Php 解析从API返回的json
API返回以下jsonPhp 解析从API返回的json,php,Php,API返回以下json $jsonData = '{"ResponseCode":200, "ResponseDetail":"Success", "AccessToken":"kksjfdlk"}{"ResponseCode":400, "ResponseDetail":"False"}'; 如何访问ResponseCode的值?考虑到发布的字符串不是有效的JSON,您可以使用preg\u match\u all函数提取ResponseCode值: $jsonData = '{"Respon
$jsonData = '{"ResponseCode":200, "ResponseDetail":"Success", "AccessToken":"kksjfdlk"}{"ResponseCode":400, "ResponseDetail":"False"}';
如何访问
ResponseCode
的值?考虑到发布的字符串不是有效的JSON,您可以使用preg\u match\u all
函数提取ResponseCode
值:
$jsonData = '{"ResponseCode":200, "ResponseDetail":"Success", "AccessToken":"kksjfdlk"}{"ResponseCode":400, "ResponseDetail":"False"}';
preg_match_all("/\"ResponseCode\"\s?:\s?(\d+)/", $jsonData, $m);
$response_codes = [];
// if there are matches
isset($m[1]) && $response_codes = $m[1];
print_r($response_codes);
输出:
Array
(
[0] => 200
[1] => 400
)
过滤-你想过滤什么?我试图输出一个响应代码$json=json\u decode($jsonData)$json->ResponseCode;这个问题太宽泛,不清楚。