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php解析json数据

json php

3

user756659 · 技术社区 · 9 年前

从来没有这样做过,并且对格式以及如何从中访问特定变量感到困惑。

结果:

$result = json_decode($result);
print_r($result);

stdClass Object
(
    [input] => stdClass Object
        (
            [address_components] => stdClass Object
                (
                    [number] => 1109
                    [predirectional] => N
                    [street] => Highland
                    [suffix] => St
                    [formatted_street] => N Highland St
                    [city] => Arlington
                    [state] => VA
                    [zip] => 22201
                    [country] => US
                )

            [formatted_address] => 1109 N Highland St, Arlington, VA 22201
        )

    [results] => Array
        (
            [0] => stdClass Object
                (
                    [address_components] => stdClass Object
                        (
                            [number] => 1109
                            [predirectional] => N
                            [street] => Highland
                            [suffix] => St
                            [formatted_street] => N Highland St
                            [city] => Arlington
                            [county] => Arlington County
                            [state] => VA
                            [zip] => 22201
                            [country] => US
                        )

                    [formatted_address] => 1109 N Highland St, Arlington, VA 22201
                    [location] => stdClass Object
                        (
                            [lat] => 38.886672
                            [lng] => -77.094735
                        )

                    [accuracy] => 1
                    [accuracy_type] => rooftop
                    [source] => Arlington
                )

            [1] => stdClass Object
                (
                    [address_components] => stdClass Object
                        (
                            [number] => 1109
                            [predirectional] => N
                            [street] => Highland
                            [suffix] => St
                            [formatted_street] => N Highland St
                            [city] => Arlington
                            [county] => Arlington County
                            [state] => VA
                            [zip] => 22201
                            [country] => US
                        )

                    [formatted_address] => 1109 N Highland St, Arlington, VA 22201
                    [location] => stdClass Object
                        (
                            [lat] => 38.886665
                            [lng] => -77.094733
                        )

                    [accuracy] => 1
                    [accuracy_type] => rooftop
                    [source] => Virginia Geographic Information Network (VGIN)
                )

        )

)

如何获取lat和lng值?我感到困惑,因为其中混合了对象和数组。我从未以这种方式在php中使用json。。。这是一个以json格式返回的地理编码api。

$lat = $result['results'][0]['location']['lat']; 等等

4 回复 | 直到 9 年前

1

3

Death-is-the-real-truth 9 年前

你必须使用 true 作为中的第二个参数 json_decode 因此,输出将是正常的php数组,而不是 stdclass 数组(以便您可以使用您正在尝试的方法)

因此,请执行以下操作:-

$result = json_decode($result,true);

echo $lat = $result['results'][0]['location']['lat'];
echo $lng = $result['results'][0]['location']['lng'];

注:-

1. 使用当前格式,您还可以获得如下数据:-

echo $lat = $result->results[0]->location->lat; //using object approach 
echo $lng = $result->results[0]->location->lng; //using object approach

2. 使用 foreach() 获取所有 lat-lng 记录。

foreach($result->results as $val) { //if using stdclass array
   echo "Latitude = ".$val->location->lat;
   echo "Longitude = ".$val->location->lng;
}

或:-

foreach($result['results'] as $val) {//if using normal array
   echo "Latitude = ".$val['location']['lat'];
   echo "Longitude = ".$val['location']['lng'];
}

2

2

Subodh Ghulaxe 9 年前

$result = json_decode($result);
echo $result->results[0]->location->lat;
echo $result->results[0]->location->lng;

3

1

Duke 9 年前

从您的示例中可以看出,您需要获取多条记录。使用 foreach() 打印详细信息

$result = json_decode($result);
foreach($result->results as $value) {
   echo "Lat--",$value->location->lat; //using object
   echo "Long--",$value->location->lng;
}

4

1

Death-is-the-real-truth 9 年前

将第二个参数设为true,以将所有对象转换为数组。否则你可以得到这样的值。

$result->results[0]->location->lat;
$result->results[0]->location->lng;

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