我很难从MySQL数据库中显示一些信息。我创建了一个销售页面,它有一个显示广告的前GRID系统,当点击每个广告时,你会得到详细内容。
我在MySQL中创建了两个表,一个存储了所有文本信息,另一个显示了一系列多个图像链接到ID值。 (见下文)
无论如何显示其中一张图片而不是全部显示在该ID内?填充网格前端图像?
Aircraft Sales - Images are stored here
飞机代码。
代码是:
<div id="home" class="main-container">
<div class="content-container">
<div class="container">
<div class="bs-example" data-example-id="thumbnails-with-custom-content">
<h1 style="text-align:left;"><font color="#D60808">Aero-Agent</font>
<small><font color="#909090">Current Stock List</small></font></h1>
<?php
// connect to database + run query to grab all aircraft in database.
/*
include_once('dbconnection.php');
$str = "SELECT * FROM `aircraftsales`";
$rs = mysql_query($str);
if(!$rs )
{
die('Could not get data: ' . mysql_error());
}
if (mysql_num_rows($rs) == 0)
echo '<h4><font color="#909090">No current aircraft for sale.</font></h2>';
else {
while($row = mysql_fetch_assoc($rs)) {
*/
include_once('dbconnection.php');
$str="SELECT * FROM `aircraftsales`";
$rs = mysql_query($str);
if(!$rs )
{
die('Could not get data: ' . mysql_error());
}
$imagesstr="SELECT * FROM `aircraftimages`";
$imagesrs = mysql_query($imagesstr);
if(!$imagesrs )
{
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_assoc($rs)) {
?>
<div class="vehicle-blob list FORSALE">
<a href="/content.php?id=<?php echo $row['id']?>"><div class="vehicle-photo photo-list selected">
<img src="exam/upload/aircraftsales/<?php echo $imagerow['image'];?>" /></di
<div class="status-graphic"></div>
<div class="overlay list">
<div class="inner-tube">
<h1><?php echo $row['title']?> <?php echo $img;?> <br /><font color="#D60808"><b>£<?php echo $row['price']?></b></font></h1>
</div>
</div>
</a>
<?php
}
?>
</div>
</div>
</div>
</div>
<?php include_once('ppls_css/includes/ppl-footer.php'); ?>
答案 0 :(得分:0)
使用左连接查询。
SELECT * FROM aircraftsales left join aircraftimages on aircraftsales.aircraft_id = aircraftimages.aircraft_id
试试吧