我如何將每個從 mysql 獲取的數據發送到另一個頁面

Question

我有 2 頁“customize.php”和“deepview.php”
這是 customiz.php：

<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.4/angular.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>

<?php
session_start();
?>
<style>
.site {
/* width: 100%; */
height: 100px;
border: 1px solid #ccc; 
text-align: center;
padding: 25px;
}

.site:hover {
    box-shadow: 0 4px 8px 0 rgba(0, 0, 0, 0.2), 0 6px 20px 0 rgba(0, 0, 0, 0.19);
}
</style>

<div id="vcs"> <!-- view of customize site -->

<?php
include ('connection.php');

$sql = "SELECT site_name FROM sites";

$result = mysqli_query($con, $sql);

if(mysqli_num_rows($result) > 0) {
    while($row = mysqli_fetch_assoc($result)) {
        echo "<a href='#' id='custlink' data-target='deepview'><div class='col-sm-4 site'>";
        echo $row['site_name'];
        echo "</div></a>";
        $_SESSION['session_site_name'] = $row['site_name'];   
    }
}

?>
</div>

<script>
$(document).ready(function() {
// custlink is the id of the link
$("a, #custlink").on("click", function() {
     var target = $(this).data('target');
     $("#vcs").load(target + '.php');
     return true;
})
})
</script>

這是 deepview.php：

<?php
session_start();
?>

<style>
.deepsite {
/* width: 100%; */
height: 100px;
border: 1px solid #ccc; 
text-align: center;
padding: 25px;
}

.deepsite:hover {
    box-shadow: 0 4px 8px 0 rgba(0, 0, 0, 0.2), 0 6px 20px 0 rgba(0, 0, 0, 0.19);
}

</style>

<?php
include ('connection.php');

$clicked_site_name = $_SESSION['session_site_name'];

$sql = "SELECT date from  tables WHERE site_name='".$clicked_site_name."'";

$result = mysqli_query($con, $sql);

if(mysqli_num_rows($result) > 0) {
    while($row = mysqli_fetch_assoc($result)) {
        echo "<a href='#'><div class='col-sm-4 deepsite'>";
        echo $row['date'];
        echo "</div></a>";
    }
}
?>

這里我想要的是當我單擊customize.php 上的鏈接以加載deepview.php 並從mysql 表中選擇“date”，其中“site_name”是我在customize.php 中單擊的站點

我嘗試了上面的代碼，它成功加載了 deepview.php，但它總是加載相同的“日期”，即使我在customize.php 中單擊了不同的“站點”。

這是表“表”

這是來自 mysql 的表“站點”

這是customize.php

這是 deepview.php

Answer 1

在customize.php我改變了以前的代碼

$sql = "SELECT site_name FROM sites";

$result = mysqli_query($con, $sql);

if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
    $rr = $row['site_name'];

?>

<a href='#' data-target="database/deepview.php?lo=<?php echo "$rr"; ?>">

<?php    
    echo "<div class='col-sm-4 site'>";
    echo $rr;
    echo "</div></a>";
}
}

我在 deepview.php 中使用全局變量 $_GET 提到了“lo”

$sql = "SELECT date from  tables WHERE site_name='".$_GET['lo']."'";

$result = mysqli_query($con, $sql);

if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
    $rr = $row['date'];
?>

<a href='' id='sslink' data-target='database/viewsinglesite?lo=<?php echo "$rr"; ?>'>

<?php

    echo "<div class='col-sm-4 deepsite'>";
    echo $row['date'];
    echo "</div></a>";
}
}