我使用json_decode从JSON响应中获取数组：
$result = (json_decode($trends,true));
这给了我以下内容（我没有包括所有的建立详细信息数组结果）
Array ( [FHRSEstablishment] => Array ( [Header] => Array ( [#text] => [ExtractDate] => 2012-05-28 [ItemCount] => 5 [ReturnCode] => Success ) [EstablishmentCollection] => Array ( [EstablishmentDetail] => Array ( [0] => Array ( [FHRSID] => 248659 [LocalAuthorityBusinessID] => INS/06/06179 [BusinessName] => Ancient Raj [BusinessType] => Restaurant/Cafe/Canteen [BusinessTypeID] => 1 [AddressLine1] => 26 North Lane, Canterbury, [PostCode] => CT2 7EE [RatingValue] => 3 [RatingKey] => fhrs_3_en-GB [RatingDate] => 2010-11-18 [LocalAuthorityCode] => 180 [LocalAuthorityName] => Canterbury City [Scores] => [SchemeType] => FHRS [Geocode] => )
以为我可以使用foreach来获得业务名称：
foreach ($result->FHRSEstablishment->EstablishmentCollection->EstablishmentDetail as $detail){
echo $detail['BusinessName'];
}
但我没有得到任何结果。
问题是您正在将$result作为对象访问：
$result->FHRSEstablishment
但是当你调用json_decode时，第二个参数设置为 true ，它会返回一个关联数组，你应该访问它：
$result['FHRSEstablishment']['EstablishmentCollection'] //...
如果您希望能够使用对象表示法访问$result，则应将其定义为：
$result = json_decode($trends) //without 2nd parameter = true