Question with open file 2

[zhenning]

Hi,

I can open a file using "open(F,"19150009212005.txt");"
I also one variable : $date=09212005
How can I open the file using this variable? I tried:
open(F,"191500/$date/.txt"); It did not work.

Thanks
zhenning

 

[duncdude]

it's because it starts with a zero! if it didn't it would work.

you will have to enclose in speech marks (either double or single) $date = '09212005';


Kind Regards
Duncan

 

[TrojanWarBlade]

Your string would interpolate as: "191500/9212005/.txt" so not only are you miss the zero but you have a filename of ".txt". Is that really what you want?
I think not.

my $date     = "09212005";
my $pathname = "191500/$date.txt";
open F, $pathname or die "Failed to open $pathname";

Trojan.

 

[zhenning]

Sorry I did not address clearly. Folloing is my code:

my $date=@data[1]; (@data[1] contains "09222005" got from web CGI)
my $pathname = "191500$date.txt";
print "$date<br>\n";
print "$pathname<br>\n";
open(F,$pathname);

It did not work. But I got the print outputs;
09222005
19150009222005.txt
Why I still can not open this file?

But the following codes worked:
my $date = "09222005";
my $pathname = "191500$date.txt";
open(F,$pathname);

The only different is at "my $date=@data[1];"

thanks

 

[zhenning]

The file I want to open is 19150009222005.txt

 

[zhenning]

Sorry it was some problem with the file. the code from Trojan worked. Thanks!!!