我正在为 friend 收拾残局。他的网站过去可以正常工作并从他的数据库中提取特色产品,但现在似乎不起作用。它只是显示错误!!
任何帮助将不胜感激......我不太了解 SQL。
代码如下:
<?php
require_once('const.php');
$link = dbConnect();
$query = "SELECT *
FROM vehicle_tbl, manufacturer_tbl
LEFT JOIN image_tbl ON vehicle_tbl.vehicle_id = image_tbl.vehicle_id
WHERE vehicle_tbl.manufacturer_id = manufacturer_tbl.manufacturer_id AND
vehicle_tbl.vehicle_feature2 = '1'
GROUP BY vehicle_tbl.vehicle_id
ORDER BY RAND()
LIMIT 1";
$result = false;
$result = @mysql_query($query, $link);
$fmain = false;
if (($result) && (@mysql_num_rows($result) > 0)) {
$fmain = @mysql_fetch_array($result, MYSQL_ASSOC);
@mysql_free_result($result);
}
$query = "SELECT *
FROM vehicle_tbl, manufacturer_tbl
LEFT JOIN image_tbl ON vehicle_tbl.vehicle_id = image_tbl.vehicle_id
WHERE vehicle_tbl.manufacturer_id = manufacturer_tbl.manufacturer_id AND
vehicle_tbl.vehicle_feature1 = '1'
GROUP BY vehicle_tbl.vehicle_id
ORDER BY RAND()
LIMIT 6";
$offers = false;
$offers = @mysql_query($query, $link);
function nextOffer() {
global $offers;
if ($offers && ($row = mysql_fetch_array($offers))) {
if (! isset($row['image_name'])) { // no image
$image = 'images/noimagesml.jpg';
} else {
$image = 'images/vehicles/sml/'.stripslashes($row['image_name']);
}
$title = stripslashes($row['manufacturer_name']).' '.stripslashes($row['vehicle_model']);
$price = number_format((float) $row['vehicle_price_pcm'], 2);
$id = (int) $row['vehicle_id'];
echo '<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td class="contenthead"><table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="11" height="40" align="left" valign="top"><img src="images/featre_left_hd.gif" width="11" height="23"></td>
<td width="100%" align="left" valign="middle" class="contenthead">'.$title.'</td>
<td width="11" height="40" align="right" valign="top"><img src="images/featre_rght_hd.gif" width="11" height="23"></td>
</tr>
</table></td>
</tr>
<tr>
<td class="contentpane"><table width="100%" border="0" cellspacing="5" cellpadding="0">
<tr>
<td align="center" valign="middle"><img src="'.$image.'" width="100" height="58" class="bordered" alt="'.$title.'"></td>
</tr>
<tr>
<td align="center" valign="top" class="princing">from just £'.$price.' pcm</td>
</tr>
<tr>
<td align="right" valign="middle"><a href="cardeal.php?vehicle='.$id.'"><img src="images/more_butt.gif" width="54" height="20" border="0"></a></td>
</tr>
</table></td>
</tr>
</table>';
} else {
echo 'Error!!';
}
}
?>
最佳答案
显示 mysql 错误如下:
mysql_query($query, $link) or die(__FILE__ . ' Line ' . __LINE__ . ': ' . mysql_error());
这应该可以帮助您进行调试。
附注mysql_* 函数已弃用 - http://php.net/manual/en/mysqlinfo.api.choosing.php
关于php - SQL 帮助 - 新手,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20189794/