PHP / MYSQL Echo Mysql數據帶有超鏈接到下一頁？

Question

在我的表格中，我有各種要求，每個要求都有自己的參考號。 用戶可以在頁面上看到請求數，但我希望用戶能夠分別單擊每個請求，然后轉到新頁面以查看該請求的特定詳細信息。

我試圖在單擊請求時在超鏈接中回顯引用，而當我運行查詢以檢索該請求的名稱等情況下，在下一頁上，它將僅顯示與該引用號匹配的信息，即'SELECT * WHERE reference = $row['reference'].

但是我不確定如何做到這一點，我也擔心這種安全性，如果我的查詢是根據我的url中的引用（即mypage.php?reference=1234 ，那么如何阻止用戶手動輸入具有不同參考編號的網址？

'mypage.php?reference=2468

用戶僅應在實際單擊請求時才能夠查看該頁面，而應該永遠不能將其直接輸入到url中，因為這會帶來安全風險。

我可以將我的引用屏蔽為字符串嗎？ 並回聲出mypage.php?reference=$someString

page1.php中

<div class="results_area">
                <h44>Existing Supplier Request's</h44>  
    <?php 
    $conn = new mysqli($host, $username, $password, $db_name);
    // Check connection
    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error); } 
    $sql = "select * from new_supplier_request where status!= 'complete' AND action_taken='actioned'";
                $result = $conn->query($sql);
                if ($result->num_rows > 0) {
                echo '<div class="table_header"><p>Request By</p><p>Date Requested</p><p>Status</p><p>Supplier Name</p><p>Description</p><p>Action</p></div>'; ?> 
                </div>

                <div class="results_area"><?php
                while($row = $result->fetch_assoc()) { 


                echo '<div class="request"><a href="ns_application.php?ns_request='.$row['reference'].'/">';

mypage.php

$reference = $row['reference'];

<?php 
$conn = new mysqli($host, $username, $password, $db_name);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error); } 
$sql = "select * from new_supplier_request where status!= 'complete' WHERE reference = $reference Limit 1";

$result = $conn->query($sql);
            if ($result->num_rows > 0) {
            while($row = $result->fetch_assoc()) { 
            echo '<h44>New Supplier Request: &nbsp;' .$reference. ' </h44><br>'; 
            echo 'The name of this supplier is, '.$reference['name'].';
} } ?>
            </div>

Answer 1

You can use JavaScript to solve this problem.
I don't think that this will be 100% secure but actually it's more secure way than you are using now.
echo "<div .... onclick='showuniquereference('".$row['reference']."')'>"."</div>";

And Function like this
function showuniquereference(code)
{
var reference = code;
var form = document.createElement("form");
form.method = "POST";
form.action = "mypage.php";
var input = document.createElement("input");
input.name = "reference";
input.value = reference;
input.type = "hidden"; 
form.appendChild(input);
document.body.appendChild(form);
form.submit();
}

You can use this posted data to show unique reference to users.