使用PhP使用SQL搜索工具
使用PhP使用SQL搜索工具
提问于 2015-11-20 10:33:19
我正在尝试创建一个搜索工具，在两个不同的表中搜索两个不同的列，因此，如果酒店表中的名称匹配或来宾表中的guest_id匹配，则输出while循环中的任何内容。然而，其中一项工作是：
$stmt = $conn->prepare("SELECT * FROM hotel
INNER JOIN booking
ON hotel.hotel_id=booking.hotel_id
INNER JOIN guest
ON guest.guest_id=booking.guest_id
WHERE name LIKE :search_term");但是这个不起作用它给了我一个空的结果
$stmt = $conn->prepare("SELECT * FROM guest
INNER JOIN booking
ON guest.guest_id=booking.guest_id
INNER JOIN hotel
On booking.hotel_id=hotel.hotel_id
WHERE guest_id LIKE :search_term");
$stmt->bindValue(':search_term','%'.$search_term. '%');我的完整代码是：
<!DOCTYPE html>
<html>
<head>
<title>Database</title>
<link href="style.css" rel="stylesheet" type="text/css"> <!-- This is linking style sheet (css)into this HTML page-->
<link href='https://fonts.googleapis.com/css?family=PT+Serif:400italic' rel='stylesheet' type='text/css'>
</head>
<body>
<div class="navigation">
<form action = "index.php" method="get">
<input type = "submit" name = "mainpage" value = "Main Page" class = "submitbut" id = "but1" />
</form>
</div>
<form action="index.php" method="post">
<input type = "text" name = "search" id = "searching" />
<input type = "submit" name = "data_submit" value = "Search" id = "scan" />
</form>
<?php
if(isset($_GET['mainpage'])){
header("Location:mainpage.php");
exit;
}
if (isset($_POST["data_submit"])){
$search_term = $_POST['search'];
$conn = new PDO(
'mysql:host=localhost;dbname=u1358595',
'root'
);
$stmt = $conn->prepare("SELECT * FROM hotel
INNER JOIN booking
ON hotel.hotel_id=booking.hotel_id
INNER JOIN guest
ON guest.guest_id=booking.guest_id
WHERE name LIKE :search_term");
$stmt->bindValue(':search_term','%'.$search_term. '%');
$stmt->execute();
echo
"<table><tr>
<th>Results</th>
</tr>";
while($hotel = $stmt->fetch())
{
echo
"<tr>"."<td>"."<a href='details.php?name=".$hotel['name']."'>".$hotel['name']."</a>"."</td>"."</tr>";
}
echo "</table>";
$stmt = $conn->prepare("SELECT * FROM guest
INNER JOIN booking
ON guest.guest_id=booking.guest_id
INNER JOIN hotel
On booking.hotel_id=hotel.hotel_id
WHERE guest_id LIKE :search_term");
$stmt->bindValue(':search_term','%'.$search_term. '%');
$stmt->execute();
echo
"<table><tr>
<th>Results</th>
</tr>";
while($hotel = $stmt->fetch())
{
echo
"<tr>"."<td>"."<a href='details.php?name=".$hotel['first_name']."'>".$hotel['last_name']."</a>"."</td>"."</tr>";
}
echo "</table>";
$conn = NULL;
}
?>
</body>
</html>回答 2
Stack Overflow用户
回答已采纳
发布于 2015-11-20 11:04:55
你对WHERE条款有异议。您没有定义从哪个表列比较guest_id。我认为guest_id应该被guest.guest_id取代。
$stmt = $conn->prepare("SELECT * FROM guest
INNER JOIN booking
ON guest.guest_id=booking.guest_id
INNER JOIN hotel
On booking.hotel_id=hotel.hotel_id
WHERE guest.guest_id LIKE :search_term");
$stmt->bindValue(':search_term','%'.$search_term. '%');编辑：
此外，可以将搜索部分替换为：
... '...WHERE guest.guest_id LIKE "%":search_term"%"');
$stmt->bindValue(':search_term', $search_term);但是，您需要在查询中使用单引号。
Stack Overflow用户
发布于 2015-11-20 11:01:51
考虑使用对：文档的匹配
SELECT * FROM hotel
INNER JOIN booking
ON hotel.hotel_id=booking.hotel_id
INNER JOIN guest
ON guest.guest_id=booking.guest_id
WHERE MATCH (col1, col2)
AGAINST ('+keyword1' IN BOOLEAN MODE);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/33824422
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