How to build my JSON data with only one MySQL request? - php

Question

Is it possible to get a JSON output looks like with one query?
[{
"name": "Date",
"data": ["2013-01-01", "2013-02-01", "2013-03-01", "2013-04-01", "2013-05-01"] //data from grouped from_date column
}, {
"name": "KD",
"data": [4, 5, 6, 2, 5] // arrays from saldo_sprzedazy for KD sales_group
}, {
"name": "SG",
"data": [5, 2, 3, 6, 7] // arrays from saldo_sprzedazy for SG sales_group
}]
My current query:
SELECT
sales_raport_all.from_date,
SUM(sales_raport_all.saldo_sprzedazy),
klienci_ax_all.sales_group,
klienci_ax_all.nazwa
FROM
sales_raport_all
INNER JOIN
klienci_ax_all
ON
sales_raport_all.konto=klienci_ax_all.konto_odbiorcy
WHERE
YEAR(from_date) = YEAR(CURDATE())
GROUP BY
sales_raport_all.from_date
I group by from_date but I need to also group by sales_group...
Is it possible to do this on the mysql table with I have?
i try to prepare data to Column Highcharts highcharts.com/demo/column-basic
EDIT:
OK, maybe this will clarify my earlier question :)
This is my PHP code:
$query = mysql_query("SELECT
sales_raport_all.from_date,
sales_raport_all.to_date,
sales_raport_all.konto,
SUM(sales_raport_all.saldo_sprzedazy),
SUM(sales_raport_all.wartosc_kosztowa),
SUM(sales_raport_all.marza),
klienci_ax_all.sales_group,
klienci_ax_all.nazwa
FROM
sales_raport_all
INNER JOIN
klienci_ax_all
ON
sales_raport_all.konto=klienci_ax_all.konto_odbiorcy
WHERE
YEAR(from_date) = YEAR(CURDATE())
GROUP BY
sales_raport_all.from_date");
$category = array();
$category['name'] = 'Data';
while($r = mysql_fetch_array($query)) {
$category['data'][] = $r['from_date'];
}
$querySG = mysql_query("SELECT
sales_raport_all.from_date,
sales_raport_all.to_date,
sales_raport_all.konto,
SUM(sales_raport_all.saldo_sprzedazy),
SUM(sales_raport_all.wartosc_kosztowa),
SUM(sales_raport_all.marza),
klienci_ax_all.sales_group,
klienci_ax_all.nazwa
FROM
sales_raport_all
INNER JOIN
klienci_ax_all
ON
sales_raport_all.konto=klienci_ax_all.konto_odbiorcy
WHERE
klienci_ax_all.sales_group = 'SG'
AND
YEAR(from_date) = YEAR(CURDATE())
GROUP BY
sales_raport_all.from_date");
$series1 = array();
$series1['name'] = 'SG';
while($r = mysql_fetch_array($querySG)) {
$series1['data'][] = intval($r['SUM(sales_raport_all.saldo_sprzedazy)']);
}
....
My question is: do I have to write a separate query for every specific sales_group or is there a simpler way?
Now I use json like this:
$result = array();
array_push($result,$category);
array_push($result,$series1);
array_push($result,$series2);
array_push($result,$series3);
print json_encode($result);

Answer 1

If you want to do this in one go, you have to group by both sales_group and from_date:
$query = mysql_query("SELECT
sales_raport_all.from_date,
sales_raport_all.to_date,
sales_raport_all.konto,
SUM(sales_raport_all.saldo_sprzedazy),
SUM(sales_raport_all.wartosc_kosztowa),
SUM(sales_raport_all.marza),
klienci_ax_all.sales_group,
klienci_ax_all.nazwa
FROM
sales_raport_all
INNER JOIN
klienci_ax_all
ON
sales_raport_all.konto=klienci_ax_all.konto_odbiorcy
WHERE
YEAR(from_date) = YEAR(CURDATE())
GROUP BY
sales_raport_all.from_date,
klienci_ax_all.sales_group
ORDER BY
sales_raport_all.from_date,
klienci_ax_all.sales_group");
Then collect all the possible dates and all the data in a raw array.
$raw = array();
$dates = array();
while ($r = mysql_fetch_array($query)) {
$date = $r['from_date'];
if (!in_array($date, $dates)) $dates[] = $date;
$sales_group = $r['sales_group'];
$raw[$sales_group][$date] = intval($r['SUM(sales_raport_all.saldo_sprzedazy)']);
}
Lastly go through the raw data, and check if on a given date the sales_group has relevant data or set it to zero.
$data = array();
$data[0] = array('name' => "Date", 'data' => $dates);
foreach ($raw as $name => $d) {
$new_data = array('name' => $name, 'data' => array());
foreach ($dates as $date) {
$new_data['data'][] = isset($d[$date]) ? $d[$date] : 0;
}
$data[] = $new_data;
}
The final $data will have the desired structure and all the data you need.