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PHP how to stop for statement after value is hit

Hello,
I have a calendar controls that correctly assigns a week (1-52) number to the date that is picked in a start date and end date control.

And in my database I have 52 price fields (one for each week).

I am trying to display the pricing for the weeks that are picked.
For example if the start week is 40 and the end week is 45 it displays the pricing for weeks 40, 41, 42, 43, 44 and 45.

I have that all working accurately through the code:
for($i=$startweek;$i<=$endingweek;$i++)
	{
 echo "<td>Week ", $i, ": $<b>", $row["price$i"], "</b></td>";
 }

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The problem is when the start date is larger then the end date.
For example if the start week is 52 (last week of the year) and the end week is 1 (first week of the next year).

I have the code:
 for($i=$startweek;$i>=$endingweek;$i++)
	{
 echo "<td>Week ", $i, ": $<b>", $row["price$i"], "</b></td>";
 }

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It will bring back the correct information for week 52 but then goes into a loop adding week 53, 54, 55, etc... with no values.

How do I get it to stop at week 52 and also bring back the values for weeks lower then 52 (ie week #1 in the example).

I hope this makes sense.

Thank you in advance for your help.
0
Razzmataz73
Asked:
Razzmataz73
2 Solutions
 
Ray PaseurCommented:
There is an ISO standard for week numbers.  You can use a combination of date() and strtotime() to get the week number.  These functions are documented in the PHP.net web site, and also written about in this article.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_201-Handling-date-and-time-in-PHP-and-MySQL.html

Try printing out the week number according to the PHP date() function.  You will find that a "homegrown" week number is not a useful way of dealing with this information.  If you still have trouble with this, please post back with some test data and I'll be glad to show you a solution.  It may take a day or two because I am teaching full time right now, but there is a good solution and I will be happy to show you how we would do it.

Best regards, ~Ray
0
 
Razzmataz73Author Commented:
The week number comes back with no problem.

My issue is how do I write a query if the start week number is less then the end date number.

For example 49 and 1.
0
 
Ray PaseurCommented:
You do not write a query that way.  You use the week number and the year to produce a lower DATETIME value and an upper DATETIME value. Then you write a query using the BETWEEN subclause of the WHERE clause.  The article either tells it explicitly or give strong suggestions about it.  You want to convert the year + week number to an ISO-8601 DATETIME string, and your query will use the resulting DATETIME strings to look into the data base table.
0
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Razzmataz73Author Commented:
Duh!  Don't know why that didn't occur to me.
That sounds perfect!
I will try it out first thing tomorrow and let you know how it goes.
0
 
Slick812Commented:
Not sure if you can apply this, but the Math of having a higher number first, is that you roll (for loop) the High to the Highest first and then roll from the lowest to the Lower number as this code which I ran , and it does what you describe-

$array52 = range(0, 53);
$startweek = 50;
$endingweek = 2;

if ($startweek == $endingweek)
{
	echo "Week ", $startweek, ': $<b>price', $array52[$startweek], "</b><br />";
} elseif ($startweek < $endingweek)
{
	for($i=$startweek; $i<=$endingweek; $i++) echo "Week ", $i, ': $<b>price', $array52[$i], "</b><br />";
} else
{
	for($i=$startweek; $i<=52; $i++) echo "Week ", $i, ': $<b>price', $array52[$i], "</b><br />";
	for($i=1; $i<=$endingweek; $i++) echo "Week ", $i, ': $<b>price', $array52[$i], "</b><br />";
}

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0
 
sivagnanam chandrakanthCommented:
simple

if($i==52) exit;
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rinfoCommented:
Firstly you should not allow  to select next week which is less then start week if you are taking into consideration weeks for your query and it is assumed that you need data for a
particular year only.
In case you intend to extend the query range spread over multiple and i assume consecutive
years then you should set a variable for currYear which should be year for the start week.
And if start week is more then end week it would be assumed that those weeks belong to
next year.
In such eventualities it would be advisable to do it in multiple steps. One each for each year.
Like one starting form 50th week and going upto 52nd week and then another query which starts at 1st week of the next year and going up to end week.
Naturally you would need to involve year in your query to get correct results.
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Ray PaseurCommented:
Please see http://www.laprbass.com/RAY_week_number.php (and give yourself some time to read the code and output carefully -- it's subtle).

This program shows how to use the week number.  You can use it in combination with strtotime() and date() to come up with the appropriate DATETIME values to use in the query.

<?php // RAY_week_number.php
error_reporting(E_ALL);
echo "<pre>";

// REQUIRED SINCE PHP VERSION 5.1+
date_default_timezone_set('America/Chicago');

// WHAT IS THE WEEK NUMBER?
echo "WE ARE NOW IN WEEK " . date('W');
echo PHP_EOL;

$now = date('c', strtotime( date('Y-01-01') . ' + ' . date('W') . ' WEEKS' ));
var_dump($now);

// SHOW SOME "INTERESTING" THINGS ABOUT WEEK NUMBER
$years = array
( 2006
, 2007
, 2008
, 2009
, 2010
, 2011
, 2012
, 2013
, 2014
, 2015
)
;

foreach ($years as $y)
{
    $c = date('c', strtotime("$y-01-01"));
    $w = date('W', strtotime("$y-01-01"));
    echo PHP_EOL . "ON $c IT IS WEEK NUMBER $w";
    if ($w > '01')
    {
        $str = "$c NEXT MONDAY";
        $tsp = strtotime($str);
        $nur = date('c', $tsp);
        $nuw = date('W', $tsp);
        echo " AND ON $nur IT IS WEEK NUMBER $nuw";
    }
}

foreach ($years as $y)
{
    $str = "$y" . '-01-01' . ' NEXT MONDAY';
    $tsp = strtotime($str);
    $nur = date('r', $tsp);
    echo PHP_EOL . "$str = $nur";
}

// NO DIFFERENCE BETWEEN "NEXT" AND "FIRST"
foreach ($years as $y)
{
    $str = "$y" . '-01-01' . ' FIRST MONDAY';
    $tsp = strtotime($str);
    $nur = date('r', $tsp);
    echo PHP_EOL . "$str = $nur";
}

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Best regards, ~Ray
0

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