Hello everyone,
I have a line like the following:
my $line = 31232000 07/28/04 DUC000 NET 60 DAYS RD64
UPSGNDSVR PREPAID;
What I'm looking for in lines like this are the customer number
(DUC000) and a code that starts with UPS, or if the code doesn't
start with UPS idealy
On Thursday, Nov 4, 2004, at 10:49 US/Central, Scott Pham wrote:
Why don't you just split the line and use the whitespace as separator?
$line =~ s/\s+/ /;
my
($record,$date,$cust,$temp1,$temp2,$temp3,$temp4,$temp5,$shipping,$paid
)
= split (/ /,$line);
As I was reminded the other day, split
Kevin Old wrote:
Hello everyone,
Hello,
I have a line like the following:
my $line = 31232000 07/28/04 DUC000 NET 60 DAYS RD64
UPSGNDSVR PREPAID;
What I'm looking for in lines like this are the customer number
(DUC000) and a code that starts with UPS, or if the code doesn't
Hello,
I have as the output of an unzip command called from a script the following:
unzip test.zip
Archive: test.zip
inflating: arch1.txt
inflating: arch2.txt
inflating: arch3.txt
inflating: arch4.txt
inflating: arch5.txt
inflating: arch6.txt
The same I have from a pkware execution :
[EMAIL PROTECTED] wrote:
So, I need some help on how to write a regexp to get the values
after inflating, that is the filenames, and put them in an array.
Both outputs above are stored on a $stdout php variable.
Well, this list is for discussing Perl, not PHP. If you want a regexp
in PHP, you'd
Gareth Segree wrote:
I have a directory of files that I want to move to another directory.
(eg. ALLY20030111W.eps
TEST W20030122
HELP WANTED20030901WW.eps
GIRL WATCH BIRD 20030101
etc..)
I want to be able to parse the filename and replace the date portion with
any date
MAN 2 MAN18800101.eps
The filenames will not have this sort of naming convention.
Files will look like the following.
ALLY20030111W.eps
TEST W20030122
HELP WANTED20030901WW.eps
GIRL WATCH BIRD 20030101
HELPER 20030121.CW.eps
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I have a directory of files that I want to move to another directory.
(eg. ALLY20030111W.eps
TEST W20030122
HELP WANTED20030901WW.eps
GIRL WATCH BIRD 20030101
etc..)
I want to be able to parse the filename and replace the date portion with
any date
(eg $1=ALLY $2=20030111
hi, i have this regex:
\.(?!.png|.log)[^.]*$
how can i replace the .before png and log with nothing? the problem is,
the alternation can be longer, like that:
\.(?!.png|.log|.txt|.c|.cpp and so on )[^.]*$
how could i do that?
THANKS:-)
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