【问题标题】:How to reference a json array having key value pairs using php如何使用php引用具有键值对的json数组
【发布时间】:2018-09-05 09:43:55
【问题描述】:

我已经阅读了类似的帖子,但在我的情况下仍然可以使用一些帮助...我正忙着从包含键值对的 json 数组中获取值:

{"address": "4 Ficticious Ave", 
"city": "Miami", 
"country": "United States", 
"email": "jane_doe@gmail.com", 
"first_name": "Jane", 
"last_name": "Doe", 
"state": "FL", 
"zip_code": "03423", 
"response_data": 
"[{"key":"7122", "value":"37-52"},
{"key":"7123","value":"Female"},
{"key":"7124","value":"$35,000 to $50,000 USD"},
{"key":"6176","value":"Miami"},
{"key":"6177","value":"FL"},
{"key":"6179","value":"United States"}]"}

我尝试获取所有值,但没有成功,尤其是使用 response_data 键|值对:

`// 从 $data 数组中获取 JSON 数据

// Identify the content as json
header("Content-Type: application/json; charset=UTF-8");

// get the contents of the JSON file
$data = file_get_contents("php://input");

//decode JSON data to PHP array
$content = json_decode($data, true);

//Fetch the details of customer
$Cust_Fname = $content['first_name'];
$Cust_Lname = $content['last_name'];
$Cust_Email = $content['email'];
$Street_Address = $content['address'];
$City = $content['city'];
$State = $content['state'];
$Country = $content['country'];
$Postal_Code = $content['zip_code'];

//also fetch the appended "array" of key/value fields...
$Response_AgeKey = $content['reponse_data'][0]['key'];
$Response_GenderKey = $content['reponse_data'][1]['key'];
$Response_IncomeKey = $content['reponse_data'][2]['key'];
$Response_CityKey = $content['reponse_data'][3]['key'];
$Response_StateKey = $content['reponse_data'][4]['key'];
$Response_CountryKey = $content['reponse_data'][5]['key'];
$Response_Age = $content['reponse_data'][0]['value'];
$Response_Gender = $content['reponse_data'][1]['value'];
$Response_Income = $content['reponse_data'][2]['value'];
$Response_City = $content['reponse_data'][3]['value'];
$Response_State = $content['reponse_data'][4]['value'];
$Response_Country = $content['reponse_data'][5]['value']

`

【问题讨论】:

    标签: php json multidimensional-array keyvaluepair


    【解决方案1】:

    "response_data" 不是有效的 JSON。有一对额外的双引号将数组括起来。删除封闭的双引号,然后它应该可以工作。

    {
      "address": "4 Ficticious Ave",
      "city": "Miami",
      "country": "United States",
      "email": "jane_doe@gmail.com",
      "first_name": "Jane",
      "last_name": "Doe",
      "state": "FL",
      "zip_code": "03423",
      "response_data": [
        {
          "key": "7122",
          "value": "37-52"
        },
        {
          "key": "7123",
          "value": "Female"
        },
        {
          "key": "7124",
          "value": "$35,000 to $50,000 USD"
        },
        {
          "key": "6176",
          "value": "Miami"
        },
        {
          "key": "6177",
          "value": "FL"
        },
        {
          "key": "6179",
          "value": "United States"
        }
      ]
    }
    

    【讨论】:

    • 谢谢你……它帮助我获得了@Elementary 提供的删除代码。
    【解决方案2】:

    正如有人所说,“response_data”不是有效的 JSON。数组中有一对额外的双引号。但我认为,如果它是响应而不是手动输入的 json,最好在使用 Json_decode 解码之前自动规范化接收到的字符串,主要是通过删除封闭的双引号。你可以使用下面的代码来实现它:

    <?php
     // Identify the content as json
    header("Content-Type: application/json; charset=UTF-8");
    
    // get the contents of the JSON file
    $data = file_get_contents("php://input");
    
    //normalize the json in order to be properly decoded
    
    $start=strpos($data,':',strpos($data,'response_data'));
    $get=substr($data,$start+1,strrpos($data,'"')-$start);
    $data=str_replace($get,trim(trim($get),'"'),$data);
    
    //decode JSON data to PHP array
    $content = json_decode($data, true);
    
    //Fetch the details of customer
    
    $Cust_Fname = $content['first_name'];
    $Cust_Lname = $content['last_name'];
    $Cust_Email = $content['email'];
    $Street_Address = $content['address'];
    $City = $content['city'];
    $State = $content['state'];
    $Country = $content['country'];
    $Postal_Code = $content['zip_code'];
    
    //also fetch the appended "array" of key/value fields...
    $Response_AgeKey = $content['reponse_data'][0]['key'];
    $Response_GenderKey = $content['reponse_data'][1]['key'];
    $Response_IncomeKey = $content['reponse_data'][2]['key'];
    $Response_CityKey = $content['reponse_data'][3]['key'];
    $Response_StateKey = $content['reponse_data'][4]['key'];
    $Response_CountryKey = $content['reponse_data'][5]['key'];
    $Response_Age = $content['reponse_data'][0]['value'];
    $Response_Gender = $content['reponse_data'][1]['value'];
    $Response_Income = $content['reponse_data'][2]['value'];
    $Response_City = $content['reponse_data'][3]['value'];
    $Response_State = $content['reponse_data'][4]['value'];
    $Response_Country = $content['reponse_data'][5]['value']
    
    ?>
    

    【讨论】:

    • 谢谢,多一双眼睛总是有帮助的!好的,所以代码运行没有错误,并且数据作为新记录插入 MySQL [未提供代码]。奇怪的是,response_data 键值对中的任何数据都没有显示……所有这些列都是空白的。猜猜我得玩弄那个检索语法。非常感谢...我现在走得更远了。
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