2012-11-05 17 views
0

我已經編寫了一個代碼,該代碼嘗試使用perl腳本存儲用戶的上次登錄時間,並將其存儲在使用dbmopen的散列中,但它不會將我更新登錄時間: - 我的代碼去如下: -嘗試使用perl腳本將用戶的上次登錄時間存儲在數據庫中

#!/usr/bin/perl -w 
use DB_File; 
use Time::localtime; 
%last_good =(); 
%words =() ; 
init_words(); 
print " What is your name ? \n"; 
$name=<STDIN>; 
chomp($name); 
print "Hi $name , please Enter your secret password: \n" ; 
$passwd = <STDIN>; 
chomp($passwd); 
$i=1; 
while(! good_password($name,$passwd)) 
{ 
    print " Wrong Password , Please Try Again : \n"; 
    $passwd=<STDIN>; 
    chomp($passwd); 
} 
print "assigning the last login time \n "; 
dbmopen (%last_good,"lastdb",0777) || die "can't dbmopen lastdb: $!"; 
$last_good{$name} = localtime; // update last login time if password correct 
dbmclose (%last_good) || die "can't dbmclose lastdb: $!"; 
print "the last login time for : $name is $last_good{$name} "; 

sub init_words 
{ 
open(WORDLIST,"password.txt") || die " Can not Read the password.txt : $! \n"; 
while(defined ($name = <WORDLIST>)) 
{ 
    chomp($name); 
    $name =~ s/\W.*// ; 
    $name =~ tr /A-Z/a-z/ ; 
    $password=<WORDLIST>; 
    chomp($password); 
    $words{$name}=$password; 
    if(! defined $last_good{$name}) 
    { 
    dbmopen (%last_good,"lastdb",0777) || die "can't dbmopen lastdb: $!"; 
    $last_good{$name} = localtime; // update login time for 1st login 
    dbmclose (%last_good) || die "can't dbmclose lastdb: $!"; 
    } 
    write; 
} 
close(WORDLIST) || die " Could not close the file : $! \n "; 
} 

什麼,我要做的是更新或將在數據庫中的第一次登錄時間如果不是他們 使用if(! defined $last_good{$name}),然後在情況下,它已經定義當我第二次運行prog時,我希望它從數據庫中讀取,而不是嘗試再次創建它。但是,當我第一次分配後嘗試打印$last_good{$name} = localtime;時,它不打印登錄localtime,但它表示它未定義。

+0

'good_password'是否也使'$ name'小寫?另外,我不明白你用'$ name =〜s/\ W。* //'完成了什麼。嘗試使用'Data :: Dumper'類轉儲'%last_good'來驗證它看起來像你期望的。 – ElPaco

回答

0

我想這只是你打電話給dbmclose之前,試圖打印價值。所以當然散列是解開的。

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