11-25-2009
Automatic file input to code
i have got many files like this in my folder temp(say)
imp_02042008.txt for date 02-04-2008
imp_03092009.txt for date 03-09-2009
imp_25112009.txt for date 25-11-2009
...................
........
in some folder.
and one of my shell code uses one of the above files based on date.
for example for todays date it should pick the file imp_25112009.txt
for tomorrows date it should pick the file like imp_26112009.txt
My code is like below one:
print substr($1,4,3),"",substr($2,22,8) inputfile.txt>outputfile.txt
This code should run on day to day basis.
What i want is a code to pick up the file (eg..imp_26112009.txt) automatically from folder temp based on day.. i.e, for a given day particular file should be taken as input in my code...
pls help me in this regard..
---------- Post updated at 07:14 AM ---------- Previous update was at 07:02 AM ----------
pls any coders do look into matter... m novice to shell scripting...
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LEARN ABOUT PHP
date_parse_from_format
DATE_PARSE_FROM_FORMAT(3) 1 DATE_PARSE_FROM_FORMAT(3)
date_parse_from_format - Get info about given date formatted according to the specified format
SYNOPSIS
array date_parse_from_format (string $format, string $date)
DESCRIPTION
Returns associative array with detailed info about given date.
PARAMETERS
o $format
- Format accepted by DateTime.createFromFormat(3).
o $date
- String representing the date.
RETURN VALUES
Returns associative array with detailed info about given date.
EXAMPLES
Example #1
date_parse_from_format(3) example
<?php
$date = "6.1.2009 13:00+01:00";
print_r(date_parse_from_format("j.n.Y H:iP", $date));
?>
The above example will output:
Array
(
[year] => 2009
[month] => 1
[day] => 6
[hour] => 13
[minute] => 0
[second] => 0
[fraction] =>
[warning_count] => 0
[warnings] => Array
(
)
[error_count] => 0
[errors] => Array
(
)
[is_localtime] => 1
[zone_type] => 1
[zone] => -60
[is_dst] =>
)
SEE ALSO
DateTime.createFromFormat(3), checkdate(3).
PHP Documentation Group DATE_PARSE_FROM_FORMAT(3)