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我还不太擅长 JSON，所以如果这是一个主要的新手错误，请原谅。我正在向本地文件发送查询，该文件对外部站点的 API 执行 cURL 并获取 JSON 对象。因为我必须为 API 的 x 查询付费，所以我只是复制并粘贴了一个，并用它来代替 cURL。我有以下脚本：

$.ajax({
             type: 'GET',
             url: 'ajax.php?v='+value, //with this being an input value, which is totally irrelevant because I'm not actually doing the cURL query anyway
             dataType: 'json',
             success:function(json){

                 var o_response = json;
                 json = $.parseJSON(json);

                 alert(o_response.toSource());
                 alert(json.toSource());

             },
             error: function (xhr, ajaxOptions, thrownError) {
                alert('There appears to be a problem with the information you submitted. Please try again or contact us.');
              }
        });

ajax.php 中的 PHP 如下所示：

<?

if (isset($_GET['v']) && $_GET['v'] != '') {

$response = '[{"query":"14-22-25-02-W5","response":{"status":"ok","err":[],"lat":51.152259,"lng":-114.202199,"country":"Canada","province":"AB","city":"Calgary","street":"49 Royal Vista Drive NW","street_prox":78,"address":"49 Royal Vista Drive NW, Calgary, AB","lsd":"14-22-25-2 W5","lsd_border":[[51.150459,-114.199327],[51.150447,-114.205067],[51.154059,-114.205071],[51.154072,-114.199332],[51.150459,-114.199327]],"uwi":"","nts":"","nts_border":[],"utm":"11S 695661E 15670479N","utm_v":"Zone 11, 695661 meters easting, 15670479 meters northing (Southern Hemisphere)"}}]';

echo json_encode($response);

}


?>

$response 与 API 给我的完全一样。

我想做的是从中获取“lat”和“lng”值。我的 JavaScript 文件中的第一个示例“alert(o_response.toSource());” bit 使它变成一个字符串，这很好，但我想要一个对象。第二个例子“alert(json.toSource());” 使其成为一个对象，但删除所有键周围的引号。例如，它这样做：

[{query:"14-22-25-02-W5", response:{status:"ok", err:[], lat:51.152259, lng:-114.202199, country:"Canada", province:"AB", city:"Calgary", street:"49 Royal Vista Drive NW", street_prox:78, address:"49 Royal Vista Drive NW, Calgary, AB", lsd:"14-22-25-2 W5", lsd_border:[[51.150459, -114.199327], [51.150447, -114.205067], [51.154059, -114.205071], [51.154072, -114.199332], [51.150459, -114.199327]], uwi:"", nts:"", nts_border:[], utm:"11S 695661E 15670479N", utm_v:"Zone 11, 695661 meters easting, 15670479 meters northing (Southern Hemisphere)"}}]

注意“query”、“response”、“status”、“lat”、“lng”等如何不再有引号。我想这是它应该工作的方式。那么，如果我尝试通过执行以下操作来获得“响应”：

alert(json.response);
alert(json['response']);
alert(json[1]);

我得到的只是 3 个未定义的警报。

我显然错过了一些东西。是不是格式不对？我是否在解析或编码不应该的东西？

任何帮助将不胜感激。

谢谢你。

4

1 回答 1

3

您的响应字符串已经是 JSON，因此无需通过json_encode. 只需使用以下

<?php
header('Content-type: application/json');

if (empty($_GET['v'])) {
    http_response_code(400);
    echo json_encode(['error' => 'Missing "v" parameter']);
    exit;
}

echo '[{"query":"14-22-25-02-W5","response":{"status":"ok","err":[],"lat":51.152259,"lng":-114.202199,"country":"Canada","province":"AB","city":"Calgary","street":"49 Royal Vista Drive NW","street_prox":78,"address":"49 Royal Vista Drive NW, Calgary, AB","lsd":"14-22-25-2 W5","lsd_border":[[51.150459,-114.199327],[51.150447,-114.205067],[51.154059,-114.205071],[51.154072,-114.199332],[51.150459,-114.199327]],"uwi":"","nts":"","nts_border":[],"utm":"11S 695661E 15670479N","utm_v":"Zone 11, 695661 meters easting, 15670479 meters northing (Southern Hemisphere)"}}]';

在 JS 方面，jQuery 已经知道响应负载是 JSON，所以再次，无需json运行$.parseJSON. 您可以直接访问对象文字属性，例如

json[0].response.lat
于 2013-10-29T21:56:38.340 回答