PHP Posting last dynamic表格

Question

我使用了一個while循環來顯示Dropdown yes和no，它顯示每個頁面的用戶權限，所以頁面名稱和是或否沒有爲每個頁面下拉。

表單顯示正確，併爲每個用戶顯示正確的數據庫值作爲默認值。但是，如果我將頁面權限之一從「是」更改爲「否」或僅更新數據庫中的最後一個下拉列表，而其他頁面權限則更改。任何想法我在這裏失蹤？

<?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

原因是因爲您不能在表單元素中使用多次相同的名稱。否則，只會使用最後一個。

您將需要爲它們提供所有不同的名稱或模仿如下的數組語法。例如，而不是：

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

用途：

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

然後在你的PHP腳本，你可以通過它們在foreach循環引用這些：

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

您需要對循環內的其他名稱屬性也這樣做。

最後，您會注意到我從代碼中刪除了反斜槓。這是因爲PHP和HTML都可以使用單引號或雙引號。儘管PHP在兩者之間存在一些差異，但在大多數情況下，您只需交換它們即可。所以，如果PHP使用單引號，那麼對於HTML屬性使用雙引號，反之亦然。