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我正在尝试创建一个系统，当我提交表单时，在页面刷新后它应该显示我从数据库中获取的新值。这些值在进入数据库时​​运行良好，但在提交后不显示，只有在我再次刷新时才会显示。感谢您的帮助

<?php

include("connect.php");

    $query = "SELECT * FROM `laliga`";
    $result = mysql_query($query);

    while($row = mysql_fetch_assoc($result)){
        $id = $row['id'];
        $home = $row['home'];
        $away = $row['away'];
        $win = $row['win'];
        $draw = $row['draw'];
        $lose = $row['lose'];
    }

    echo "<h2>La Liga</h2>";

    echo $home, " - ", $away;

    if (isset($_POST) && $_POST['laliga'] == 1){
        $select = mysql_real_escape_string($_POST['laliga']);
        $select = $win + $select;
        mysql_query("UPDATE laliga SET win='$select'");
    }else if (isset($_POST) && $_POST['laliga'] == 'X'){
        $select = mysql_real_escape_string($_POST['laliga']);
        $select = '1';
        $select = $draw + $select;
        mysql_query("UPDATE laliga SET draw='$select'");
    }else if (isset($_POST) && $_POST['laliga'] == 2){
        $select = mysql_real_escape_string($_POST['laliga']);
        $select = '1';
        $select = $lose + $select;
        mysql_query("UPDATE laliga SET lose='$select'");
    } 

    header('Location: ../laliga.php');


    ?>

    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

        <input type="radio" name="laliga" value="1">1
        <input type="radio" name="laliga" value="X">X
        <input type="radio" name="laliga" value="2">2

        <input type="submit" name="submit" value="Submit"/>

    </form>

    <br/>
    <?php

    echo $home, " -> ", $win;
    echo "<br/>Barazim  -> ", $draw,"<br/>";
    echo $away, " -> ", $lose;

    ?>
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2 回答 2

1

您应该在 PHP 文件的顶部处理所有发布数据，而 header 函数将解决您的问题，这是一种愚蠢且低效的处理方式。通过首先处理发布数据并更新数据库，当您查询数据库时，数据就在那里！目前您正在尝试查找数据然后添加它。这有意义吗？祝你好运！

于 2013-06-13T23:49:58.980 回答
0

添加：

header('Location: <mypage.php>');

之后mysql_query。

于 2013-06-13T23:12:43.990 回答