How to use a variable as a command option? Post: 302603484

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Top Forums Shell Programming and Scripting How to use a variable as a command option? Post 302603484 by quinestor on Thursday 1st of March 2012 06:20:26 AM

03-01-2012

Registered User

How to use a variable as a command option?

I am just learning shell scripting and already I found out I have the bad habit of thinking that it is similar to php or c.

I learned some basics and now encountered this problem:

On shell it is possible to type:

$ date --date="2009-10-10 09:08:34"

Sat Oct 10 09:08:34 CEST 2009

This tells date to output the date in that string in that default format. But now, I can't pass that string as a variable:

$ dateLabel="2009-10-10 09:08:34"

$ date --date=$dateLabel

date: the argument `09:08:34' lacks a leading `+';

when using an option to specify date(s), any non-option

argument must be a format string beginning with `+'

Try `date --help' for more information.

I thought the problem could be the quotes but I have the same result if I first set the variable to have quotes:

$ dateLabel="\"2009-10-10 09:08:34\""

$ date --date=$dateLabel

date: the argument `09:08:34"' lacks a leading `+';

when using an option to specify date(s), any non-option

argument must be a format string beginning with `+'

Try `date --help' for more information.

I just don't get it, is it like there is a different encoding for that blank space?

Anyone can help me?

quinestor

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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)

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