未获取mysql数据库数据..用php转换成html页面(被难住了)


Not get mysql db data .... into html page with php (stumped)

我正在处理一项任务,它要求我从3aStudent_slip.php中选择一个"slip_id",并将其传递给4aservice_request.php,并填充php代码中正在构建的表。我没有任何php类,所以我真的很难理解为什么它没有从服务器上的"ProgrammingDatabase"中获取任何数据库。

正在使用以下代码。。。

<?php
    require_once('auth.php');
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Service Requests</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
    <div id="innerWrapper">
<h1>Service request by <?php echo $_SESSION['SESS_FIRST_NAME'];?></h1>
<a href="index.php">Login Page</a> | 
<a href="amenu.php">Menu Page</a> | 
<a href="logout.php">Logout</a>
<?php
$slip_id = strtoupper($_POST['slip_id']);
echo("<h2>Services for Slip ID $slip_id</h2>");
//Verify Password
$vlogin=$_SESSION['vlogin'];
$vpassword=$_SESSION['vpasswd'];
//Connection String
$con=mysql_connect("localhost", $vlogin, $vpasswd);
if(!$con)
{
    die("Could not connect".mysql_error());
}
//Select Database
mysql_select_db("ProgrammingDatabase", $con);
//The actual SQL code goes below into the structured variable $result
$result=mysql_query("SELECT * FROM service_request");
//Constructing the table and column names
echo "<table border='1'>
<tr>
<th>Service ID</th>
<th>Description</th>
</tr>";
//Looping until there are no more records from $result
//If there are records, print the column for that row
//do the while loop below with the variables from $result
while($row=mysql_fetch_array($result))
{
    echo "<tr>";
    echo "<td>".$row['service_id']."</td>";
    echo "<td>".$row['description']."</td>";
    echo "</tr>";
}
echo "</table>";
//Close the SQL connection string
mysql_close($con);
?>
<br />
<form action="a4Services_Student.php " method="post">
<br />
</form>
</div>
</body>
</html>

正如一些评论已经指出的,您正在使用的函数是不安全的,而且还会贬值。最好的方法是使用PDO。我这里有一个例子https://snippetbox.xyz/5c3db100112bca204643/

<?php 
    /** How to get information out a database securely **/
    $id = 6; // example value 
    //connect to mysql database using pdo
    $conn = new PDO('mysql:host=localhost;dbname=someDatabase', $username, $password);
    $query = "SELECT * FROM myTable WHERE id = :id";
    //prepare the statement to avoid sql injection
    $stmt = $conn->prepare($query);
    //load variable into the statement and execute
    $stmt->execute(array('id' => $id));
    //fetch the results
    $rows = $stmt->fetchAll(PDO::FETCH_OBJ);
    //loop through all the lines
    foreach ($rows as $row){
        //loop through results here
        //example
        //echo $row->value;
    }
?>