Php JSON：以不同格式显示JSON结果

我只是JSON的新手。根据我的问题，下面是我目前的代码广告结果

代码1

<?php 

    require_once '../config/configPDO.php';

    header('Content-Type: application/json');

    $response = array();

    $badgeid = '10010080';
    $pwd = '10010080';

    $stmt = $conn->prepare("SELECT * FROM ot_users WHERE badgeid = '$badgeid' AND pwd = '$pwd' AND roles_id = 7 AND team_id <> 1");
    $stmt->execute();
    $result = $stmt->fetch(PDO::FETCH_ASSOC);

    if (!empty($result)) {

        $response['error'] = false; 
        $response['message'] = 'Login successfull'; 
        $response['user'] = $result;  

    }else{
        $response['error'] = false; 
        $response['message'] = 'Invalid username or password';
    }


    echo json_encode($response);

?>

---

只是php变量将$result更改为$result[0]试试这个。无法执行您的“$badgeid&pwd=$pwd”链接，因为passowrd不正确
    header('Content-Type: application/json');

    $response = array();

    $badgeid = '10010080';
    $pwd = '10010080';

    $url = "http://172.20.0.45/TGWebService/TGWebService.asmx/ot_displayUser?badgeid=$badgeid&pwd=$pwd";
    $data = file_get_contents($url);
    $json = json_decode($data);
    $result = $json->otUserList;

        if (!empty($result)) {

            $response['error'] = false; 
            $response['message'] = 'Login successfull'; 
            $response['user'] = $result[0];  

        }else{


       $response['error'] = false; 
        $response['message'] = 'Invalid username or password';
    }

    echo json_encode($response);

欢迎，@PeterSondak。你已经看过电影了吗

我假设如果您使用
var\u dump
测试
$result
变量，您的结果将如下所示：

object(stdClass)#1 (3) {
  ["error"]=>
  bool(false)
  ["message"]=>
  string(17) "Login successfull"
  ["user"]=>
  array(1) {
    [0]=>
    object(stdClass)#2 (7) {
      ["badgeid"]=>
      string(8) "10010080"
      ["email"]=>
      string(0) ""
      ["pwd"]=>
      string(0) ""
      ["fullname"]=>
      string(17) "AZWAN BIN SANIMIN"
      ["roles_id"]=>
      string(1) "7"
      ["team_id"]=>
      string(1) "2"
      ["users_id"]=>
      string(0) ""
    }
  }
}

如您所见，
user
属性的值是一个数组。因此，您应该将代码更改为以下内容，以获得第一个索引：

$result = $json->otUserList[0]

您将得到：

"{"error":"false","message":"Login successfull","user":{"badgeid":"10010080","email":"","pwd":"","fullname":"AZWAN BIN SANIMIN","roles_id":"7","team_id":"2","users_id":""}}"

第一个Json返回完整的对象数据可以用于每次更改。你能编辑答案吗？第二个是多维的，更改结果只得到一个列表并将其放入
用户
@Kevin你能编辑我的答案吗。请
"{"error":"false","message":"Login successfull","user":{"badgeid":"10010080","email":"","pwd":"","fullname":"AZWAN BIN SANIMIN","roles_id":"7","team_id":"2","users_id":""}}"