如何从经纬度逆地理编码API返回结果中提取州名与州代码？

问题描述 #

我购买了一款支持经纬度逆地理编码的API，原本可从中提取州名与州缩写，但现在出现PHP报错。当前使用代码如下：

$json = @file_get_contents($url);
$result = array_values(json_decode($json, true));
print_r($result);

调用后返回数组结构如下，请问如何设置变量分别获取州名（State Name）和州代码（State Code）？

Array
(
    [0] => FeatureCollection
    [1] => Array
        (
            [0] => Array
                (
                    [type] => Feature
                    [properties] => Array
                        (
                            [datasource] => Array
                                (
                                    [sourcename] => openstreetmap
                                    [attribution] => © OpenStreetMap contributors
                                    [license] => Open Database License
                                    [url] => https://www.openstreetmap.org/copyright
                                )

                            [country] => United States
                            [country_code] => us
                            [state] => North Carolina
                            [county] => Cumberland County
                            [city] => Hope Mills
                            [hamlet] => Timberlake
                            [postcode] => 28348
                            [street] => Labonte Drive
                            [housenumber] => 5941
                            [lon] => -78.976173283181
                            [lat] => 34.974986403843
                            [state_code] => NC
                            [distance] => 8.514796509686
                            [result_type] => building
                            [formatted] => 5941 Labonte Drive, Hope Mills, NC 28348, United States of America
                            [address_line1] => 5941 Labonte Drive
                            [address_line2] => Hope Mills, NC 28348, United States of America
                            [timezone] => Array
                                (
                                    [name] => America/New_York
                                    [offset_STD] => -05:00
                                    [offset_STD_seconds] => -18000
                                    [offset_DST] => -04:00
                                    [offset_DST_seconds] => -14400
                                    [abbreviation_STD] => EST
                                    [abbreviation_DST] => EDT
                                )

                            [rank] => Array
                                (
                                    [importance] => -1.15
                                    [popularity] => 1.8893922812046
                                )

                            [place_id] => 51809f819f79be53c0596246bf5acc7c4140f00102f901f3abfb0000000000c00203
                        )

                    [geometry] => Array
                        (
                            [type] => Point
                            [coordinates] => Array
                                (
                                    [0] => -78.976173283181
                                    [1] => 34.974986403843
                                )

                        )

                    [bbox] => Array
                        (
                            [0] => -78.976223283181
                            [1] => 34.974936403843
                            [2] => -78.976123283181
                            [3] => 34.975036403843
                        )

                )

        )

)

解决方案 #

第一步：排查报错原因 #

先去掉 @ 符号屏蔽错误，查看具体报错信息，确定是API请求失败还是JSON解码异常：

$json = file_get_contents($url);
if (!$json) {
    die("API请求失败");
}

$result = json_decode($json, true);
if (json_last_error() !== JSON_ERROR_NONE) {
    die("JSON解码失败：" . json_last_error_msg());
}

第二步：提取州名与州代码 #

根据返回的数组结构，通过层级索引直接访问目标字段，同时添加容错处理避免键不存在时抛出警告：

// 基于当前array_values转换后的数组结构提取
$stateName = $result[1][0]['properties']['state'] ?? '未知州名';
$stateCode = $result[1][0]['properties']['state_code'] ?? '未知州代码';

// 更推荐的方式：不使用array_values，保留原API返回的键名（可读性更强、容错性更高）
// $result = json_decode($json, true); // 去掉array_values
// $features = $result['features'] ?? [];
// if (!empty($features)) {
//     $properties = $features[0]['properties'] ?? [];
//     $stateName = $properties['state'] ?? '未知州名';
//     $stateCode = $properties['state_code'] ?? '未知州代码';
// }

// 输出结果
echo "州名：{$stateName}\n";
echo "州代码：{$stateCode}";

关键说明 #

• 使用 ?? 空合并运算符，确保当目标字段不存在时返回默认值，避免PHP警告；
• 优先保留原API返回的键名（如features），而非用array_values转为索引数组，这样代码更易维护，适配API结构的细微变动。

内容的提问来源于stack exchange，提问作者user3384413