Question

我正在开发一个使用mysql数据库存储信息的Android应用程序。当我的PHP代码中有以下功能时：

/*
 * Get User details
 * 
 */
public function getUserDetailsEse($email)
{
   $stmt = $this->conn->prepare("
SELECT u.title
     , u.name
     , u.surname
     , s.reg_num
     , s.barcode
     , s.prog
     , s.yos
     , s.sem
     , s.nat_id
     , s.dob
     , s.mobile
     , s.email
     , s.addr
     , s.registad_courses 
  FROM `users` u
     , `students` s 
 WHERE u.id=s.uza_id 
   AND s.email= ?
");

$stmt->bind_param("s", $email);

$stmt->execute();

if ($stmt->num_rows > 0) {
        // user existed 
        echo 'Query ran but nun';
        $stmt->close();
        return true;
    } else {
        // user not existed
        echo 'Query ddnt run';
        $stmt->close();
        return false;
    }

$stmt->close();
}

我想从mysql数据库中获取这些细节，然后在我的sqlite数据库中显示它们以在应用程序内使用。然后我使用：

调用该函数

$details = $db->getUserDetailsEse($email);

if ($details!=FALSE){
    $response1["error"] = FALSE;
    $response1["id"] = $details["id"];
    $response1["details"]["title"] = $details["title"];
    $response1["details"]["name"] = $details["name"];
    $response1["details"]["surname"] = $details["surname"];
    $response1["details"]["reg_num"] = $details["reg_num"];
    $response1["details"]["barcode"] = $details["barcode"];
    $response1["details"]["prog"] = $details["prog"];
    $response1["details"]["yos"] = $details["yos"];
    $response1["details"]["sem"] = $details["sem"];
    $response1["details"]["nat_id"] = $details["nat_id"];
    $response1["details"]["dob"] = $details["dob"];
    $response1["details"]["mobile"] = $details["mobile"];
    $response1["details"]["email"] = $details["email"];
    $response1["details"]["addr"] = $details["addr"];
    $response1["details"]["registad_courses"] = $details["registad_courses"];
    echo json_encode($response1);
}  else {
    $response1["error"] = TRUE;
    $response1["error_msg"] = "Failed to fetch details. Login credentials are wrong. Please try again!";
    echo json_encode($response1);
}

当我在phpmyadmin中运行查询时，它工作正常：

但是当我尝试在Postman上运行它时，我收到以下错误：

以下是我查询的2个表格：

编辑：尝试打印$ DETAILS

Answer 1

正如您在postman响应中看到的那样，登录凭据是错误的。您是否在PDO上配置了连接？

如果您正在使用PDO，请在下面查看连接。

<?php
$dbh = new PDO('mysql:host=localhost;dbname=test', $user, $pass);
?>

编辑：在我的环境上我尝试了这个

    public function getUserDetailsEse($email)
{
    $stmt = $this->conn->prepare("
  SELECT u.title
 , u.name
 , u.surname
 , s.reg_num
 , s.barcode
 , s.prog
 , s.yos
 , s.sem
 , s.nat_id
 , s.dob
 , s.mobile
 , s.email
 , s.addr
 , s.registad_courses 
 FROM `users` u
 , `students` s 
WHERE u.id=s.uza_id AND u.email = :email
");

    $stmt->bindParam("email", $email);

    $stmt->execute();

    if ($stmt->rowCount() > 0) {
        // user existed
        echo 'Query ran but nun';
        $stmt->closeCursor();
        return true;
    } else {
        // user not existed
        echo 'Query ddnt run';
        $stmt->closeCursor();
        return false;
    }

    $stmt->closeCursor();
}

如果仍然无法解决问题，则不在您的查询中。

查询mysql数据库时，PDO查询不返回任何结果

1 个答案: