如何回显数据库中的项目答案

【问题标题】：How to echo an item from a database如何回显数据库中的项目
【发布时间】：2020-06-09 14:19:14
【问题描述】：

我还在学习 PHP，所以请原谅我糟糕的代码。我正在尝试在用户登录后输出他们的名字，但是没有任何返回，请我帮忙。

   <?php

session_start();

$DATABASE_HOST="localhost";
$DATABASE_USER="root";
$DATABASE_PWORD="";
$DATABASE_NAME="registration";

$connection=mysqli_connect($DATABASE_HOST, $DATABASE_USER, $DATABASE_PWORD, $DATABASE_NAME);
if (mysqli_connect_errno()){
    //if there is an issue with connecting to the database, ends code and displays the error
    die("failed to connect to server: " .mysqli_connect_error()); //kills program
}

if (!isset($_POST['email'], $_POST['pswd'])){ //checking if both fields were inputted into on the form, isset()checks if data exists
    //unable to get data
    die("please fill in both email and password"); //kills program
}

$email = mysqli_real_escape_string($connection, $_POST['email']); //saves input as string, preventing misinterpretation
$password = mysqli_real_escape_string($connection, $_POST['pswd']);//saves input as string, preventing misinterpretation

$SQLstatement = "SELECT * FROM users WHERE email='$email' and password='$password'"; //querys the database for match

$Queryresult = mysqli_query($connection, $SQLstatement) or die(mysqli_error($connection)); //runs the query 

$rowsQueryResult = mysqli_num_rows($Queryresult);//number of 'emails' in database where the emails match
$dbFirstName=$rowsQueryResult ['firstName'];
    if ($rowsQueryResult==1){//if the number of emails where a match is made, is 1 
        echo "Welcome $dbFirstName <br/> ";
        echo "successful login. <a href='accountPage.php'>Click</a> here to access the accounts page";      //successful login, links to accounts page
        $_SESSION['firstName']=$dbFirstName;

    }else{ //if matches are 0 or >=2 
        die ('unsuccessful login'); //kills program
    }

?>

感谢您的宝贵时间和帮助

【问题讨论】：

mysqli_num_rows 应该是mysqli_fetch_assoc
非常感谢，问题已解决
我强烈建议切换到PDO，而不是使用简单的mysqli。

标签： php html sql xampp

【解决方案1】：

这个问题可以通过使用mysqli_fetch_assoc()函数代替mysqli_num_rows()来解决。但是，我建议您使用 PDO，因为它更易于实现且更具可读性。

【讨论】：

【解决方案2】：

mysqli_num_rows() 函数返回结果集中的行数。

$rowsQueryResult = mysqli_num_rows($Queryresult);`

将在电子邮件匹配的数据库中提供“电子邮件”的数量。

你需要使用mysqli_fetch_assoc()作为

$row = mysqli_fetch_assoc($Queryresult);
$dbFirstName=$row['firstName'];

【讨论】：