php - 在表单中打印 mySQL 查询

Question

抱歉，新手问题我知道，但是我目前只有在通过 PHP 等使用 echo 命令时才显示 mySQL 结果。

现在，我能够很好地获取信息，但是我希望查询结果以某种形式显示，例如

而不是:-

    $loggedinuser = $session->username;

    //$query = "SELECT surname, firstname FROM PASSENGER WHERE username = '$jared'";
    $query2 = "SELECT p.surname, f.destination FROM PASSENGER p, FLIGHT_INFO f WHERE p.username = '$loggedinuser' AND p.flightNo = f.flightNo";
    $result = mysql_query($query2);  
    while($row = mysql_fetch_array($result))
{
echo $row['surname'] ." ". $row['destination'];
}

    echo mysql_error();

我想以 HTML 格式打印 SQL 结果...例如一个遵循以下约定的:-

    Surname: <input type="text" name="firstname" value="echo $row['surname']/>

如果你看到我来自哪里。这种使用表单显示 mySQL 结果的做法真的很糟糕吗？还是可以这样做？如果不是真的，你会推荐我用什么方法打印数据？在某种表格中，但我该如何使用字段？

问候，

汤姆。

类似这样的工作:-

                            <?php

                    $loggedinuser = $session->username;

            //$query = "SELECT surname, firstname FROM PASSENGER WHERE username = '$jared'";
            $query2 = "SELECT p.surname, f.destination FROM PASSENGER p, FLIGHT_INFO f WHERE p.username = '$loggedinuser' AND p.flightNo = f.flightNo";
        $result = mysql_query($query2);  
    while($row = mysql_fetch_array($result))
    {

  "<form>"
  <input type="text" name="firstname" value="<?php echo {$row['surname']} ?>" disabled       />
     "</form>"
           }
               echo mysql_error();
       ?>

请问如何在以下代码中的 echo 之间添加一个新行？

                echo ' Surname: <input type="text" name="surname" value="'.$row['surname'].  "\" disabled /> ";
        echo 'Forename: <input type="text" name="surname" value="'.$row['forename']. "\" disabled />";
        echo 'Email Address: <input type="text" name="surname" value="'.$row['emailAddress']. "\" disabled />"; 
        echo 'Preference ID: <input type="text" name="surname" value="'.$row['dob']. "\" disabled />"; 
        echo 'Seat Number: <input type="text" name="surname" value="'.$row['seatNo']. "\" disabled />"; 
        echo 'Group ID: <input type="text" name="surname" value="'.$row['groupID']. "\" disabled />"; 

Answer 1

<input value="<?php echo $row["firstname"]; ?>">