从URL中提取JSON格式的数据
问题描述:
Hi i have been trying to pull data in JSON format. Here is my code
$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);
However when i run the code it does not return anything. When i try to access the source URL via firefox, it asks me where to save the JSON data as *.php, however when i access the source URL from Chrome it displays the JSON data as normal.
What do i have to do here? Can anyone point me to a solution? Thx
答
running the returned json on http://jsonformatter.curiousconcept.com/
{
"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"
}
]
}
it found invalid characters on this line
"vendor_name":"Nanny\'s Pavillion (Central Park)"