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我有一个已经设置了一些图像的数据库，我想让 url 在用户点击下一个时显示每个查询的 ID。用户应该能够共享 URL 并将其粘贴到他们的浏览器中，该 url 应该从查询中提取该唯一 ID。我遇到的问题是每次我粘贴一个网址时，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，我不确定该怎么做:(这是我到目前为止的代码。

<?php
if (isset($_GET['id'])) {
    include("PHP/db.php");

    echo $where = $_GET["id"];
    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>
4

2 回答 2

1

它会进入内部状态(next == true)。

确保您的变量在使用前已初始化。

于 2013-06-19T05:53:14.187 回答
1

尝试这个：

<?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>

我变了：

- 指向下一个的链接现在是 id=rand
- 更改了您的代码以给我一个“rand ID”，并且它已经在您加载的页面的 href 上定义

于 2013-06-19T05:50:10.510 回答