使用 JSON 输出列表内容的正确方法是什么？

Question

我正在使用 yelp API 来为我的网站随机抽取附近的列表。来自 yelp 在其网站上的文档：

rating                 number   Rating for this business (value ranges from 1, 1.5, ... 4.5, 5)
rating_img_url         string   URL to star rating image for this business (size = 84x17)
rating_img_url_small   string   URL to small version of rating image for this business (size = 50x10)
rating_img_url_large   string   URL to large version of rating image for this business (size = 166x30)
snippet_text           string   Snippet text associated with this business
snippet_image_url      string   URL of snippet image associated with this business
location               dict     Location data for this business
location.address       list     Address for this business. Only includes address fields.
location.display_address list   Address for this business formatted for display. Includes all address fields, cross streets and city, state_code, etc.
location.city          string   City for this business
location.state_code    string   ISO 3166-2 state code for this business
location.postal_code   string   Postal code for this business
location.country_code  string   ISO 3166-1 country code for this business
location.cross_streets string   Cross streets for this business

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如何输出 location.display_address 等位置变量？我下面的代码通过回显“$response->businesses->rating”正确输出字符串和数字。但是，我的代码的最后一行不起作用。

// Handle Yelp response data
$response = json_decode($data);
$business = $response->businesses;

$numbers = range(0, 19);
shuffle($numbers);
echo "<img src='".$business[$numbers[$ran]]->image_url."'><br/>";
echo $business[$numbers[$ran]]->name."<br/>";
echo "<img border=0 src='".$business[$numbers[$ran]]->rating_img_url_large."'><br/>";
echo "<br/>";

echo $business[$numbers[$ran]]->location[display_address]."<br/>";

我得到的错误是

“致命错误：无法在 /home/content/38/11397138/html/yelpcall.php 第 51 行使用 stdClass 类型的对象作为数组”

编辑： var dump 返回：

object(stdClass)#14 (6) { ["city"]=> string(11) "Chino Hills" ["display_address"]=> array(2) { [0]=> string(14) "2923 Chino Ave" [1]=> string(21) "Chino Hills, CA 91709" } ["postal_code"]=> string(5) "91709" ["country_code"]=> string(2) "US" ["address"]=> array(1) { [0]=> string(14) "2923 Chino Ave" } ["state_code"]=> string(2) "CA" }

Answer 1

试试var_dump($business[$numbers[$ran]]->location);，这会告诉你更多关于你正在使用的变量的信息，并帮助你公开它的属性。

在您的情况下，位置属性的类型似乎是 StdClass。因此访问属性如下：$business[$numbers[$ran]]->location->display_address

编辑：感谢您的更新。 StdClass 的 display_address 属性是一个数字索引数组，因此要访问数组的元素，您可以这样做： $business[$numbers[$ran]]->location->display_address[0] 但它看起来更像是这样使用的：

$address = $business[$numbers[$ran]]->location->display_address;
foreach( $address as $line ){
    echo $line.'<br />';
} 

Answer 2

尝试 echo $business[$numbers[$ran]]->location["display_address"]."
"; （在“display_address”键名周围添加引号。