沒有排在表中的數據庫信息（php / sql）
[英]Database information not lining up in a table (php/sql)
問題描述
當我的DetailsTable使用CSS時，我從數據庫中獲取的信息未在表中正確顯示。 所有這些信息都聚集在一起，因此沒有作為一個表格。
如果有人能夠建議為什么我想要得到的信息沒有以表格形式列出，我將非常感激
</head>
<body>
<div id="DetailsTable" >
<table>
<tr>
<th>Name</th>
<th>TypeOfShoe</th>
<th>Description</th>
<th>Price(£)</th>
<th>Fabric</th>
<th>Colour</th>
<th>Brand</th>
</tr>
</table>
<?php
$shoesID=$_GET['id'];
$stmt = $conn->prepare("SELECT shoes.name, shoes.images, shoes.description, shoes.price, types.typeOfShoe, types.fabric, types.colour, brands.name AS bname FROM shoes INNER JOIN types on types.types_id = shoes.type_id INNER JOIN brands on brands.brands_id = shoes.brands_id
WHERE shoes.id = :id");
$stmt->bindValue(':id',$shoesID);
$stmt->execute();
if ($shoes=$stmt->fetch()){
echo "<td>".$shoes['name']."</td>";
echo "<td>".$shoes['typeOfShoe']."</td>";
echo "<td>".$shoes['description']."</td>";
echo "<td>".$shoes['price']."</td>";
echo "<td>".$shoes['fabric']."</td>";
echo "<td>".$shoes['colour']."</td>";
echo "<td>".$shoes['bname']."</td>";
}
$conn=NULL;
?>
<br> <br/>
<div id="Image" >
<img src="<?php echo $shoes['images']; ?>" height="500px" width="500px" />
</div>
</div>
CSS代碼如下。
body {
background-color: lemonchiffon;
}
#DetailsTable {
border-collapse: collapse;
width: 100%;
}
th, td {
text-align: left;
padding: 8px;
}
#Image{
display: block;
margin: auto;
width: 50%;
height: 10px;
clear: both;
}
2 個解決方案
解決方案1
0 已采納 2016-11-24 20:44:05
您已經在PHP代碼之前關閉了</table >標記，這可能是問題所在
解決方案2
0 2016-11-24 21:06:50
</head>
<body>
<div id="DetailsTable" >
<table>
<tr>
<th>Name</th>
<th>TypeOfShoe</th>
<th>Description</th>
<th>Price(£)</th>
<th>Fabric</th>
<th>Colour</th>
<th>Brand</th>
</tr>
<?php
$shoesID=$_GET['id'];
$stmt = $conn->prepare("SELECT shoes.name, shoes.images, shoes.description, shoes.price, types.typeOfShoe, types.fabric, types.colour, brands.name AS bname FROM shoes INNER JOIN types on types.types_id = shoes.type_id INNER JOIN brands on brands.brands_id = shoes.brands_id
WHERE shoes.id = :id");
$stmt->bindValue(':id',$shoesID);
$stmt->execute();
if ($shoes=$stmt->fetch()){
echo '<tr>';
echo "<td>".$shoes['name']."</td>";
echo "<td>".$shoes['typeOfShoe']."</td>";
echo "<td>".$shoes['description']."</td>";
echo "<td>".$shoes['price']."</td>";
echo "<td>".$shoes['fabric']."</td>";
echo "<td>".$shoes['colour']."</td>";
echo "<td>".$shoes['bname']."</td>";
echo '</tr>';
}
$conn=NULL;
?>
</html>
<br> <br/>
<div id="Image" >
<img src="<?php echo $shoes['images']; ?>" height="500px" width="500px" />
</div>
</div>
將信息發送到數據庫SQL PHP
聲明:本站的技術帖子網頁，遵循CC BY-SA 4.0協議，如果您需要轉載，請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.