Php URL中显示的用于共享的下一个和上一个链接和ID
我有一个已经设置了一些图像的数据库,我想让url在用户点击下一个查询时显示每个查询的ID。用户应该能够共享URL并将其粘贴到浏览器中,URL应该从查询中提取唯一的ID。我遇到的问题是,每次我粘贴一个url,我都会得到一个随机图像,而不是ID中的图像。我在这里不知所措,我不知道该怎么做:(这是我到目前为止的代码Php URL中显示的用于共享的下一个和上一个链接和ID,php,html,mysql,sql,button,Php,Html,Mysql,Sql,Button,我有一个已经设置了一些图像的数据库,我想让url在用户点击下一个查询时显示每个查询的ID。用户应该能够共享URL并将其粘贴到浏览器中,URL应该从查询中提取唯一的ID。我遇到的问题是,每次我粘贴一个url,我都会得到一个随机图像,而不是ID中的图像。我在这里不知所措,我不知道该怎么做:(这是我到目前为止的代码 <?php if (isset($_GET['id'])) { include("PHP/db.php"); echo $where = $_GET["id"];
<?php
if (isset($_GET['id'])) {
include("PHP/db.php");
echo $where = $_GET["id"];
echo $query = "SELECT * FROM images WHERE ID =" . $where;
$result = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($result);
$ID = $row['ID'];
$title = $row['name'];
$image = "<img height=500 width=600 src=http://www.goupics.com/img/" . $row['name'] . " >";
}
if($_GET['next4']) {
echo 'HELLO THIS IS THE NEXT IF METHOD';
include("PHP/db.php");
$query = "SELECT * FROM images ORDER BY RAND() LIMIT 1";
$result = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($result);
$ID = $row['ID'];
$title = $row['name'];
$image = "<img height=500 width=600 src=http://www.goupics.com/img/" . $row['name'] . " >";
}
?>
<body>
</div>
<div id="title"> <?php echo $title ?> </div>
<div id="mainpic">
<?php echo $image ?>
</div>
<div id="prevnext">
<div id="next">
<a href="?id=<?php echo $ID ?>" name="name4" >Next</a>
</div>
<div id="prev">
<a href="?id=<?php echo $ID ?>">Previous</a>
</div>
</div>
试试这个:
<?php
include("PHP/db.php");
$query2 = "SELECT * FROM images ORDER BY RAND() LIMIT 1";
$result2 = mysqli_query($dbc, $query2);
$rand_row = mysqli_fetch_array($result2);
$rand_id = $rand_row ['ID'];
if (!isset($_GET['id'])) {
$_GET['id'] = $rand_id;
}
if (isset($_GET['id'])) {
echo $where = $_GET["id"];
echo $query = "SELECT * FROM images WHERE ID =" . $where;
$result = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($result);
$ID = $row['ID'];
$title = $row['name'];
$image = "<img height=500 width=600 src=http://www.goupics.com/img/" . $row['name'] . " >";
}
?>
<body>
</div>
<div id="title"> <?php echo $title ?> </div>
<div id="mainpic">
<?php echo $image ?>
</div>
<div id="prevnext">
<div id="next">
<a href="?id=<?php echo $rand_id; ?>" name="name4" >Next</a>
</div>
<div id="prev">
<a href="?id=rand">Previous</a>
</div>
</div>
我改变了:
-到下一个的链接现在是id=rand-更改您的代码以给我一个“rand ID”,它已经在您加载的页面的href上定义了它将进入条件
(next==true)
确保变量在使用前已初始化。如果用户单击“下一步”,if语句只会自动关闭,但即使我放置了一个ID为的自定义URL,if语句也会自动关闭。$dbc是到mysql数据库的连接,它是从包含的文件include(“PHP/db.PHP”)中提取的;