Question

因此，我一直在尝试创建一个嵌套的JSON，其中的数据将来自MYSQL。

写了很长的查询后就得到了这个SQL数据-

 +-------------+--------+-------+-------+
 | Type        | month  | Year  | Total | 
 +-------------+--------+-------+-------+
 | AR          | April  | 2018  | 23443 |
 +-------------+--------+-------+-------+
 | AP          | April  | 2018  | 11456 |
 +-------------+--------+-------+-------+
 | AR          | May    | 2018  | 26483 |
 +-------------+--------+-------+-------+
 | AR          | May    | 2018  | 14442 |
 +-------------+--------+-------+-------+

需要创建此JSON-

    [
       {
        "categorie": "April 2018", 
         "values": [
             {
               "value": 23443, 
               "rate": "AR"
             }, 
             {
               "value": 11456, 
               "rate": "AP"
             }
          ]
       }, 
  .
  .
  .
  ]

从早上开始，我就一直为此head之以鼻，但没有解决方案。在SO中得到了这个答案- Create nested json object using php mysql，但是它使用来自SQL的2个查询来获取数据。

需要帮助来创建将生成JSON的PHP文件。

include '../config/config.php';
if(isset($_GET['sub_cat_id']))
{
         $sub_cat_id = $_GET['sub_cat_id']; 
        $result = mysql_query("SELECT 'AR' as Type,month(DocumentDate) as PeriodM, year(DocumentDate) as PeriodY, sum(Amount) as Total from custledgerentry group by PeriodY,PeriodM union all select 'AP' as Type,month(DocumentDate) as PeriodM, year(DocumentDate) as PeriodY, sum(Amount) as Total from vendledgerentry group by PeriodY,PeriodM"); 
        $json_response = array();
        $i=1;
                        while ($row = mysql_fetch_array($result))
                        {
                        $row_array['categorie'] = $row['month'];        
                        $row_array['value'] = $row['question']; 



        echo json_encode($row_array);
}

Answer 1

当您使用要分组的值作为数组键时，这变得很简单：

$results = [];

foreach ($databaseResult as $row) {
    $category = "$row[month] $row[Year]";

    if (!isset($results[$category])) {
        $results[$category] = ['category' => $category, 'values' => []];
    }

    $results[$category]['values'][] = ['rate' => $row['Type'], 'value' => $row['Total']];
}

echo json_encode(array_values($results));

使用mysql和PHP创建嵌套的JSON

1 个答案: