PHP发布最后一个动态表单

Question

I have used a while loop to display Dropdown yes and no which displays a users permissions for each page, so page name and Yes or no drop down for each page.

The form appears correct and displaying correct database value for each user as default. However if I change one of the page permission from yes to no or vice verser its only updating the last of the dropdowns in the database and not the rest the rest thats changed. Any Ideas what I am missing here?

<?php

$sql = "SELECT DISTINCT username FROM users ORDER BY username";
$resultusers = $connect->query($sql);
?>
<?php
 require ("cw/connect.php");


?>
<?php
if(isset($_POST['selectbutton']))
{
    $username = $_POST['selectuser'];

$query = "SELECT
users.username, users.first, users.last, users.id, permissions.PermID, permissions.PermUserID, permissions.PermPagesID, permissions.view, pages.PagesName, pages.PagesLink, pages.PagesID
FROM users INNER JOIN permissions ON users.id = permissions.PermUserID INNER JOIN pages ON permissions.PermPagesID = pages.PagesID WHERE username = ?";

$stmt = mysqli_prepare($connect, $query);

if($stmt){
    //Put whats to be binded from the statement so in this case we want data for this username selected
    mysqli_stmt_bind_param($stmt,"s",$username);

    // here add all the varibles to be pulled from database
    mysqli_stmt_bind_result($stmt, $username, $first, $last, $id, $PermID, $PermUserID, $PermPagesID, $view, $PagesName, $PagesLink, $PagesID);


    mysqli_stmt_execute($stmt);

    mysqli_stmt_fetch($stmt);




?>


<b>First:</b><?php echo $first; ?><br>
<b>Last:</b><?php echo $last; ?><br>
<b>User:</b><?php echo $username; ?><br>
<hr>


<form action="user-permissions.php" method="post">
<?php
while ($stmt->fetch()) {

echo $PagesName;
echo "<br>"; 
echo "<select name=\"permissionSelect\">";

?>

<option value="<?php echo $view; ?>"><?php echo $view; ?></option>
<option value="No">No</option>
<option value="Yes">Yes</option>

</select>
<br><br>
<?php
echo "<input type=\"hidden\" name=\"PermPagesID\" value=\"".$PermPagesID."\">";
echo "<input type=\"hidden\" name=\"PermUserID\" value=\"".$PermUserID."\">";
        }
        ?>
<?php
echo "<input type=\"submit\" name=\"updatep\">";
echo "<form>";


     }  

}
?>


<?php
//UPDATE PERSON
 require ("cw/connect.php");

if(mysqli_connect_error()){
    echo mysqli_connect_error();
    exit();

}
if(isset($_POST['updatep']))
{
    $view = mysqli_real_escape_string($connect,$_POST['permissionSelect']);
    $PermPagesID = mysqli_real_escape_string($connect,$_POST['PermPagesID']);
    $PermUserID = mysqli_real_escape_string($connect,$_POST['PermUserID']); 


$query = "UPDATE permissions SET view = ? WHERE PermPagesID = $PermPagesID AND PermUserID = $PermUserID";

$stmt = mysqli_prepare($connect, $query);

if($stmt){
    mysqli_stmt_bind_param($stmt, "s", $view);

    mysqli_stmt_execute($stmt);

    mysqli_stmt_fetch($stmt);
    echo "Updated to: ".$username;


    }else{

    echo "object not created";

}
}

?>

Answer 1

The reason is because you cannot have the same name used more than once in your form elements. Otherwise, only the last one will be used.

You will need to give them all a different name, or mimic array syntax as follows. For example, instead of:

echo '<select name="permissionSelect">';

Use:

echo '<select name="permissionSelect[]">';

Then in your PHP script you can reference these by looping through them with a foreach:

foreach($_POST['permissionSelect'] as $permissionSelect){
    //handle it here
}

You will need to do this for the other name attributes inside the loop as well.

Lastly, you will notice that I removed you backslashes from your code. This is because you both PHP and HTML can use either single or double quotes. And while there is some difference between the two in PHP, in most cases you can just interchange them. So, if PHP is using single quotes, then use double quotes for your HTML attributes, or vice versa.