使用数据库在PHP中创建JSON

Question

我有这个查询，我已经运行以从三个表中收集数据。我想从收集的这些数据中创建JSON。我编写了以下脚本来创建它。如果有任何快速或聪明的解决方法，请指导我。

这是我需要从以下位置创建JSON的数据库结果的快照：

我正在使用的脚本就是这个。

    $jsonarray = array();

    while ($i<count($result))
   {
        $channel_cat = Array("Channel Category" => Array("name" => $result[$i]['chanct_name']));

        $id = $result[$i]['chanct_id'];
    while($id == $result[$i]['chanct_id'])
        {
            $content[] = Array( "Channel Name" => $result[$i]['con_name'], "image" => $result[$i]['con_image']); 
            $i++;
        }

        $jsonarray[][] = Array($channel_cat, $content);
        unset($content);

   }
     echo json_encode($jsonarray);

这给了我粘贴在下面的结果。

[[[{"Channel Category":{"name":"Movie"}},
[{"Channel Name":"Channel1","image":"Thanks.jpg"},
{"Channel Name":"Channel2","image":"Thanks.jpg"},
{"Channel Name":"Channel4","image":"amazon-logo-b_tpng.png"},
{"Channel Name":"High","image":"Thanks.jpg"}]]]
,[[{"Channel Category":{"name":"Documentary"}},
[{"Channel Name":"Channel7","image":"amazon-logo-b_tpng.png"}]]]]

但我正在寻找下面的结果。

{
"channelsCategories":
    [
        {
            "name":"Movie",
            "image": "MoviePoster",
            "contents":
                [
                    {
                        "name":"Channel 1",
                        "image":"Thanks.jpg",
                    },
                    {
                        "name":"Channel 2",
                        "image":"Thanks.jpg",
                    },
                    {
                        "name":"Channel 4",
                        "image":"amazon-logo...",
                    },
                    {
                        "name":"High",
                        "image":"Thanks.jpg",
                    }
            ]

        },
        {
            "name":"Documentary",
            "image": "MoviePoster",
            "contents":
                [
                    {
                        "name":"Channel 7",
                        "image":"amazon.....",
                    }
            ]

        }


    ]
}

任何帮助或指导都会非常有帮助。

Answer 1

试试这个，它会给你你期望的结果：

$typeArr = array();
foreach($result as $a){
    $typeArr[$a['chanct_name']][] = array(
        'name'=>$a['con_name'],
        'image'=>$a['con_image']
    );
}

$jsonarray = array();
foreach($typeArr as $type=>$contents){
    $jsonarray['channelsCategories'][] = array(
        'name'=>$type,
        'contents'=>$contents
    );
}

echo json_encode($jsonarray);