無法將JSON對象轉換爲任何可用的

Question

我對JSON並不太滿意，所以請原諒這是否是一個主要的新手錯誤。我向本地文件發送查詢，該文件對外部網站的API執行cURL並返回JSON對象。因爲我必須爲API查詢x查詢，所以我只複製並粘貼了一個，而我正在使用它來替換cURL。我有以下腳本：

$.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.'); 
       } 
     }); 

而PHP在ajax.php看起來是這樣的：

<? 

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); 

} 


?> 

隨着$響應是完全一樣什麼API給我。

我想要做的是從中獲取「lat」和「lng」值。我的JavaScript文件中的第一個示例是「alert（o_response.toSource（））;」位使它成爲一個字符串，這很好，但我想要一個對象。第二個例子「alert（json.toSource（））;」使其成爲一個對象，但刪除所有鍵周圍的引號。例如，它這樣做：

[{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)"}}] 

注意如何「查詢」，「響應」，「狀態」，「緯度」，「LNG」等，不再有他們周圍的報價。我想這是它應該工作的方式。那麼，如果我嘗試通過執行以下操作來獲得「響應」：

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

我得到的只有3個未定義的提醒。

我明顯錯過了一些東西。它的格式不正確嗎？我解析或編碼的東西，我不應該？

任何幫助將非常感激。

謝謝。

Answer 1

您的響應字符串已經是JSON，因此無需通過json_encode運行。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的已經知道了響應的有效載荷是JSON如此反覆，就沒有必要通過$.parseJSON運行json。你可以直接訪問對象文字屬性，例如：

json[0].response.lat