检查哪个单选按钮？

Question

我想检查哪个单选按钮被选中.然后,在我问这个问题之前,我在这里查看了问题并且他们说了代码

        if(document.getElementById('number1').checked) 

是答案.但是,我得到错误"使用未定义的常量文档 - 假定'文档'"和

Call to undefined function getElementById().

哪里出错了？我是否必须编写函数,getElementById('number1').checked因为它说"未定义"？谢谢

Answer 1

你的代码是Javascript.要检查PHP中单选按钮的值,它需要有一个name属性,该属性是通过GET或POST以表单形式发送的.

// If form method='get'
if (isset($_GET['name_of_radio_group'])) {

  // Show the radio button value, i.e. which one was checked when the form was sent
  echo $_GET['name_of_radio_group'];
}

// If form method='post'
if (isset($_POST['name_of_radio_group'])) {

  // Show the radio button value, i.e. which one was checked when the form was sent
  echo $_POST['name_of_radio_group'];
}

Answer 2

您发布的代码是 JavaScript 代码。为了确定是将表单作为帖子提交还是获取并使用超全局变量 $_POST[]、$_GET[]、$_REQUEST[] 查询值。

您有 HTML 代码：

<input type="radio" name="radio_group1" value="rg1v1" />Radio Group 1 - Value 1<br />
<input type="radio" name="radio_group1" value="rg1v2" />Radio Group 1 - Value 2<br />
<input type="radio" name="radio_group1" value="rg1v3" />Radio Group 1 - Value 3<br />

假设您使用 post 方法将表单提交到 php 文件，以下代码将测试选择了哪个单选按钮。

<?php
    switch($_POST['radio_group1']) {
        case "rg1v1":
            $value = "Radio Group 1 - Value 1 has been selected.";
            break;
        case "rg1v2":
            $value = "Radio Group 1 - Value 2 has been selected.";
            break;
        case "rg1v3":
            $value = "Radio Group 1 - Value 3 has been selected.";
            break;
        default:
            $value = "No radio has been selected for Radio Group 1";
    }
?>