Question

您好我一直试图以JSON格式提取数据。这是我的代码

$resourse_url  = 'http://www.livepicly.com/app/api.php?method=list_vendor_name';
$json_data = file_get_contents($resourse_url);
$json_output = json_decode($json_data, TRUE);

$vendor = $json_output['vendor_name'][1];

echo "<pre>";
print_r($vendor);
exit(1);

然而，当我运行代码时，它不会返回任何内容。当我尝试通过firefox访问源URL时，它会询问我将JSON数据保存为* .php的位置，但是当我从Chrome访问源URL时，它会正常显示JSON数据。

我该怎么办？有谁能指出我的解决方案？ THX

Answer 1

在http://jsonformatter.curiousconcept.com/

上运行返回的json

{
   "result":[
      {
         "vendor_id":"726",
         "vendor_name":"Scusa"
      },
      {
         "vendor_id":"519",
         "vendor_name":"Emilie French Restaurant and Bar"
      },
      {
         "vendor_id":"482",
         "vendor_name":"Cassis French Fine Dining"
      },
      {
         "vendor_id":"435",
         "vendor_name":"Asuka Japanase Dining"
      },
      {
         "vendor_id":"12050",
         "vendor_name":"Taipan"
      },
      {
         "vendor_id":"12061",
         "vendor_name":"Social House"
      },
      {
         "vendor_id":"12103",
         "vendor_name":"Harum Manis Indonesian Restaurant"
      },
      {
         "vendor_id":"12193",
         "vendor_name":"Nanny\'s Pavillion (Central Park)"
      },
      {
         "vendor_id":"12272",
         "vendor_name":"Bistro Baron"
      },
      {
         "vendor_id":"20704",
         "vendor_name":"Kitchenette (Central Park)"
      },
      {
         "vendor_id":"21217",
         "vendor_name":"Kitchenette (Plaza Indonesia)"
      },
      {
         "vendor_id":"29859",
         "vendor_name":"Momento Restaurant And Bar"
      },
      {
         "vendor_id":"31055",
         "vendor_name":"Tortuga Kitchen And Bar"
      },
      {
         "vendor_id":"31056",
         "vendor_name":"Tuck And Chug"
      },
      {
         "vendor_id":"31060",
         "vendor_name":"AUROZ Gourmet Grill"
      }
   ]
}

它在此行找到了无效字符

"vendor_name":"Nanny\'s Pavillion (Central Park)"

从URL中提取JSON格式的数据

1 个答案: