Question

我有这段代码

    $query1 = "SELECT FLOOR(RAND() * COUNT(*)) AS `offset` FROM `participanti`";
    $offset_result = mysqli_query($connect,$query1);
    $offset_row = mysqli_fetch_object( $offset_result ); 
    $offset = $offset_row->offset;
    $query2 = "SELECT * FROM `participanti` LIMIT $offset, 1";
    $result = mysqli_query($connect,$query2);

    while( $row = mysqli_fetch_assoc($result) ){
        $winner = $row['id'];
    }

但每次重新加载页面时，我都会得到一个不同的数字。如何只获取一次随机值，并且只要我想要保持该值不变？

Answer 1

为了让它永远存在，你很可能需要将获胜者存储在数据库中，例如：

$query1 = "SELECT FLOOR(RAND() * COUNT(*)) AS `offset` FROM `participanti`";
    $offset_result = mysqli_query($connect,$query1);
    $offset_row = mysqli_fetch_object( $offset_result ); 
    $offset = $offset_row->offset;
    $query2 = "SELECT * FROM `participanti` LIMIT $offset, 1";
    $result = mysqli_query($connect,$query2);

    while( $row = mysqli_fetch_assoc($result) ){
        $winner = $row['id'];
        $insert = "INSERT INTO `winners`('winner_id') VALUES('".$winner."')";
        $save = mysqli_query($connect,$insert);
    }

Answer 2

鉴于您选择了一个随机数，实际上每次运行时都会得到一个随机数。

没有什么可以说你不能存储它：

<php
    // file 1

    session_start();
    $_SESSION['winner']=3;

?>

<php
    // file 2 (or reloaded page)

    session_start();
    echo($_SESSION['winner']);

?>

Answer 3

尝试使用会话对象以下列方式存储值：

if(!isset($_SESSION["num"]))
{
    $_SESSION["num"] = $winner;
}

如果之前未设置会话，则会设置会话。

Answer 4

将值存储在会话中（或其他位置），并在再次检查之前检查它是否已存储;

if(!isset($_SESSION['winner'])) {

    $query1 = "SELECT FLOOR(RAND() * COUNT(*)) AS `offset` FROM `participanti`";
    $offset_result = mysqli_query($connect,$query1);
    $offset_row = mysqli_fetch_object( $offset_result ); 
    $offset = $offset_row->offset;
    $query2 = "SELECT * FROM `participanti` LIMIT $offset, 1";
    $result = mysqli_query($connect,$query2);

    while( $row = mysqli_fetch_assoc($result) ){
        $_SESSION['winner'] = $winner = $row['id'];
    }

}

仅从数据库中选择一次随机条目

4 个答案: