Question

我正在设计一个房地产网站我在我的网站上有很多广告，当用户点击某个广告时，它会转到另一个页面viewmore.php，这会向用户提供有关该特定广告的更多详细信息。

现在，正如您在viewmore.php文件中看到的，我在Cookie中保存了广告的ID，并将广告的ID发送到收藏页面，用户可以随时查看该帖子页。

问题：

考虑到我访问此页面localhost/viewmore.php?ID=10，因此当您转到收藏页面时，您会看到属于此ID的广告数据，但是当您访问其他广告（例如localhost/viewmore.php?ID=11）时，您会转到收藏页面您会看到属于id=11的广告数据，之前的添加消失了。我想将它们保存在我最喜欢的页面中，或者事实上保存我访问的所有帖子。

我该怎么做？

//reviewmore.php
<!doctype html>
<?php
(is_numeric($_GET['ID'])) ? $ID = $_GET['ID'] : $ID = 1;
?>
<?php 
$cookie_name = "favoritepost";
$cookie_value ="$ID";
setcookie($cookie_name, $cookie_value, time() + (86400 * 30), "/"); // 86400 = 1 day
?>
<html>
  <body>
    <?php
 error_reporting(0);
include("config.php");
(is_numeric($_GET['ID'])) ? $ID = $_GET['ID'] : $ID = 1;
$result = mysqli_query($connect,"SELECT*FROM ".$db_table." WHERE idhome = $ID");
?>
<?php
 error_reporting(0);
include("config.php");
(is_numeric($_GET['ID'])) ? $ID = $_GET['ID'] : $ID = 1;
$result = mysqli_query($connect,"SELECT*FROM ".$db_table." WHERE idhome = $ID");
?>
<?php $row = mysqli_fetch_array($result):
$price=$row['price'];
$rent=$row['rent'];
$room=$row['room'];
$date=$row['date'];
?>
<?php 
echo"price";
echo"room";
echo"date";
?>
</body>
  </html>

favoritepage.php

<!doctype html>
<html>
  <body>
 <?php 
$cookie_name = "favoritepost";
?>
<?php
 error_reporting(0);
include("config.php");
$result = mysqli_query($connect,"SELECT*FROM ".$db_table." WHERE idhome = $_COOKIE[$cookie_name]");
?>
<?php $row = mysqli_fetch_array($result):
$price=$row['price'];
$rent=$row['rent'];
$room=$row['room'];
$date=$row['date'];
?>
<?php 
echo"price";
echo"room";
echo"date";
?>
    </body>
  </html>

Answer 1

将ID作为数组存储在您的Cookie中，例如，您必须将serialize()/unserialize()或json_encode()/json_decode()数组放入Cookie中，因为它只支持文本而非复杂数据

<?php
$ID = is_numeric($_GET['ID']) ? $_GET['ID'] : 1;

$cookie_name = "favoritepost";

if ( isset($_COOKIE[$cookie_name]) ) {
    $kookie = unserialize($_COOKIE[$cookie_name]);
} else {
    $kookie = array();
}
if ( ! in_array($ID, $kookie) ) {
    $kookie[] = $ID;
}

setcookie($cookie_name, serialize($kookie), time() + (86400 * 30), "/"); // 86400 = 1 day
?>
<html>

现在输入的最后一个cookie我假设您将在查询中使用，所以只需获取

<!doctype html>
<html>
<body>
<?php
$cookie_name = "favoritepost";
include("config.php");
if ( ! isset($_COOKIE[$cookie_name]) ) {
    echo 'NO COOKIE SET';
    exit;
}

$kookie = unserialize($_COOKIE[$cookie_name]);
$ID = $kookie[count($kookie)-1];
$result = mysqli_query($connect,"SELECT * FROM $db_table WHERE idhome = $ID");

$row = mysqli_fetch_array($result):
$price=$row['price'];
$rent=$row['rent'];
$room=$row['room'];
$date=$row['date'];

echo"price";
echo"room";
echo"date";
?>

在php中保存具有相同名称但不同值的cookie

1 个答案: