Question

获取错误消息 - 注意：尝试获取非对象的属性我尝试使用YELP API进行实验/学习，但我遇到了问题简单的错误信息，我似乎无法弄明白。在此之后：https://github.com/Yelp/yelp-api/tree/master/v2/php

function query_api($term, $location) {     
$response = json_decode(search($term, $location));
$business_id = $response->businesses[0]->id;

print sprintf(
    "%d businesses found, querying business info for the top result \"%s\"\n\n",         
    count($response->businesses),
    $business_id
);

$response = get_business($business_id);

print sprintf("Result for business \"%s\" found:\n", $business_id);
print "$response\n";

}

调用函数

$longopts  = array(
"term::",
"location::",
);

$options = getopt("", $longopts);

$term = $options['term'] ?: '';
$location = $options['location'] ?: '';

query_api($term, $location);

Answer 1

此通知表明$response不是对象，而您正在尝试访问不存在的属性businesses。
使用var_dump($response)获取有关此变量的信息。

尝试使用Yelp API PHP

1 个答案: