Question

我正在尝试创建一个搜索工具，搜索两个不同表中的2个不同列，因此如果来自hotel表的任一名称匹配或guest表中的guest_id，则输出while循环中的任何内容。但是其中一个有效：

$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");

然而，这个没有用，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>

Answer 1

你的WHERE子句有问题。您没有定义比较哪个表列guest_id。 guest_id应该替换为我相信的guest.guest_id。

 $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，搜索部分可以替换为：

... '...WHERE guest.guest_id LIKE "%":search_term"%"');
    $stmt->bindValue(':search_term', $search_term);

您需要为查询使用单引号。

Answer 2

考虑使用MATCH AGAINST：documentation

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);

使用PhP搜索SQL

2 个答案: