I am creating an output file, and I want to use the date parameter passed in
as part of the output file name.
This is how I name my output file:
1. Assign the date param to ARGV -> $active_date = $ARGV[0];
2. $outputfile = "NEW_CUSTOMERS_after_$active_date";
When I run the task I get the followi
never mind.
> -Original Message-
> From: Kim Green
> Sent: Thursday, September 13, 2001 9:11 AM
> To: '[EMAIL PROTECTED]'
> Subject: Variable interpolation in a format
>
> Below is my format:
>
> for
_name,$created_date,$active_date
.
My script is accepting a date parameter, and I want to print that date in
the title. I tried that by adding $active_date to the New Customers Since
title, and the variable was not interpolated.
What's the best way to print my param
d TO_CHAR(active_date,'MM-DD') > ?
order by 1
If I pass in 08/20/2001, a record with an active_date of 12/31/2000 is
returned. It's as if the month and day are compared , not the entire year.
Has anyone experienced this?
Kim Green
Billing Systems Developer
Madison River
Kim Green
Billing Systems Developer
Madison River Communications
919.563.8385
[EMAIL PROTECTED]
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TWEEN ph.effective_start_Date and ph.effective_end_Date
> and SYSDATE BETWEEN sh.effective_start_Date and sh.effective_end_Date
> and cnh.node_name_uppercase = ?");
>
> $customer_name = $ARGV[0];
>
> Please advise,
>
> Kim Green
> Billing Systems Developer
> Madison River Communications
> 919.563.8385
> [EMAIL PROTECTED]
>
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I don't know how or where to put & how to use this $option.
What if my Default value is a sql statement? Can I just include the name
of the file handle between the brackets?
-Original Message-
From: Brett W. McCoy [mailto:[EMAIL PROTECTED]]
Sent: Friday, June 29, 2001 6:40 PM
To: M.W
What would cause this format for the output file to result in the error
message "Format not terminated at end of line" :
format OUTPUTFORMAT =
@<<<,@<
,@<<< ,@<
$servicename,$filename
,$servi
my $node_name = $data[3];
#print " $servicename $filename
$service_type_name $node_name\n";
$each_file = "$servicename $filename
$service_type_name $node_name";
open (FILE,">>$outputfile");
What's the proper syntax to indicate that a variable is optional? The script
that I have created works great when I pass in a variable, but the script
need to execute the SQL even if I don't pass in a variable.
Thanks,
Kim
My question pertains to using command line variables in Perl.
I created a script that uses SQL and runs from an application, and the only
parameter is optional. This script works well when the parameter is
required or not used at all.
I have altered the SQL script so that it can accept a variabl
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