【发布时间】:2010-12-07 15:02:53
【问题描述】:
我在这里为我数据库中的每辆车显示一组行。 每行都有一个表单字段,登录用户可以在其中提交报价。 当用户提出任何汽车的报价时,表单字段将替换为显示所提交报价值的文本。
但是,我遇到的结果并不理想。 如果我为一排报价,很好,逻辑有效。如果我继续为另一行提供另一个报价,则逻辑有效,除了上一行现在再次显示表单这一事实。
如有必要,我可以提供更多详细信息,但也许有人已经对此很熟悉了。
提前致谢。
<?php
require("db-connect.php");
$display = "SELECT filename, car_id, make, model, year, mileage, vin, description, GROUP_CONCAT(filename) FROM scraplis_cars LEFT JOIN scraplis_images USING (car_id) GROUP BY car_id ORDER BY date_time DESC";
$dResult = mysql_query($display) or die('error:' . mysql_error());
$offer = "SELECT car_id, user_id, offer_id, value FROM scraplis_offers WHERE user_id = '".$_SESSION['user_id']."'";
$oResult = mysql_query($offer) or die('Error ' . mysql_error());
$oRow = mysql_fetch_array($oResult);
if(!isset($_SESSION['access'])){
header("location:index.php");
}
?>
<?php if($dResult): ?>
<table class="post">
<thead>
<tr>
<?php if(isset($_SESSION['email']) && $_SESSION['access'] == 0) : ?>
<th scope="col">Images</th>
<th scope="col">Make</th>
<th scope="col">Model</th>
<th scope="col">Year</th>
<th scope="col">Mileage</th>
<th scope="col">VIN #</th>
<th scope="col">Description</th>
<th scope="col">Offer</th>
</tr>
</thead>
<tbody>
<?php while($dRow = mysql_fetch_array($dResult)) : ?>
<?php $str = $dRow[8]; ?>
<?php $images = explode(',', $str); ?>
<tr>
<td>
<ul>
<?php if(!empty($str)) : ?>
<?php foreach($images as $value) :?>
<li>
<a href="images/<?php echo $value; ?>" rel="lightbox[<?php echo $row['car_id']; ?>]">
<img src="images/<?php echo $value; ?>"/>
</a>
</li>
<?php endforeach; ?>
<?php endif; ?>
<ul>
</td>
<td><?php echo $dRow['make']; ?></td>
<td><?php echo $dRow['model']; ?></td>
<td><?php echo $dRow['year']; ?></td>
<td><?php echo number_format($dRow['mileage']); ?></td>
<td><?php echo $dRow['vin']; ?></td>
<td><span><?php echo $dRow['description']; ?></span></td>
<td>
<?php if($oRow['car_id'] == $dRow['car_id']) : ?>
Offer pending approval - $<?php echo $oRow['value']; ?>
<?php else : ?>
<form id="offer" method="post" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="text" id="price" name="offer" />
<input type="hidden" name="submitted" value="<?php echo $dRow['car_id']; ?>" />
<input type="submit" name="price" value="Submit" />
</form>
<?php endif; ?>
</td>
</tr>
<?php endwhile; ?>
<?php else : ?>
<th scope="col">Delete</th>
<th scope="col">Images</th>
<th scope="col">Make</th>
<th scope="col">Model</th>
<th scope="col">Year</th>
<th scope="col">Mileage</th>
<th scope="col">VIN #</th>
<th scope="col">Description</th>
</tr>
</thead>
<tbody>
<?php while($dRow = mysql_fetch_array($dResult)) : ?>
<?php $str = $dRow[8]; ?>
<?php $images = explode(',', $str); ?>
<tr>
<td>
<form method="post" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="checkbox" name="record" value="<?php echo $row['car_id']; ?>" />
<input type="submit" name="delete-car" value="Delete" />
</form>
</td>
<td>
<ul>
<?php if(!empty($str)) : ?>
<?php foreach($images as $value) :?>
<li>
<a href="images/<?php echo $value; ?>" rel="lightbox[<?php echo $row['car_id']; ?>]">
<img src="images/<?php echo $value; ?>"/>
</a>
</li>
<?php endforeach; ?>
<?php endif; ?>
</ul>
</td>
<td><?php echo $dRow['make']; ?></td>
<td><?php echo $dRow['model']; ?></td>
<td><?php echo $dRow['year']; ?></td>
<td><?php echo number_format($dRow['mileage']); ?></td>
<td><?php echo $dRow['vin']; ?></td>
<td><span><?php echo $dRow['description']; ?></span></td>
</tr>
<?php endwhile; ?>
<?php endif; ?>
</tbody>
</table>
<?php endif; ?>
【问题讨论】:
-
为什么不离开加入您的报价表?您似乎无法控制为给定的行或循环带回哪个报价,只为给定的用户返回所有报价?在加入中,您可以获取该用户针对该车辆的最新报价。