[SOLVED] How to calculate two dates into a percentage

[DirtySnipe]

Say I have two veriables $a and $b

 

They are both type datetime format

 

 

$a is a start time and $b is a finish time.

 

Can anyone tell me how i can get a percentage 0-100 of completion?

[Daniel0]

That doesn't make any sense.

 

If you have 2009-05-27 and 2009-05-30 and you are "30% completed", what would that mean, and what data would you use to get to that number?

[Maq]

If I understand correctly and if you know where they currently are in the date range, you could use logic similar to:

 

a = start
b = end
c = current

percent = (c - a) / (b - a) * 100

[DirtySnipe]

thank you.

 

now I just gotta figure out how to put that into php...lol

[DirtySnipe]

Im a complete nubin when it comes to php. Can some one point me as to what im doing wrong (probably alot..lol)

 

 

$percent =(echo date('Y-m-d H:i:s') - $beginProductionTime)/($endProductionTime - echo date('Y-m-d H:i:s')*100);

[Maq]

You can't do it like that.  What kind of precision do you want?  Days, hours, minutes?

[DirtySnipe]

minutes would probably be the best.

[roopurt18]

Untested but should get you close...

<?php

// three inputs
$start = '2009-01-01';
$end = '2008-01-01 12:00:48';
$percentage = .3;

// order them correctly
$start = strtotime( $start );
$end = strtotime( $end );
if( !$start || !$end ) {
  throw new Exception( 'Invalid dates' );
}
$min = min( $start, $end );
$max = max( $start, $end );

// $min is the begin date in seconds, $max is the end date in seconds
echo date( 'Y-m-d H:i:s', $min + ($percentage * ($max - $min)) );
?>

[DirtySnipe]

I'd like to say thank you all for your time so far and you are all brilliant.

 

roopurt18 I tried that code but not sure what is wrong.

 

I changed the date to the following

 

$start = '2009-05-29';
$end = '2009-06-01 12:00:48';

 

The output displays

 

2009-05-30 01:12:14 and not a % ??

 

 

[roopurt18]

My code takes two dates, calculates the number of seconds between them, multiplies by a percentage, and then adds that percentage to the first date.  Finally it outputs the result as a date.

 

Otherwise what you are asking for makes no sense mathematically so you'll need to be more clear.

[Daniel0]

I think what he means that if start is 2009-05-28, current is 2009-05-29 and end is 2009-05-30 then the result should be 50%.

 

I.e. something like this:

<?php
$start = '2009-05-28';
$end = '2009-05-30';

$start = strtotime($start);
$end = strtotime($end);
if (!$end || !$start) {
throw new Exception('Invalid dates.');
}
else if ($start > $end) {
throw new Exception('Start date is larger than end date.');
}

$diff = $end - $start;

$current = time();
$cdiff = $current - $start;

if ($cdiff > $diff) {
$percentage = 1.0;
}
else if ($current < $start) {
$percentage = 0.0;
}
else {
$percentage = $cdiff / $diff;
}

printf('%.2f%%', $percentage * 100);

[DirtySnipe]

OK I have the following code:-

 

$start = '2009-05-29 12:00:48';
$end = '2009-06-01 12:00:48';
$start = strtotime($start);
$end = strtotime($end);
if (!$end || !$start) {
throw new Exception('Invalid dates.');
}
else if ($start > $end) {
throw new Exception('Start date is larger than end date.');
}

$diff = $end - $start;

$current = time();
$cdiff = $current - $start;

if ($cdiff > $diff) {
$percentage = 1.0;
}
else if ($current < $start) {
$percentage = 0.0;
}
else {
$percentage = $cdiff / $diff;
}

 

and i have placed the following in the table to show the results.

 

printf('%.2f%%', $percentage * 100);

 

But i get the following output

 

0.12%5

[Daniel0]

Well, those two date inputs are all in the past. You need to explain what you are trying to do. Otherwise we cannot really help you. We have all been trying to guess what you are trying to do, but all the times you simply just come back and say you aren't getting the correct results. That's like walking in saying that you're trying to get the number 5. I can provide you millions of ways you can get the number 5, but unless you tell me how you plan on getting to that number I can't give you the solution you're looking for.

[DirtySnipe]

sorry i missed part of the code. Please look up at editing post.

 

What im trying to do is take a start date and a finish date.

 

And work out a percentage to completion.

[DirtySnipe]

also as a note both dates are not in the past. The start date is but the end date is in june.

[Daniel0]

What did you expect the result to be?

 

also as a note both dates are not in the past. The start date is but the end date is in june.

 

It wasn't before your edit.

[DirtySnipe]

The code looks like its working but it is outputting a %5 after the number.

[Daniel0]

That's something you're outputting somewhere else.

 

Right now

<?php
$start = '2009-05-29 12:00:48';
$end = '2009-06-01 12:00:48';
$start = strtotime($start);
$end = strtotime($end);
if (!$end || !$start) {
   throw new Exception('Invalid dates.');
}
else if ($start > $end) {
   throw new Exception('Start date is larger than end date.');
}

$diff = $end - $start;

$current = time();
$cdiff = $current - $start;

if ($cdiff > $diff) {
   $percentage = 1.0;
}
else if ($current < $start) {
   $percentage = 0.0;
}
else {
   $percentage = $cdiff / $diff;
}

printf('%.2f%%', $percentage * 100);

 

outputs 1.88%, but that will of course change.

[DirtySnipe]

hmm nope.

 

I wonder if this problem i have is anything to do with me outputting the table results into a table??

 

all of my code so far.

[Daniel0]

Uh... yeah it does. You cannot do echo printf. printf already does output it and it returns the length of the outputted string, which is what you are then echoing.

[DirtySnipe]

ahhh bliss

 

Thank you soo much. You have been a great help.

[.josh]

If I understand correctly and if you know where they currently are in the date range, you could use logic similar to:

 

a = start
b = end
c = current

percent = (c - a) / (b - a) * 100

 

this is what you want. All you have to do is convert your vars into timestamps and insert where appropriate.

 

edit: oops my bad. I didn't realize there was another page to this thead.