Php 在Json数组中输出Json数组

我正试图得到下面这样的输出。它将纬度和经度合并到对象内部的数组中。这就是我想让我自己去做的。我对php和json非常陌生，已经查了好几天了，运气不好。php可以从mysql数据库中输出这个。如果可以，你怎么做

所需的Json输出

[
    {
        "business_name": "Johns Pizza",
        "latlng": [
            99.457434,
            14.349869
        ],
        "business_id": "1"
    },
        {
        "business_name": "Daves Pizza",
        "latlng": [
            19.457434,
            13.349869
        ],
        "business_id": "2"
    },
       {
        "business_name": "Jakes Pizza",
        "latlng": [
            75.457434,
            12.349869
        ],
        "business_id": "3"
    }
] 

当前Json输出

       [{"business_name":"Crush Wine Bar","latlng":"43.6894658949280400,-116.3533687591552700",
"business_id":"2","distance":"23.0298400562551"},{"business_name":"Apple Headquarters","latlng":"37.3324083200000000,-122.0304781500000000",
"business_id":"3","distance":"23.0298400562551"}]

我的php文件

<?php
if (isset($_GET['lat']) && isset($_GET['lon'])&& isset($_GET['dist'])) {
    //make a simple database connection
    $db = new mysqli();
    $lat = $_GET['lat'];
    $lon = $_GET['lon'];
    $dist = $_GET['dist'];

    $search_sql = "SELECT business_id, business_name, CONCAT_WS(\",\",business_lat, business_lon) AS latlng, ( 3959 * acos( cos( radians(37) ) * cos( radians( " . $lat . " ) ) * cos( radians( " . $lon . " ) - radians(-122) ) + sin( radians(37) ) * sin( radians( " . $lat . " ) ) ) ) AS distance
    FROM business_data HAVING distance < " . $dist . " ORDER BY distance LIMIT 0 , 20";

    $search_results = $db->query($search_sql);

    if($search_results->num_rows) {
        while ($row = $search_results->fetch_assoc()) {
            $return_data[] = array (
                'business_name' => $row['business_name'],
                'latlng'        => $row['latlng'],
                'business_id'   => $row['business_id'],
                'distance'      => $row['distance']
            );
        } //end while
    } else {
        $return_data[] = array();
    }//end if
} else {
    $return_data[] = array();
}

$bd_json = json_encode($return_data);

$db->close();

echo $bd_json;
?>

---

您可以这样做：

while ($row = $search_results->fetch_assoc())
{
    $return_data[] = array(
                            'business_name' => $row['business_name'],
                            'latlng'        => explode(',', $row['latlng']),
                            'business_id'   => $row['business_id'],
                            'distance'      => $row['distance']
                            );
}

---

我不知道该说什么你只是想要漂亮的JSON输出打印吗？或者我遗漏了什么？我正在尝试让latlong结果成为数组中自己的数组。谢谢。但是输出结果是一样的。@BillyKorsen我能看到

var\u dump（$search\u results->fetch\u assoc（））和最后一次
var\u转储（$return\u data）？它实际上没有删除重复项。我会编辑输出，谢谢你，尽管它更容易阅读。我不理解这个问题。我想看看db结果结构和
$return\u data
与我给定的代码。它不应该和旧的一样。