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MYSQL PHP: How do I calculate percentage change from mysql table?

Posted on 2008-06-11
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Last Modified: 2013-12-13
MYSQL PHP

Hello,

I have a MySQL table with 2 fields:

[ Date ] [ Value ]
2008-06-11 2200
...
2008-05-11 2000


I would like to calculate the percentage increase or decrease as at the last date (e.g. 2007-12-31), when compared to 1 month ago.

So in the example, I want to calculate (2200-2000) / 2000

How do I write this in a mysql query in a PHP script?


Thanks.
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Question by:gingera
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Accepted Solution

by:
Rob Siklos earned 500 total points
ID: 21762713
Assuming there is only one record for each day, you should be able to do something like this:
$date = "2008-06-11";
 
$sql = "SELECT (t1.value - t2.value)/t2.value AS result
FROM table t1
LEFT JOIN table t2
ON t2.`date` = DATE_SUB(t1.`date`, INTERVAL 1 MONTH)
WHERE t1.`date` = '$date'";

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Author Comment

by:gingera
ID: 21763029
Thanks Rob. Hmmm... I yielded no result with your query.

i.e. when I

echo "$result";

nothing appeared.
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LVL 9

Expert Comment

by:Rob Siklos
ID: 21763071
I just tested it and it worked for me.  Can you post the PHP code you're using to execute the query and print the result?
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Expert Comment

by:Rob Siklos
ID: 21763082
Oh, also, for robustness I would change the LEFT JOIN to INNER JOIN
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Author Comment

by:gingera
ID: 21763169
OK this is the code I tried

Table name: rice

Table fields: date, close
<?php
 
$connection = mysql_connect('localhost', 'username', 'password');
$db = mysql_select_db('prices', $connection);
$query_percentage_change = 	"SELECT ((t1.`close` - t2.`close`)/t2.`close`) AS percentage_change
					FROM `rice` t1
					INNER JOIN `rice` t2
					ON t2.`date` = DATE_SUB(t1.`date`, INTERVAL 1 MONTH)
					WHERE t1.`date` = (SELECT MAX(`date`) FROM `rice`) ";
	$result_percentage_change = mysql_query($query_percentage_change);
	while ( $row = mysql_fetch_array($result_percentage_change) )
	{
			extract($row);
			echo "$percentage_change";
	}
?>

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0
 

Author Comment

by:gingera
ID: 21763187
I got error message

PHP Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource
0
 
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Expert Comment

by:Rob Siklos
ID: 21763201
before "extract($row);" can you add "print_r($row);" and see what gets printed?
0
 

Author Comment

by:gingera
ID: 21763260
OK, did that. Nothing displayed.
0
 
LVL 9

Expert Comment

by:Rob Siklos
ID: 21763263
Seems like there's something wrong with the result of the call to mysql_query().

After that line, try adding this:

if (!$result_percentage_change) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    die($message);
}
0
 

Author Comment

by:gingera
ID: 21763294
OK I have a possible explanation.

Maybe there is no data for the date exactly 1 month before. For example, a weekend?

If that is the case, how can we find the next closes date to 1 month (if possible)?
0
 

Author Comment

by:gingera
ID: 21763340
OK I changed "interval 1 month" to "interval 2 month" and a result is displayed.

So the problem is as I have described above.

How do I solve this problem (no data for exactly 1 month ago)?
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LVL 9

Expert Comment

by:Rob Siklos
ID: 21763355
Ok, that is a different problem for sure.

At first thought, I would say that you need a routine in PHP to select the date you want to compare to.
0
 

Author Comment

by:gingera
ID: 21763376
OK I am closing this question and allocating the points to you Rob. Thanks very much for your help.

I will start another question. THANKS!
0
 

Author Closing Comment

by:gingera
ID: 31466259
Thanks for your patience for this newbie.
0

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