尝试在表单值字段中显示数据库值

Question

Brief synopsis: When I select a user from a dropdown (which populates from Users table), I want the resulting page to show the specific details for that user in a form based view with the current values already filled in based on current data in my database. Database Values are not appearing, instead showing $current_userlastname on the page. Code below:

if ($dbConnected) {
    $UserID = $_POST['UserID'];

$sql = "SELECT * FROM Users WHERE ID = '$UserID'";
$result = mysql_query($sql);

while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
    $current_userlastname = $row['User_LastName'];
    $current_userfirstname = $row['User_FirstName'];
    $current_useremail = $row['User_Email'];
    $current_username = $row['UserName'];
    $current_userpassword = $row['UserPassword'];
}

echo '<h2 style = "font-family: arial,helvetica,sans-serif;"> User EDIT </h2>';

echo '<form name="postUser" action="UserUpdate.php" method="POST">';
echo '<input type="hidden" name="UserID" value=".$UserID."/>';
echo '
        <table>
          <tr>
            <td>Last Name</td>
            <td><input type="text" name="User_LastName" value="$current_userlastname" /></td>
          </tr>
          <tr>
            <td>First Name</td>
            <td><input type="text" name="User_FirstName" value="$current_userfirstname" /></td>
          </tr>
          <tr>
            <td>Email Address</td>
            <td><input type="text" name="User_Email" value="$current_useremail" /></td>
          </tr>
          <tr>
            <td>Username</td>
            <td><input type="text" name="UserName" value="$current_username" /></td> 
          </tr>
          <tr>
            <td>User Password</td>
            <td><input type="text" name="UserPassword" value="$current_userpassword" /></td> 
          </tr>
          <tr>
            <td></td>
            <td align="right"><input type="submit" value="save" /></td>
          </tr>
        </table> 
        ';
echo '</form>';

Answer 1

You should always use mysqli_* instead of mysql_*.

<?php
$conn = mysqli_connect("hosta", "username", "password", "database");

$UserID = $_POST['UserID'];

$sql = "SELECT * FROM Users WHERE ID='$UserID'";
$query = mysqli_query($conn, $sql);

if ($row = mysqli_fetch_assoc($query)) { 
    $current_userlastname = $row['User_LastName'];
    $current_userfirstname = $row['User_FirstName'];
    $current_useremail = $row['User_Email'];
    $current_username = $row['UserName'];
    $current_userpassword = $row['UserPassword'];

echo '<h2 style = "font-family: arial,helvetica,sans-serif;"> User EDIT </h2>';

echo '<form name="postUser" action="UserUpdate.php" method="post">';
echo '<input type="hidden" name="UserID" value="'.$UserID.'"/>';
echo '<table>
          <tr>
            <td>Last Name</td>
            <td><input type="text" name="User_LastName" value="'.$current_userlastname.'" /></td>
          </tr>
          <tr>
            <td>First Name</td>
            <td><input type="text" name="User_FirstName" value="'.$current_userfirstname.'" /></td>
          </tr>
          <tr>
            <td>Email Address</td>
            <td><input type="text" name="User_Email" value="'.$current_useremail.'" /></td>
          </tr>
          <tr>
            <td>Username</td>
            <td><input type="text" name="UserName" value="'.$current_username.'" /></td> 
          </tr>
          <tr>
            <td>User Password</td>
            <td><input type="text" name="UserPassword" value="'.$current_userpassword.'" /></td> 
          </tr>
          <tr>
            <td></td>
            <td align="right"><input type="submit" value="save" /></td>
          </tr>
        </table>';
        echo '</form>';
}