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Top Forums Shell Programming and Scripting how to display an error if input value is not an IP address?? Post 302405988 by Dendany83 on Sunday 21st of March 2010 08:19:48 AM
Old 03-21-2010
Question how to display an error if input value is not an IP address??

Dears,

I need your assistance, I'm creating a shell script to read the IP address of a server and i would like to display an error message in red if the entered value is not an IP. any ideas how can i do it?Smilie

#!/bin/sh

echo "Please enter the server IP address: \c"
read IPadd
????????????????????????????
????????????????

Similarly if i have the date in format yyyymmdd and i need to convert it to yymmdd, how can i do it? and how to display error message if the entered date is not in the correct (requested) format yyyymmdd??

Thanks in advance
 

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DATEFMT_SET_PATTERN(3)							 1						    DATEFMT_SET_PATTERN(3)

IntlDateFormatter::setPattern - Set the pattern used for the IntlDateFormatter

	Object oriented style

SYNOPSIS
public bool IntlDateFormatter::setPattern (string $pattern) DESCRIPTION
Procedural style bool datefmt_set_pattern (IntlDateFormatter $fmt, string $pattern) Set the pattern used for the IntlDateFormatter. PARAMETERS
o $fmt - The formatter resource. o $pattern - New pattern string to use. Possible patterns are documented at http://userguide.icu-project.org/formatparse/datetime. RETURN VALUES
Returns TRUE on success or FALSE on failure. Bad formatstrings are usually the cause of the failure. EXAMPLES
Example #1 datefmt_set_pattern(3) example <?php $fmt = datefmt_create( 'en_US', IntlDateFormatter::FULL,IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, 'MM/dd/yyyy' ); echo 'pattern of the formatter is : ' . datefmt_get_pattern($fmt); echo 'First Formatted output with pattern is ' . datefmt_format($fmt, 0); datefmt_set_pattern($fmt, 'yyyymmdd hh:mm:ss z'); echo 'Now pattern of the formatter is : ' . datefmt_get_pattern($fmt); echo 'Second Formatted output with pattern is ' . datefmt_format($fmt, 0); ?> Example #2 OO example <?php $fmt = new IntlDateFormatter( 'en_US', IntlDateFormatter::FULL,IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, 'MM/dd/yyyy' ); echo 'pattern of the formatter is : ' . $fmt->getPattern(); echo 'First Formatted output is ' . $fmt->format(0); $fmt->setPattern('yyyymmdd hh:mm:ss z'); echo 'Now pattern of the formatter is : ' . $fmt->getPattern(); echo 'Second Formatted output is ' . $fmt->format(0); ?> The above example will output: pattern of the formatter is : MM/dd/yyyy First Formatted output with pattern is 12/31/1969 Now pattern of the formatter is : yyyymmdd hh:mm:ss z Second Formatted output with pattern is 19690031 04:00:00 PST SEE ALSO
datefmt_get_pattern(3), datefmt_create(3). PHP Documentation Group DATEFMT_SET_PATTERN(3)
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