如何从响应应用程序json获取文件代码和url？
filecode: rbukubn7al60 and url: https://example.com/rbukubn7al60个
{"server_time":"2024-01-03 02:45:55","status":200,"result":{"filecode":"rbukubn7al60","url":"https://example.com/rbukubn7al60"},"msg":"OK"}
如何从响应应用程序json获取文件代码和url？
filecode: rbukubn7al60 and url: https://example.com/rbukubn7al60个
{"server_time":"2024-01-03 02:45:55","status":200,"result":{"filecode":"rbukubn7al60","url":"https://example.com/rbukubn7al60"},"msg":"OK"}
您可以使用PHP的json_decode函数来解析JSON响应并访问filecode和url属性.以下是你如何做到这一点:
$json = '{"server_time":"2024-01-03 02:45:55","status":200,"result":{"filecode":"rbukubn7al60","url":"https://example.com/rbukubn7al60"},"msg":"OK"}';
$data = json_decode($json);
$filecode = $data->result->filecode;
$url = $data->result->url;
echo $filecode; // Outputs: rbukubn7al60
echo $url; // Outputs: https://example.com/rbukubn7al60
在此代码中，json_decode($json)将JSON字符串转换为PHP对象.$data->result->filecode和$data->result->url访问result对象的filecode和url属性.