php - 如何抓取搜索结果的id来自

Question

您看到的是我将图像显示为搜索结果的代码。我在下面有一个 anchor ，所以当您单击图片时，它会将您带到测试页面。

我想设置一个页面来显示与该图片关联的其余行条目:

(Picture)    Player: Steve Sax
             Card Type: Donruss
             Year: 1989
             Value: $2.00

如何获取搜索结果行中的“id”，然后将其回显到显示在 TEST 页面上的表格中？

<?php
$servername = "*********";
$username = "*********";
$password = "*********";
$dbname = "*********";
$username1=$_SESSION['activeusername'];
    $userid = $_SESSION['activeid'];
    $userid = $_SESSION['activeid'];
    $itemid = $_SESSION['activeid'];
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT * FROM useritems JOIN (users, iteminfo) on (users.id=useritems.UserID AND iteminfo.ID=useritems.ItemID) AND userid='2'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo "<a href='test1.php'><div id='frame'>
<div id='splash-image' style='background: url(".$row['imagename']."); background-size:cover;'></div>
<div id='text'>
<table >
<tr></tr>
</table>
</div>
</div>";
    }
} else {
    echo "You Have No Items In Your Closet!!!";
}

mysqli_close($conn);
?>

Answer 1

通过 get 传递 id 为:

echo "<a href='test1.php?id=". $row['id']."'>

并将其放在着陆页上:

$id=$_GET['id'];
echo $id;