如何使用PHP将mysql数据转换为所需的json格式

Question

我有一个关于使用MySQL查询将我的数据转换为JSON对象的问题。我的查询是转换为JSON对象，

我学会了使用array（）并尝试创建一个可以在任何移动应用程序中使用的json。

mysql数据是 -

state_name      city_name  varient_name price

Uttar Pradesh   Lucknow     varient1    1202
Uttar Pradesh   Lucknow     varient2    1203
Uttar Pradesh   Lucknow     varient3    1204
Delhi           Faridabad   varient1    1206
Delhi           Faridabad   varient2    1207
Delhi           Faridabad   varient3    1208
Delhi           Saket       varient1    1207
Delhi           Saket       varient2    1208
Delhi           Saket       varient3    1209

以及格式所需的响应 -

{
    "states": [
        {
            "state_name": "Uttar Pradesh",
            "cities": [
                {
                    "city_name": "Lucknow",
                    "pricedata": [
                        {
                            "varient_name": "varient1",
                            "varient_price": "1201"
                        },
                        {
                            "varient_name": "varient2",
                            "varient_price": "1201"
                        },
                        {
                            "varient_name": "varient3",
                            "varient_price": "1203"
                        }
                    ]
                }
            ]
        },
        {
            "state_name": "Delhi",
            "cities": [
                {
                    "city_name": "Faridabad",
                    "pricedata": [
                        {
                            "varient_name": "varient1",
                            "varient_price": "1204"
                        },
                        {
                            "varient_name": "varient2",
                            "varient_price": "1205"
                        },
                        {
                            "varient_name": "varient3",
                            "varient_price": "1206"
                        }
                    ]
                },
                {
                    "city_name": "Saket",
                    "pricedata": [
                        {
                            "varient_name": "varient1",
                            "varient_price": "1207"
                        },
                        {
                            "varient_name": "varient2",
                            "varient_price": "1208"
                        },
                        {
                            "varient_name": "varient3",
                            "varient_price": "1209"
                        }
                    ]
                }
            ]
        }
    ]
}

Answer 1

$aR = array();
$req = mysqli_query("SELECT * FROM table");
while($row = mysqli_fetch_assoc($req))
{
    $aR[] = $row;
}
echo '<pre>'.print_r(json_encode($aR),1).'</pre>';