【发布时间】：2013-11-01 09:11:09
【问题描述】：
我已成功从一张表中提取所有必填字段 ['hotel' 是我的表名]。 另外，我想同时从另一个表中提取一些字段 ['rates' 是另一个表名]。我该怎么做？
这是我的表“酒店”的工作代码：
<!--deal-->
<?php
$mysqli = new mysqli('localhost', 'root', '', 'swaminarayan');
if ($result = $mysqli->query("SELECT * FROM hotel")) {
$directory = "../administrator/images/hotels/";
while ($row = $result->fetch_assoc())
{?
<article class="one-fourth">
<figure>
<?php
$search_dir = "$directory/{$row['name']}{$row['hotel_address']}";
$images = glob("$search_dir/*.jpg");
sort($images);
//display images
//display random image
if (count($images) > 0) { // make sure at least one image exists
// Get a random index in the array with rand(min, max) which is inclusive
$randomImageIndex = rand(0, count($images)-1);
$img = $images[$randomImageIndex]; // random image
echo "<img src='$img' height='150' width='150' /> ";
}else {
// possibly display a placeholder image?
} ?
</figure>
<div class="details">
<h1><?php echo " ".$row['name']." ";?>
</h1>
<span class="price">Price per room per night from <em>$<?php echo " ".$row['defult_price']." ";?></em> </span> // This defult_price is not from table 'hotel', it is from table 'rates'
<div class="description">
<p><?php echo " ".$row['hotel_description']." ";?> <a href="<?php echo "".$row['hotel_address']."";?>.php">More info</a></p>
</div>
<a href="booking-step1.php" title="Book now" class="gradient-button">Book now</a>
</div>
</article><?php }}?>
<!--//deal-->
在上面，我想从“rates”表中提取房价字段并在此处显示{如这里：[<span class="price">Price per room per night from <em>$<?php echo " ".$row['default_price']." ";?></em> </span> }]，这只是“rates”表中的值，否则全部来自“hotel”表。
【问题讨论】：
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呃，SQL 查询中没有 JOIN 部分，所以实际上 $row 中应该没有表 'hotel' 中的数据。由于我不知道您的表结构，我很难判断哪个是正确的 JOIN。
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我有两张桌子：'hotel' 和 'rates'。 'hotel'表有以下字段：hotel_id city_id country_id name photo1 photo2 photo3 photo4 photo5 hotel_address phone1 phone2 phone3 phone4 contact_email room_description area_description travel_description food_description rates_added 和，'rates'表有以下字段：id hotel_id room_type default_price price_monday price_tuesday price_wednesday price_thursday price_friday price_saturday price_sunday
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呃，嗯，应该类似于“SELECT * FROM hotel AS h INNER JOIN rates AS r ON r.hotel_id = h.hotel_id”。
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谢谢杜克林，效果很好。非常感谢您的帮助。