Question

在下面的代码中，我没有检查任何单选按钮，也没有将文本框保持空白并单击“提交”按钮。我希望用户只看到最后的其他声明，即; “在文本框中输入一些值”。

当文本框为空白但选择了单选按钮时，代码可以正常工作。

请告知我应该在代码中进行哪些更改，以便在文本框为空并且未选中单选按钮时输出“在文本框中输入一些值”作为输出。

<html>
 <body>

<form action = "Convertor.php" method = "post">

    Enter Number <input type = "text" name = "n1" placeholder = "Enter only 
    number"><br><br>

    <input type = "radio" name = "r" value = "rs"> Convert USD to Rs <br>
    <input type = "radio" name = "r" value = "m"> Convert Meter To 
 Centimeter <br>
    <input type = "radio" name = "r" value = "d"> Convert Degree to 
 Fahrenheit <br>

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

</body>
</html> 

<?php

if(isset($_POST['n1']))
{
    $no1 = $_POST['n1'];
    $result = 0;
    $condition = $_POST['r'];

    if($no1 != "" ||  $condition == "") //Also tried !isset($_POST['r']) in 
    place of $condition == ""
    {

        if(is_numeric($no1))
        { 
            switch($condition)
            {
                case "rs":
                    $result = $no1 * 65;
                    break;

                case "m":
                    $result = $no1 * 100;
                    break;

                case "d":
                    $result = $no1 * 1.8 + 32;
                    break;                  
            }

            echo $result;
        }

        else
        {
            echo "Enter only numbers";
        }
    }

    else
    {
        echo "Enter some value in textbox";
    }
    }
 ?>

Answer 1

您可以使用isset函数检查是否选择了无线电广播

<?php
if (isset($_POST['n1'])) {
    $no1 = $_POST['n1'];
    $result = 0;
    $condition = isset($_POST['r']) ? $_POST['r'] : "";

    if(!$condition) {
        echo "Please select the Conversion Type";
    } else {
        if ($no1) //Also tried !isset($_POST['r']) in
        {
            if (is_numeric($no1)) {
                switch ($condition) {
                    case "rs":
                        $result = $no1 * 65;
                        break;

                    case "m":
                        $result = $no1 * 100;
                        break;

                    case "d":
                        $result = $no1 * 1.8 + 32;
                        break;
                }

                echo $result;
            } else {
                echo "Enter only numbers";
            }
        } else {
            echo "Enter some value in textbox";
        }
    }
}
?>

如果你愿意，你可以使用必需的属性来确保用户输入值（或检查收音机），你也可以为输入编号而不是文本类型

检查一下：

        <input type="number" name="n1" placeholder="Enter only number" required="required"><br><br>
        <input type="radio" name="r" value="rs" required="required"> Convert USD to Rs <br>
        <input type="radio" name="r" value="m" required="required"> Convert Meter To Centimeter <br>
        <input type="radio" name="r" value="d" required="required"> Convert Degree to

未选中单选按钮时收到错误消息

1 个答案: