2014-01-22 75 views
0

我想從屬性文件打開讀取文件路徑,但我無法這樣做。不過,我嘗試了一些硬編碼,它工作正常。無法從屬性文件中以Perl打開文件

Error: Could not open file '/home/ad/prod/logs/test.log' 
    No such file or directory at binaryImageTransfer.pl 

在下面的代碼中,當我嘗試直接使用tempVar時,它工作。我也嘗試打印logFileName。它的工作原理

my $config = Config::Tiny->new; 
$config = Config::Tiny->read('myconfig_imagetransfer'); 
$logFileName = $config->{TRUNK}->{LOG_FILE}; 
chomp $logFileName; 
#my $tempVar = '/home/ad/prod/logs/test.log'; 
my $tempVar = $logFileName; 
open($logFile, '>>', $tempVar) or die "Could not open file $tempVar $!"; 


myconfig_imagetransfer 
TRUNK SOURCE_DIR = /home/ad/prod/war 
TARGET_DIR = /app/tomcat6/trunkcurrent/webapps 
LOG_FILE = '/home/ad/prod/logs/test.log' 

在這方面的任何幫助將不勝感激。預先感謝您

+0

真棒!它的作品!!謝謝!! – sudhir

回答

0

圍繞LOG_FILE值的單引號是不必要的。此外,在你的配置文件中,'TRUNK'應該是段名,所以試試這個配置文件:

[TRUNK] 
SOURCE_DIR = /home/ad/prod/war 
TARGET_DIR = /app/tomcat6/trunkcurrent/webapps 
LOG_FILE = /home/ad/prod/logs/test.log 
+0

真棒!有用!!謝謝!! – sudhir