使用 PHP 和 JavaScript 动态搜索 SQL 表并在 HTML 上显示

Question

我有一个表单，我希望用户从 SQL 表中选择一个组织，当提交表单时，应将所选组织的 ID 保存到另一个表中。我在网上和 SO 上进行了研究，这就是我现在所拥有的。它不起作用。怎么了？ Newbrand.php：

<form action="newbrand.php" method="post">

        Brand Name: <input type="text" name="bname" /><br><br>
        Ogranization: <input type="text" name="searchbar" id="searchbar"><br><br>
        <script>
        $("#searchbar").keyup(function(){
        var searchTerm = $(this).val();

        $.post('search.php', { search_term: searchTerm}, function(data){

            $(".searchResults").html(data);
            $("#searchUl").css("display", "block");
            });
        });
        </script>
        Organization ID: <input type="hidden" name="gid" value="" /><br><br>
        Gallery ID: <input type="text" name="gid" /><br><br>
</form>

搜索.php：

<?php
include 'db_connect.php';
$link = mysqli_connect($host, $username, $password, $db);

$search_term = sanitize(htmlentities($_POST['search_term']));

if (!empty($search_term)){

$search = "(SELECT `Organization_Name` FROM `organizations_info` WHERE `Organization_Name` LIKE '%$search_term%'  LIMIT 0, 5) ";
$query = mysqli_query($link, $search);
$result = mysqli_num_rows($query);

while ($row = mysqli_fetch_assoc($query)){
    #$user_id = $row['user_id'];
    #$username = $row['username'];
    $orgname = $row['Organization_Name'];
    $check = mysqli_num_rows($query);



    if ($check != 0){
        echo "<a style='text-decoration: none; color: black;' href='newbrand.php?band=$orgname'><li class='searchResults'>" . ucfirst($orgname) . "</li></a>";
    } else {
        echo "<li class='searchResults'>No Results Found</li>";
    }
}
}
?>

Answer 1

我在网上搜索了这个功能并找到了这个。我复制了你的源代码并让它像这样工作：

$search = "(SELECT Organization_Name FROM organizations_info WHERE Organization_Name LIKE '%". $search_term . "%' LIMIT 0, 5) ";

如您所见，我更改了 '%". $search_term ."%' ，因为您不能只将字符串放在“”之间的 sql 查询中，您必须将其切割成碎片，希望您现在明白了.我不经常上这个网站，所以如果它解决了问题，如果你给我发邮件到 sceptdeckheroes@gmail.com 会很酷，希望收到你的来信，祝你好运。

Answer 2

问题在于您的查询。它没有传递$search_term 的值，而是将其作为字符串传递。您可能希望使用准备好的语句并首先绑定参数：

$stmt = mysqli_prepare($link, "SELECT `Organization_Name` FROM `organizations_info` WHERE `Organization_Name` LIKE ? LIMIT 0, 5");
mysqli_stmt_bind_param($stmt, "s", "%{$search_term}%");

mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $organizaton_name);

while (mysqli_stmt_fetch($stmt)) {

}

参考：mysqli SELECT With Prepared Statements