Question

我正在以JSON格式获得以下响应，我希望它将其转换为PHP变量。

JSON：

{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900"
,"ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}

输出应为 PHP：

$transkey = aa900d54-7bfb-47e9-a5de-e423ec34a900;
$vkey = fbb28b32-f439-4801-a434-99c70aa388ca

请告诉我如何做。

Answer 1

如果你想访问你的json，请先尝试解码它：

$result = json_decode($yourJSON, true);

foreach($result['CreateTransactionResponse'] as $key => $val){
   echo $transkey = 'TransportKey= ' . $val['TransportKey'] . '<br/>;
   echo $vkey = 'ValidationKey= ' . $val['ValidationKey'];
}

或者它是否是JSON的数组

$result = json_decode($yourJSON, true);

$data = [];
foreach($result['CreateTransactionResponse'] as $key => $val){
   $data[] = [
        'TransportKey' => $val['TransportKey'],
        'ValidationKey' => $val['ValidationKey']
   ];
}
print_r($data);

Answer 2

只需使用json_decode（）;

from __future__ import unicode_literals

import webbrowser

from bs4 import BeautifulSoup
import requests


RANDOM_WIKI_URL = "https://en.wikipedia.org/wiki/Special:Random"


def get_user_input():
    user_input = ''
    while user_input not in ('a', 'b', 'q'):
        print '-' * 79
        print "(a): Open in new browser tab"
        print "(b): Get new random article"
        print "(q): Quit"
        print '-' * 79
        user_input = raw_input("[a|b|q]: ").lower()
    return user_input

def main():
    while True:
        print "=" * 79
        print "Retrieving random wikipedia article..."
        response = requests.get(RANDOM_WIKI_URL)
        data = response.content
        url = response.url

        soup = BeautifulSoup(data, 'html.parser')
        title = soup.select('#firstHeading')[0].get_text()

        print "Random Wikipedia article: '{}'".format(title)
        user_input = get_user_input()
        if user_input == 'q':
            break
        elif user_input == 'a':
            webbrowser.open_new_tab(url)


if __name__ == '__main__':
    main()

Answer 3

json到数组（json_decode），然后是数组中的extract。

$arr = json_decode($json, true);
extract($arr);
var_dump($CreateTransactionResponse);

输出：

array (size=1)
  'CreateTransactionResult' => 
    array (size=3)
      'TransportKey' => string 'aa900d54-7bfb-47e9-a5de-e423ec34a900' (length=36)
      'ValidationKey' => string 'fbb28b32-f439-4801-a434-99c70aa388ca' (length=36)
      'Messages' => 
        array (size=0)
          empty

有关extract

的更多信息

使用$CreateTransactionResult['TransportKey']从JSON访问传输密钥。类似地，$CreateTransactionResult['ValidationKey']用于验证密钥。

Answer 4

    try this code it will work  
$JSON='{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900" ,"ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}';    

    $arr=json_decode($JSON, TRUE);
    foreach ($arr as  $value) {

    foreach ($arr['CreateTransactionResponse'] as $key => $var) {
        echo 'TransportKey = '.$var['TransportKey'].'<br>';
        echo 'ValidationKey = '.$var['ValidationKey'].'<br>';
        foreach ($var['Messages'] as $key => $msg) {


        echo 'Messages = '.$msg.'<br>';
    }

        }
    }

Answer 5

在这种情况下，如果一次只有一个TransportKey和一个ValidationKey值（不是数组/对象），这是最简单的 。否则，如果对象包含我们想要使用的对象或内部对象或转换为变量，则应使用foreach循环遍历该对象。

//Debuggig
//The string you provided is converted to a json object 
//In your case if it is a json object already pass directly it to $j
//below is just for debugging and understanding
//$json='{"CreateTransactionResponse":{"CreateTransactionResult":{"TransportKey":"aa900d54-7bfb-47e9-a5de-e423ec34a900","ValidationKey":"fbb28b32-f439-4801-a434-99c70aa388ca","Messages":{}}}}';
//$j=json_decode($json);

$transkey=$j->CreateTransactionResponse->CreateTransactionResult->TransportKey;
$vkey=$j->CreateTransactionResponse->CreateTransactionResult->ValidationKey;

echo $transkey."</br>";
echo $vkey."<br/>";
/*result as said: 
aa900d54-7bfb-47e9-a5de-e423ec34a900
fbb28b32-f439-4801-a434-99c70aa388ca
*/

如何在PHP变量中获得JSON响应

5 个答案: