php - 不会查询程序的更新部分

Question

我很难确定我的代码问题的原因，也就是说，它不会查询“UPDATE”部分，但“SELECT”部分确实有效。当我尝试使用 print_r 函数时，它给出了一个错误/警告:

"Warning: mysqli_query(): Couldn't fetch mysqli"** and **"Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given"

if(!isset($_POST['n_pass'])&&!isset($_POST['n_pass'])){

    if(!isset($_POST['password'])||$_POST['password']==""){
            echo 'enter current password';
            die;
    } else {
            include 'include/database.php';
            $fname = mysqli_real_escape_string($conn, $_POST['fname']);
            $lname = mysqli_real_escape_string($conn,$_POST['lname']);
            $email = mysqli_real_escape_string($conn,$_POST['email']);
            $username = mysqli_real_escape_string($conn,$_POST['uname']);
            $password = mysqli_real_escape_string($conn,$_POST['password']);

            //Check if the password is equal to the password inside database
            $sql = "SELECT password FROM users where id = $id";
            $result = mysqli_query($conn, $sql);
            $row = mysqli_fetch_assoc($result);
            $cpass = $row['password'];
            $verify_pass = password_verify($password,$cpass); //check if current password is equal to the existing password

            if($verify_pass != 1){
                echo 'incorrect password';
                die;
            } else {
                **//Update Data
                $sql="UPDATE users SET firstname=$fname, lastname=$lname, email=$email, username=$username where id=$id";
                $result = mysqli_query($conn, $sql);
                $row = mysqli_fetch_assoc($result);

                print_r($row['firstname']);
                die;
                header("Location: profile.php?successfullyupdated");
            }
        }   
    }

Answer 1

您的查询缺少字符串周围的引号:

$sql="UPDATE users SET firstname='$fname', lastname='$lname', email='$email', username='$username' where id=$id";

您只能跳过 id 字段，因为它是一个整数。

旁注:您对 SQL 注入(inject)持开放态度。您应该使用准备好的语句。google 上有很多资源可以从这个主题开始

最后，请注意

$row = mysqli_fetch_assoc($result); 
print_r($row['firstname']);

完全没用，因为您没有从 UPDATE 查询返回任何内容。你可以这样做:

if(mysqli_query($conn, $sql)){
    //query was successful - run your code here for success
}else{
    //query failed - run your code here for fail
}