2017-03-06 103 views
1

我正在使用PHP和MySQL的網站上工作,並有一些問題,從數據庫行生成網頁網址。動態Html頁面從PHP與PHP

我只有3頁connection.php(MySQL連接)的index.php(其中與產品的詳細信息URL按鈕顯示人的產品/內容縮略圖)和details.php我想要顯示的單品信息。

從的index.php我添加一個鏈接重定向與這details.php頁:

<a href="details.php?id=<?php echo $row['ID']; ?>"

它的工作,但最大的問題是在details.php因爲劇本不顯示單產品細節,但顯示所有產品,請有人可以幫助我?謝謝

的index.php代碼

......other html code...... 
<div class="row"> 
<?php 

require_once 'connection.php'; 

$query = "SELECT * FROM campi_name"; 
$stmt = $DBcon->prepare($query); 
$stmt->execute(); 
while($row=$stmt->fetch(PDO::FETCH_ASSOC)){ 
?> 


<div class="col-sm-4 col-md-3"> 
<div class="thumbnail"> 
<img src="<?php echo $row['Thumbnail']; ?>" alt="<?php echo $row['Title']; ? >"> 
<div class="caption"> 
<h4><?php echo substr($row['Title'], 0, 30); ?></h4> 
<p><?php echo $row['Brand']; ?></p> 
<?php echo $row['ID']; ?> 
<p><a href="#" class="btn btn-primary btn-lg" role="button">Cofronta</a> <a href="dettagli.php?id=<?php echo $row['ID']; ?>" class="btn btn-default btn-lg" role="button">Dettagli</a></p> 
</div> 
</div> 
</div> 


<?php 
} 
?> 
......other html code...... 

connection.php代碼

$DBhost = "localhost"; 
$DBuser = "root"; 
$DBpass = ""; 
$DBname = "prodotti"; 



try { 
$DBcon = new PDO("mysql:host=$DBhost;dbname=$DBname",$DBuser,$DBpass); 
$DBcon->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
} catch(PDOException $ex){ 
die($ex->getMessage()); 
} 

?> 

details.php代碼

......other html code...... 
<div class="container"> 


<div class="row"> 

<?php 

require_once 'connection.php'; 

$query = "SELECT * FROM campi_name"; 
$stmt = $DBcon->prepare($query); 
$stmt->execute(); 
while($row=$stmt->fetch(PDO::FETCH_ASSOC)){ 
?> 


<div class="col-sm-4 stylerow">    
<a href="<?php echo $row['AffiliateLink']; ?>" class="thumbnail"> 
<img src="<?php echo $row['Thumbnail']; ?>" alt="<?php echo $row['Title']; ? >"> 
</a> 

</div> 


<div class="col-sm-8 stylerow"> 
<h2><?php echo $row['Title']; ?></h2> 
<p><?php echo $row['Brand']; ?></p> 
<button type="button" class="btn btn-primary btn-lg">Amazon</button> 
</div> 

</div> 

</div><!-- /.container --> 
......other html code...... 
+2

看'SELECT * FROM campi_name',你需要告訴它你想要什麼。 'SELECT * FROM campi_name where id =?'然後綁定GET'ID'。 '$ stmt-> execute(array($ _ GET ['id']));' – chris85

+0

你將需要使用details.php上的GET,就像chris85說的 –

回答

3

Add $ id = $ _ GET ['id'];

編輯下面在代碼

$query = "SELECT * FROM campi_name"; 

線要

$query = "SELECT * FROM campi_name where id='$id' "; 
+0

很酷謝謝你! – Stew

+0

@Raman你的代碼很容易出現sql注入攻擊'$ id'應該綁定 –