将Google API JSON响应转换为单个php变量
我有一个网页,它包含两个邮政编码,并将距离与谷歌发布的其他信息分开 守则:将Google API JSON响应转换为单个php变量,php,json,api,Php,Json,Api,我有一个网页,它包含两个邮政编码,并将距离与谷歌发布的其他信息分开 守则: $url = "http://maps.googleapis.com/maps/api/distancematrix/json?origins=$postcode1&destinations=$postcode2&mode=bicycling&language=en-EN&sensor=false"; $data = @file_get_contents($url); $result =
$url = "http://maps.googleapis.com/maps/api/distancematrix/json?origins=$postcode1&destinations=$postcode2&mode=bicycling&language=en-EN&sensor=false";
$data = @file_get_contents($url);
$result = json_decode($data, true);
print_r($result);
结果是:
{
"destination_addresses" : [ "Pound Rd, Oldbury B68 8NE, UK" ],
"origin_addresses" : [ "Vyse St, Birmingham B18 6NE, UK" ],
"rows" : [
{
"elements" : [
{
"distance" : {
"text" : "10.4 km",
"value" : 10384
},
"duration" : {
"text" : "36 mins",
"value" : 2133
},
"status" : "OK"
}
]
}
],
"status" : "OK"
}
我对JSON还不熟悉,所以如果这听起来很愚蠢,请原谅。我试了不少。最接近我的是
echo $result['destination_addresses'][0];
返回目标地址。我想知道如何单独获取所有内容,尤其是距离文本
提前谢谢
Jake32以下是解决方案:-
echo "Destination Address=".$result['destination_addresses'][0]."<br>";
echo "Origin Address=".$result['origin_addresses'][0]."<br>";
echo "Distance text=".$result['rows'][0]['elements'][0]['distance']['text']."<br>";
echo“Destination Address=”.$result['Destination_addresses'][0]。“
”;
echo“Origin Address=“.result['Origin_addresses'][0]。”
;
echo“Distance text=”.$result['rows'][0]['elements'][0]['Distance']['text'].“
”;