Question

我试图使用PHP循环json响应，但是我收到了这个错误：注意：尝试在第45行获取非对象的属性这是一行：echo $my_value->website; 这是我的完整代码：

$place = file_get_contents('https://maps.googleapis.com/maps/api/place/details/json?placeid=ChIJjSse9T9644kRo9eQsaidHMQ&key=[my key]');

$place_details = json_decode($place);
echo'<table id="table2" class="display" cellspacing="0" width="100%">';
echo'<thead>';
echo '<tr>';`
echo '<th>Website</th>';
echo '</tr>';
echo '</thead>';`
echo '<tbody>';
foreach ($place_details->result as $my_value)
{ 
echo '<tr>';
echo '<td>';
echo $my_value->website;
echo '</td>';
}
echo '</tbody>';
echo '</table>';

提前致谢

Answer 1

每website只有一个 result，基于此，您可以使用：

$place = file_get_contents('https://maps.googleapis.com/maps/api/place/details/json?placeid=ChIJjSse9T9644kRo9eQsaidHMQ&key=AIzaSyDLBcG7vbaj4HmsE6gwSnNDCD0yHUNb9y4');
$place_details = json_decode($place, true);
echo $place_details['result']['website'];
# https://www.bmc.org/

注意：尝试在google中获取非对象的属性api

1 个答案: