Mysql select into array

Question

I'm basically trying to simply select from a table into an array...and it's not working correctly.

I have the following query

$graph = mysql_query("SELECT MONTHNAME(dateadded) MONTH, COUNT(*) COUNT
FROM products
WHERE ((YEAR(dateadded)=2012) && (site_url = '$_GET[site_url_graph]'))
GROUP BY MONTH(dateadded)",$db);

and I need to the results to be in an array like this (can be long or short month name thats not the issue):

$data = array(
    'Jan' => 12,
    'Feb' => 25,
    'Mar' => 0,
    'Apr' => 7,
    'May' => 80,
    'Jun' => 67,
    'Jul' => 45,
    'Aug' => 66,
    'Sep' => 23,
    'Oct' => 23,
    'Nov' => 78,
    'Dec' => 6
);

I'm trying this but getting message that not an array:

$data = array();
while($graphData = mysql_fetch_array($graph)){
    $data[] = $graphData;
    }

I'm sure this is a simple fix but tearing hair out here!

have you tryed running the query in phpmyadmin? what error do you get? — lvil
– lvil, Commented Sep 30, 2012 at 8:19

GBD · Accepted Answer · 2012-09-30 08:26:27Z

2

To get your expected array, you need to change your code following way.

$data = array();
while($graphData = mysql_fetch_array($graph)){
    $data[$graphData['MONTH']] = $graphData['COUNT'];
}

answered Sep 30, 2012 at 8:26

GBD

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Comments

bms · Accepted Answer · 2012-09-30 12:06:15Z

0

$graph = mysql_query("SELECT MONTHNAME(dateadded) MONTH, COUNT(*) COUNT
                      FROM products
                      WHERE ((YEAR(dateadded)=2012) && (site_url = '$_GET[site_url_graph]'))
                      GROUP BY MONTH(dateadded)",$db);

$data = array();
while(list($month, $count) = mysql_fetch_array($graph)) {
    $data[$month] = $count; 
}

So, basically this is the fast and breezy version of what you want. Notice the usage of list. Nice and readable.

answered Sep 30, 2012 at 12:06

bms

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Comments

lvil · Accepted Answer · 2012-09-30 08:20:32Z

-1

Maybe this is the problem:
"COUNT(*) COUNT"? Try to change it to COUNT(*)

answered Sep 30, 2012 at 8:20

lvil

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2 Comments

StudioTime Over a year ago

No, the query is fine - the error I think is in the last part of question

lvil Over a year ago

What are you getting when 'print_r($graphData)'

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