Question

是否可以通过一个查询获得JSON输出？

[{
    "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
}]

我当前的查询：

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

我按from_date分组，但我还需要按sales_group分组... 我可以在mysql表上执行此操作吗？我尝试为Column Highcharts highcharts.com/demo/column-basic

准备数据

修改

好的，也许这会澄清我之前的问题:) 这是我的PHP代码：

$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)']);
}
....

我的问题是：我是否必须为每个特定的sales_group编写单独的查询，还是有更简单的方法？

现在我像这样使用json：

$result = array();
array_push($result,$category);
array_push($result,$series1);
array_push($result,$series2);
array_push($result,$series3);

print json_encode($result);

Answer 1

如果您想一次性完成此操作，则必须按sales_group和from_date进行分组：

$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");

然后收集原始数组中的所有可能日期和所有数据。

$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)']);
}

最后浏览原始数据，检查sales_group在给定日期是否有相关数据或将其设置为零。

$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;
}

最终$data将具有所需的结构和您需要的所有数据。

如何只用一个MySQL请求构建我的JSON数据？

1 个答案: