Multidimensional array output

Question

So I have this array data which i want to be coded somewhat like [lojas, raparacoes, valor],[nome_1, count_1, val_1],[nome_2, count_2, val_2], etc, etc...

lojas, reparacoes and valor are like headers

nome_* comes from $row['nome']

count_* comes from intval($row['COUNT( DISTINCT id_reparacao )'])

val_* comes from intval($row2['SUM(valor)'])

$data = array(array('Lojas'), array('Reparacoes'), array('Valor'));
$qry=mysql_query ('SELECT COUNT( DISTINCT id_reparacao ) , lojas.nome, lojas.id
FROM reparacoes
INNER JOIN lojas ON lojas.id = id_loja 
GROUP BY lojas.id ');

        while($row = mysql_fetch_array($qry))
        {
            $qry2=mysql_query ('SELECT SUM(valor) FROM re_servicos where id_reparacao=(select id_reparacao from reparacoes where id_loja='.$row['id'].' and estado="Fechada")');
            while($row2 = mysql_fetch_array($qry2))
                {
                    $data=[$row['nome'],intval($row['COUNT( DISTINCT id_reparacao )']), intval($row2['SUM(valor)'])];   
                }
        }

However, with this code I'm not getting the desired output in the array, I guess the problem is the way i fill it but I don't know how to properly fill it so it gets the output I posted in the first paragraph.

PS: I don't know if it matters but for better understanding, I need this array to build a google bar chart

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Madara's Ghost
– Madara's Ghost, Commented Jul 12, 2013 at 10:18

Hieu Le · Accepted Answer · 2013-07-12 10:17:36Z

1

You can try this code.

 $data = array();
$data[] = array('Lojas', 'Reparacoes', 'Valor');
$qry=mysql_query ('SELECT COUNT( DISTINCT id_reparacao ) , lojas.nome, lojas.id
FROM reparacoes
INNER JOIN lojas ON lojas.id = id_loja 
GROUP BY lojas.id ');

        while($row = mysql_fetch_array($qry))
        {
            $qry2=mysql_query ('SELECT SUM(valor) FROM re_servicos where id_reparacao=(select id_reparacao from reparacoes where id_loja='.$row['id'].' and estado="Fechada")');
            while($row2 = mysql_fetch_array($qry2))
                {
                    $data[]=array($row['nome'],intval($row['COUNT( DISTINCT id_reparacao )']), intval($row2['SUM(valor)']));   
                }
        }

answered Jul 12, 2013 at 10:17

Hieu Le

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1 Comment

Correia JPV Over a year ago

Working perfectly, Thank you. (I'm accepting this answer as soon as possible)

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