Question

我正在尝试从数据库中名为usuario的表中获取id值，将$ username作为参数传递，函数$ conexion-＆gt; connect（）返回一个mysqli对象。这些函数没有给我任何错误但它没有返回数据库中的值。我错过了什么吗？或者犯了什么错误。谢谢你的帮助。

public function checkUserNameExists($username){
    $conexion = new Connection();
    $conexion->connect();
    $query = "select id from usuario where username = ?";
    $reg = 0;
    $stmt= $conexion->connect()->prepare($query);
    $stmt->bind_param('s',$username);

    $stmt->execute();
    $stmt->bind_result($id);
    while($stmt->fetch()){
        $reg = $id;
    }
    $stmt->close();
    return $reg;
}

这是函数connect（），它位于类文件“Connection”

中

public function connect(){
    $mysqli = new mysqli($this->db_host,$this->db_user,$this->db_pass,$this->db_name);
    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    return $mysqli
}

Answer 1

public function checkUserNameExists($username){
    $conexion = new Connection();
    $conn = $conexion->connect();
    $query = "select id from usuario where username = ?";
    $reg = 0;
    $stmt= $conn->prepare($query);
    $stmt->bind_param('s',$username);

    $stmt->execute();
    $stmt->bind_result($id);
    while($stmt->fetch()){
        $reg = $id;
    }
    $stmt->close();
    return $reg;
}

您应该将新mysqli的返回值存储在变量中，然后使用该变量进行查询或准备。

检索值mysqli_fetch

1 个答案: