如何将名称放入选项并发布?

时间:2014-02-04 23:27:19

标签: php html mysql

我想知道如何从选项标签获取VIA信息,这是我的代码到目前为止,我不知道该怎么做,谢谢。

<html>
<body>
    <form method = "post" action="<?php echo $_SERVER['PHP_SELF'];?>">
        <select>
            <option name = "rock">Rock</option>
            <option name = "paper">Paper</option>
            <option name = "scissors">Scissors</option>
        </select>
        <button type = "submit">Submit</button>
    </form>
    <?php
    $rock = $_POST["rock"];
    $paper = $_POST["paper"];
    $scissors = $_POST["scissors"];





    ?>
</body>
</html>

1 个答案:

答案 0 :(得分:1)

您想要为select命名,并为每个选项设置值,如下所示:

<select name="choice">
    <option value= "rock">Rock</option>
    <option value= "paper">Paper</option>
    <option value= "scissors">Scissors</option>
</select>

$pick = $_POST['choice']

另请注意,将操作字段留空会自动回发到同一页面

<form method = "post" action="">