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我使用表格從我的數據庫中獲取汽車信息,並且此表中的每列包含汽車信息,但是在最後一列中我想查看其中的汽車圖片,保存在我的數據庫中,我已經擁有我的代碼從我的數據庫中獲取圖片,並可直接在瀏覽器中查看它。但我如何直接在表格列中查看圖片而無需在新窗口中打開圖片。 視圖代碼查看錶中提取的圖片
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td class="style1"><?php echo $rows['branch']; ?></td>
<td class="style1"><?php echo $rows['model']; ?></td>
<td class="style1"><?php echo $rows['doors']; ?></td>
<td class="style1"><?php echo $rows['fuel']; ?></td>
<td class="style1"><?php echo $rows['engine']; ?></td>
<td class="style1"><?php echo $rows['colo']; ?></td>
<td class="style1"><?php echo $rows['wheel']; ?></td>
<td class="style1"><?php echo $rows['condit']; ?></td>
<td class="style1"><?php echo $rows['warranty']; ?></td>
<td class="style1" style="width: 47px"><?php echo $rows['price']; ?></td>
<td class="style1"><?php echo $rows['CarOwner']; ?></td>
<td align="center" class="style3"><a href="info.php?id=<?php echo $rows['CarOwner']; ?>"><?php echo $rows['CarOwner']; ?></a></td>
<td align="center" class="style3"><?php echo "download.php"; ?></td>
,並從MySQL
此下載圖像<?php
//$id = $_GET['id'];
include_once 'D_B.php';// Connect to server and select database.
$query = "SELECT `name`, `type`, `size`, `content` FROM `upload` WHERE `id`='1'";
$result = mysql_query($query) or die('Error, query failed');
list($name, $type, $size, $content) =mysql_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
echo $content;
exit;
?>
你不知道圖像的HTML標籤? –