Question

$query = "INSERT INTO 'users' (username, password, conirmpwd, gender) VALUES ('$username','$email','$password','$confirmpwd','$gender')";
    $result = mysqli_query($query);
    if($result){
        $msg="User Created Successfully";
    }else{
        $msg="Username is already taken";
    }
}

任何想法该错误有什么问题？成功连接

警告：mysqli_query（）需要至少2个参数，1中给出1 第12行的D：\ XAMPP \ htdocs \ codeinventor \ register.php

Answer 1

两件事： -

1.查看mysqli_query（）手册： - http://php.net/manual/en/mysqli.query.php

mysqli_query()问两个参数，第一个是连接对象，第二个是查询。因此，请在此处提供您的连接对象，如下所示： -

 $result = mysqli_query($con,$query);// here $con is database connection object(change with yours object)

2. INSERT查询中使用的序列也不正确应该是： -

$query = "INSERT INTO `users` (username, password, conirmpwd, gender) VALUES ('$username','$password','$confirmpwd','$gender')";

或者如果您的表格中也有email列：

$query = "INSERT INTO `users` (username, password,email,conirmpwd, gender) VALUES ('$username','$password','$email','$confirmpwd','$gender')";

Answer 2

您需要指定对数据库建立的连接。正如警告所说，mysqli是一种需要两个参数的方法。所以，你应该传递两个变量mysqli（）as：

$con = mysqli_connect('localhost','your_db','db_username','db_pass');

$query = "INSERT INTO 'users' (username, password, conirmpwd, gender) VALUES ('$username','$email','$password','$confirmpwd','$gender')";
    $result = mysqli_query($con,$query);
    if($result){
        $msg="User Created Successfully";
    }else{
        $msg="Username is already taken";
    }
}

Answer 3

您需要将mysqli资源传递给mysqli_query。像这样：

$con = mysqli_connect($host, $username, $password, $dbName);
$res = mysqli_query($con, "INSERT INTO 'users' (username, password, conirmpwd, gender) VALUES ('$username','$email','$password','$confirmpwd','$gender')");

您需要将$host，$username，$password，$dbName替换为您的数据库信息！

localhost中的SQL错误

3 个答案: