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php loops

Posted on 2013-06-12
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Last Modified: 2013-06-18
On the attached page I have the following recordset:
record_set('artGallery',"SELECT * FROM art_gallery WHERE artgal_Name = '$art_nme' ");

I am trying to get the following code to iterate through the array such that the "$sum_total" and "$sum_total2" produce the product of the two numbers in rows 1,2 3 etc as a list
This needs to be echoed in the body

I presume I need some type of loop ?
test.txt
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Question by:doctorbill
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by:Gary
Gary earned 300 total points
ID: 39242122
$rows = mysql_num_rows($artGallery);

if ($rows > 0) {
	$sum_total=0;
	$sum_total2=0;

	while ($rowData = mysql_fetch_assoc($rows)) {
		$first_number = $row_artGallery[artgal_sp];
		$second_number = "2.54";
		$sum_total += $second_number * $first_number;
		$third_number = $row_artGallery[artgal_sl];
		$forth_number = "2.54";
		$sum_total2 += $forth_number * $third_number;

	}
}

print ($sum_total);
print ($sum_total2);

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Author Comment

by:doctorbill
ID: 39242201
Is this in the top section of the page, above the head tag, or in the body where I can use echo commands ?
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by:Gary
Gary earned 300 total points
ID: 39242247
I don't know where you want it printing out.
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Author Comment

by:doctorbill
ID: 39242315
I wanted the results echoed in the body section as a list
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Gary earned 300 total points
ID: 39242346
Place this in the body wherever you need it displayed

$rows = mysql_num_rows($artGallery);

if ($rows > 0) {
	while ($rowData = mysql_fetch_assoc($rows)) {
		$first_number = $row_artGallery[artgal_sp];
		$second_number = "2.54";
		$sum_total = $second_number * $first_number;
		$third_number = $row_artGallery[artgal_sl];
		$forth_number = "2.54";
		$sum_total2 = $forth_number * $third_number;
		print ($sum_total);
		print ($sum_total2); 
	}
}

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LVL 109

Assisted Solution

by:Ray Paseur
Ray Paseur earned 200 total points
ID: 39242707
Consider these good learning resources for PHP and MySQL.  The professional approach might be to write a query that will aggregate and sum the data.  In any case, SELECT * would not be the right approach if you can find out the names of the columns.  You would want to select by column name instead.

Beginner: Yank Ignore the hokey title -- its a good book.
Beginner to intermediate: Welling/Thompson
Intermediate to Advanced: Powers
Advanced: Zandstra
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Author Comment

by:doctorbill
ID: 39243687
I am getting this:

PHP Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in I:\lbcLive\artistInfo.php on line 53
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Assisted Solution

by:Ray Paseur
Ray Paseur earned 200 total points
ID: 39244567
That usually means your query failed and returned FALSE instead of a MySQL result resource.  This article shows how to run queries and check for success or failure.  Check the code samples - they are well-commented.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/PHP_Databases/A_11177-PHP-MySQL-Deprecated-as-of-PHP-5-5-0.html
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Author Closing Comment

by:doctorbill
ID: 39255844
thanks
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