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ENTRY AGE IN A REPEATING REGION, DREAMWEAVER

Posted on 2013-12-17
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Last Modified: 2014-02-04
I am working in a php/mysql -dreamweaver environment.  I want to calculate the age in days of each entry.  I have the date of the entry saved in the mysql table.  

Date format  is  " mm/dd/yyyy "

How do I add an age field to show how old the entry is in days?
0
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Question by:jws2bay
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10 Comments
 
LVL 35

Expert Comment

by:gr8gonzo
ID: 39725013
Here's my own comment. Getting age by years is pretty easy. Getting it by days is a little more complex, since there can be more than 365 days in a year (for leap years). Here's a little PHP class I just whipped up that should give you the age in either years or total # of days:

<?php
date_default_timezone_set("UTC");

class Birthday
{
	// DOB parts
	public  $originalFormat;
	private $tsDOB;	// Timestamp
	private $mm;		// DOB month with leading zeros
	private $dd;		// DOB day with leading zeros
	private $yyyy;	// DOB year
	
	// Some calculated values
	public $hasHadBirthdayThisYear;
	public $lastBirthday;
	private $tsLastBirthday;
	
	// Our constructor
	public function __construct($birthday)
	{
		// Store original value
		$this->originalFormat = $birthday;
		
		// Get the parts necessary for DOB calculate
		$this->tsDOB = strtotime($birthday);
		list($this->mm,$this->dd,$this->yyyy) = explode("/", date("m/d/Y",$this->tsDOB));
		
		// Has the birthday passed this year?
		$this->hasHadBirthdayThisYear = (date("md") >= ($this->mm.$this->dd));
		
		// Calculate last birthday
		$this->lastBirthday = $this->mm."/".$this->dd."/".(date("Y") - ($this->hasHadBirthdayThisYear ? 0 : 1));
		$this->tsLastBirthday = strtotime($this->lastBirthday);
	}
	
	public function GetAgeInYears()
	{
		$ageInYears = (date("Y")-$this->yyyy) + ($this->hasHadBirthdayThisYear ? 1 : 0);
		return $ageInYears;
	}
	
	public function GetAgeInDays()
	{
		// Calculate age in years
		$ageInYears = $this->GetAgeInYears();
		
		// Now convert to days
		$daysInAYear = 365.242;
		$ageInDays = $ageInYears * $daysInAYear; // Total of 365.242 days in a year, so we'll end up with a fraction most likely
		
		// Increase by # of days since last birthday
		$daysSinceLastBirthday = ((time() - $this->tsLastBirthday) / 86400);
		$ageInDays += $daysSinceLastBirthday;
		
		// Round down
		$ageInDays = floor($ageInDays);
		
		return $ageInDays;
	}
}

$Birthday = new Birthday("04/22/1980");
echo "Age is: ".$Birthday->GetAgeInYears() . " years or " . $Birthday->GetAgeInDays() . " days\n";
echo "\n";

$Birthday = new Birthday("03/21/1983");
echo "Age is: ".$Birthday->GetAgeInYears() . " years or " . $Birthday->GetAgeInDays() . " days\n";
echo "\n";

$Birthday = new Birthday("12/18/2012");
echo "Age is: ".$Birthday->GetAgeInYears() . " years or " . $Birthday->GetAgeInDays() . " days\n";
echo "\n";
?>

Open in new window

0
 
LVL 54

Expert Comment

by:Scott Fell, EE MVE
ID: 39725021
I would store only the birthdate in the db.  Then calculate the age in days using js/jquery

http://jsbin.com/UXIMeVA/1/edit?html,js,output

<!DOCTYPE html>
<html>
<head>
<link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/smoothness/jquery-ui.min.css" rel="stylesheet" type="text/css" />
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1/jquery-ui.min.js"></script>
<meta charset=utf-8 />
<title>padas</title>
</head>
<body>
  <input id="date" name="date">
  <div id="result">Please endter a date</div>
</body>
</html>

Open in new window

$('#date').change(function(){
  var date_enter=$(this).val();
  
  var days = (new Date() - new Date(date_enter)) / (1000 * 60 * 60 * 24);
  var  days_round_down=Math.floor(days);
  $('#result').html('You are '+days_round_down+' days old');
  
});

Open in new window

0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 39725412
Here is some information about how to handle DATETIME values in PHP and MySQL.  Please read it over and post back if you still have questions.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_201-Handling-date-and-time-in-PHP-and-MySQL.html
0
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Author Comment

by:jws2bay
ID: 39725480
I have a working jquary function to calculate the difference between two dates dates.

What I  want is to the calculate  and show the age for each item in a repeated region on the page.

 I am using dreamweaver on this page, but don't know how to impliment this function.
0
 
LVL 54

Expert Comment

by:Scott Fell, EE MVE
ID: 39725716
Dreamweaver is not the issue.  You can use dreamweaver or notepad to edit your page.  What matters is the final rendered html.  If you can show us your html code (view source) we can help update your jquery.  Or just provide a link to your page.
0
 
LVL 111

Accepted Solution

by:
Ray Paseur earned 1500 total points
ID: 39726560
This is computationally less efficient, (not that it matters) but precise to the day.
http://www.laprbass.com/RAY_temp_jws2bay.php
Outputs:
December 1 => 17
Yesterday => 1
Today => 0
Tomorrow => 0
May 12, 1993 => 7525
Sep 15, 1950 => 23105
Mar 24, 1917 => 35333

<?php // RAY_temp_jws2bay.php
error_reporting(E_ALL);

date_default_timezone_set("UTC");

function age_in_days($birthday, $currently='TODAY')
{
    $tsb = strtotime($birthday);
    $tsc = strtotime($currently);
    if ($tsb)
    {
        if ($tsc)
        {
            $alpha = date('Y-m-d', $tsb);
            $omega = date('Y-m-d', $tsc);
            $kount = 0;
            while ($alpha < $omega)
            {
                $kount++;
                $alpha = date('Y-m-d', strtotime("$alpha + 1 DAY"));
            }
            return $kount;
        }
        else
        {
            // CURRENTLY IS WRONG
            return FALSE;
        }
    }
    else
    {
        // BIRTHDAY IS WRONG
        return FALSE;
    }
}

// TEST CASES
$arr = array
( 'December 1'
, 'Yesterday'
, 'Today'
, 'Tomorrow'
, 'May 12, 1993'
, 'Sep 15, 1950'
, 'Mar 24, 1917'
)
;
foreach ($arr as $d)
{
    echo PHP_EOL . "<br>$d => ";
    echo age_in_days($d);
}

Open in new window

You probably want to rethink your date format of mm/dd/yyyy.  There is an ISO-8601 standard for computer representations of date/time values, and it exists for a variety of reasons.  One of most important reasons is that the standard disambiguates dates like 03/05/yyyy.  Is that March 5 or May 3?  If you follow the standard, there is no ambiguity.  This is all explained in the article.

Best of luck with your project, ~Ray
0
 

Author Comment

by:jws2bay
ID: 39732129
I got pulled away from this issue.   Shown below is the page code for the repeating section.

The function to calculate the age is called displayDate(<?php echo $row_RFQ['Date']; ?>), and it will display the age at id="AgeII" in the repeating region.

How do you call the function from within the repeating region without an action to attach it?


*******************

<div id="StatusTbl" style="overflow:auto; height:650px"><?php do { ?>
                              <table width="100%" border="0" cellpadding="2">
                                <tr class="style25">
                                  <td width="3"  align="left">|</td>
                                  <td width="25"  align="left" ><a onclick="showDetails(<?php echo $row_RFQ['RFQ_No']; ?>)"><?php echo $row_RFQ['RFQ_No']; ?></a></td>
                                  <td width="2"  align="center">&nbsp;</td>
                                  <td width="30"  align="center"><?php echo $row_RFQ['Status']; ?></td>
                                  <td width="2"  align="center">&nbsp;</td>
                                  <td width="50"  align="center"><?php echo $row_RFQ['Date']; ?></td>
                                  <td width="2"  align="center" id="AgeII">&nbsp;</td>
                                  <td  align="center"><?php echo $row_RFQ['Rep_name']; ?></td>
                                  <td width="2"  align="center">&nbsp;</td>
                                  <td width="150" align="center"><?php echo $row_RFQ['Name']; ?></td>
                                  <td width="2"  align="center">&nbsp;</td>
                                  <td width="120"  align="left"><?php echo $row_RFQ['Phone']; ?></td>
                                  <td  width="2" align="center">&nbsp;</td>
                                  <td  width="300" align="left"><?php echo $row_RFQ['Message']; ?></td>
                                  <td  width="2" align="center">|</td>
                                  </tr>
                                <tr class="style25">
                                  <td colspan="15"  align="center"><img src="Assets/dividerA.gif" width="100%" height="3"></td>
                                  </tr>
                                </table>
                              <?php } while ($row_RFQ = mysql_fetch_assoc($RFQ)); ?></div>
0
 

Author Closing Comment

by:jws2bay
ID: 39834280
I apologize for letting this question hang.  Had a little crisis here.
0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 39834318
What was the reason for giving a marked-down grade?  Please see the grading guidelines and give us the explanation, thanks.
http://support.experts-exchange.com/customer/portal/articles/481419
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