如何在 php 中回显嵌套的 JSON 对象

Question

我正在尝试弄清楚如何从下面的 JSON 中回显address1。我已经试过这个 - echo "$arr->location[1]->address1<br>"; ，但它返回此错误

可捕获的致命错误：类 stdClass 的对象无法转换为第 202 行/home/benrud/public_html/student/webdesign/2016/02_benrud/tinker/data/index.php 中的字符串。

echo $arr;返回下面的 JSON。

{
  "photos": [
    "https://s3-media2.fl.yelpcdn.com/bphoto/37El1q8mqM_1tKtQugncZQ/o.jpg",
    "https://s3-media1.fl.yelpcdn.com/bphoto/GLsNPPz5do-_NJktIQvz6w/o.jpg",
    "https://s3-media3.fl.yelpcdn.com/bphoto/Z4rdHERgb10MZgDXnct5lA/o.jpg"
  ],
  "coordinates": {
    "latitude": 33.0479031276,
    "longitude": -117.256002333
  },
  "image_url": "https://s3-media1.fl.yelpcdn.com/bphoto/37El1q8mqM_1tKtQugncZQ/o.jpg",
  "is_claimed": false,
  "id": "oscars-mexican-seafood-encinitas-2",
  "review_count": 48,
  "rating": 4.5,
  "hours": [
    {
      "hours_type": "REGULAR",
      "is_open_now": true,
      "open": [
        {
          "is_overnight": false,
          "end": "2100",
          "day": 0,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2100",
          "day": 1,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2100",
          "day": 2,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2100",
          "day": 3,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2200",
          "day": 4,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2200",
          "day": 5,
          "start": "0800"
        },
        {
          "is_overnight": false,
          "end": "2100",
          "day": 6,
          "start": "0800"
        }
      ]
    }
  ],
  "display_phone": "[phone]",
  "categories": [
    {
      "alias": "seafood",
      "title": "Seafood"
    },
    {
      "alias": "mexican",
      "title": "Mexican"
    }
  ],
  "price": "$",
  "phone": "[phone]",
  "name": "Oscars Mexican Seafood",
  "location": {
    "zip_code": "92024",
    "address3": null,
    "address1": "115 N El Camino Real",
    "country": "US",
    "city": "Encinitas",
    "state": "CA",
    "cross_streets": "Via Molena & Encinitas Blvd",
    "display_address": [
      "115 N El Camino Real",
      "Encinitas, CA 92024"
    ],
    "address2": ""
  },
  "transactions": [],
  "url": "https://www.yelp.com/biz/oscars-mexican-seafood-encinitas-2?adjust_creative=YqqOIA_bNY3Qb_A1TRMMUg&utm_campaign=yelp_api_v3&utm_medium=api_v3_business_lookup&utm_source=YqqOIA_bNY3Qb_A1TRMMUg",
  "is_closed": false
}

Answer 1

位置不

是一个数组，所以我想象只是$arr->location->address1.

Answer 2

$arr = json_decode($json, true);

true 参数确保它是一个数组而不是一个对象

Answer 3

您必须首先解码然后调用密钥，以便：

// Decode the JSON STRING
$arr = json_decode($arr, true); 
/*
 * first argument is the JSON STRING, 
 * Second sets the flag that the string is a dictionary 
 * (associative array)
 */

现在是时候调用该元素了。我把它放在一个条件中，以防止错误

if (array_key_exists('address1', $arr['location'])) {
    echo $arr['location']['address1'];
}
else {
    echo "Array element Not Found. Here is what I have:nr";
    print_r($arr);
}

这应该返回元素的值或转储解析的 PHP 数组以供查看，以便您可以编辑 if 语句以获取正确的位置。