我正在做一个小的Web应用程序,其工作原理就像一个booking.com。
我已经编写了用于搜索城市的代码,并且可以在表格中看到结果,如何使所有结果(例如酒店名称)可链接,并重定向到正确的酒店页面每个结果?
我已经尝试过,但是重定向到正确的页面时遇到了一些问题,它也给我带来了一些代码错误
<?php
$con= new mysqli("localhost","root","","registration");
$name = $_POST['search'];
//$query = "SELECT * FROM hotels
// WHERE city LIKE '%{$name}%'";
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con, "SELECT * FROM hotels
WHERE city LIKE '%{$name}%'");
echo "
<table border='1' cellpadding='0' cellspacing='0' style='border-collapse: collapse' bordercolor='#111111' width='532' height='23' id='AutoNumber1'>
<tr>
<td width='120' height='23' align='center'>Name</td>
<td width='179' height='23' align='center'>Address</td>
<td width='100' height='23' align='center'>Phone number</td>
<td width='150' height='23' align='center'>E-mail</td>
<td width='50' height='23' align='center'>Stars</td>
<td width='100' height='23' align='center'>Price single room</td>
<td width='100' height='23' align='center'>Price double room</td>
</tr>";
while ($row = mysqli_fetch_array($result))
{
echo "
<tr>
<td width='120' height='23'>$row[namehotel]</td>
<td width='179' height='23'>$row[address]</td>
<td width='100' height='23'>$row[phonenumber]</td>
<td width='150' height='23'>$row[email]</td>
<td width='50' height='23'>$row[stars]</td>
<td width='100' height='23'>$row[pricesingle]</td>
<td width='100' height='23'>$row[pricedouble]</td>
</tr>";
echo "<br>";
}
mysqli_close($con);
?>
同时,我无法创建链接,例如,如果我有希尔顿的话,该名称应该是可单击的,并且应重定向到希尔顿酒店的页面(由我在html中创建)
谢谢!
答案 0 :(得分:0)
我不确定我是否理解正确。您是否需要html超链接?
echo '<tr>
<td width="120" height="23">
<a href="your/target/with'.$row[hotelID].'" >'.$row[namehotel].'</a>
</td>
<td width="179" height="23">'.$row[address].'</td>
<td width="100" height="23">'.$row[phonenumber].'</td>
...
...
</tr>';
答案 1 :(得分:0)
好,我能够创建超链接,但是是否可以为每个结果创建一个不同的链接?
我使用
创建了超链接<a href="pagehotel.php"></a>
创建此页面时,我无法带上$_POST['namehotel']
,但出现未定义错误
我发布代码:
<?php
session_start();
$con= mysqli_connect("localhost","root","","registration");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con, "SELECT * FROM hotels
WHERE namehotel LIKE '%{$_POST['namehotel']}%'");
mysqli_close($con);
?>
如何从上次搜索中获得$ _POST ['namehotel']?