($anslist);
my %list = map { split(/\t/, $_, 2) } @a_list;
}
else {
my @u_list = $pagev->listFile($unanslist);
chomp @u_list;
my %questions = map { $_ => '' } @u_list;
my $question = $cgi->param('question');
my $answer = $cgi->param('answer');
chomp($question, $answer);
open(my $ANS, '>>', $anslist) or die "Can't open file $anslist: $!";
print $ANS "$question\t$answer\n";
close($ANS) or die "Can't close file $anslist: $!";
delete $questions{$question};
open(my $UNANS, '>', $unanslist) or die "Can't open file $unanslist: $!";
print $UNANS "$_\n" foreach keys %questions;
close($UNANS) or die "Can't close file $unanslist: $!";
my $questionslist = join("<br>", @u_list);
my $site = $pagev->{'site'};
my $xs = $pagev->{'xs'};
my $username = $pagev->{'username'};
}
my @varList = ('questionslist', $questionslist, 'action', $action, 'site', $site, 'xs', $xs, '$username', $username);
$pagev->pageHeader($usePage, @varList);
答案 0 :(得分:6)
您的问题在于范围界定。在第46行,当您引用$ questionlist等时,它们已经超出了范围,因为您在else块中定义了它们。
你有(简化):
if ($blah) {
# do stuff
} else {
# do other stuff
my $variable = "something";
}
doSomethingWith($variable); # illegal, $variable is not in scope.
$variable
未在else块之外定义,而else块甚至可能不会被执行。您需要将函数放在else块中,或者找到一些方法来赋予这些变量默认值,并在if语句之前声明它们。
答案 1 :(得分:1)
理想情况下,您的变量声明应该具有尽可能最小的范围,同时仍然在需要访问变量的代码范围内。
my $var1 = 'aBcD';
my $var2;
if ($uc) {
$var2 = uc($var1);
}
else {
$var2 = lc($var1);
}
当然,这可以写成
my $var1 = 'aBcD';
my $var2 = $uc ? uc($var1) : lc($var1);