如何上传HTML表单中输入一个单选按钮值

Question

我所做的形式，使其获得所选择的单选按钮的值，它被传递到PHP部分，并在database.But在数据库中它显示为“开”无论选择是什么。

我不知道在哪里我已经错了

HTML表格：

<form action="Database.php" name="register" method="post">
<div>
    First Name   <input type="text"  name="fname"/><br>
    Last Name    <input type="text"  name="lname"/><br>
    Email        <input type="email" name="email"/><br>
    Contact No.  <input type="text"  name="num"/><br>
    Gender <br>  <input type="radio" name="g1" value="Male"/>Male

                     <input type="radio" name="g1" value="Female"/>Female
            <br>
            <br>
</form>

PHP：

$fname      = $conn->real_escape_string($_POST['fname']);
$lname      = $conn->real_escape_string($_POST['lname']);
$email      = $conn->real_escape_string($_POST['email']);
$cnumber    = $conn->real_escape_string($_POST['num']);
$gender     = $conn->real_escape_string($_POST['g1']);

$sql="INSERT INTO data (fname,  lname, email, cnumber, gender) 

VALUES ('".$fname. "','".$lname."','".$email."', '".$cnumber."', '".$gender."')";

我预计输出为男/女，但“上说：”

Answer 1

使用表单我使用了一个小的代码，发现这个工作

<form action="Database.php" name="register" method="post">
    <div>
        First Name   <input type="text"  name="fname"/><br>
        Last Name    <input type="text"  name="lname"/><br>
        Email        <input type="email" name="email"/><br>
        Contact No.  <input type="text"  name="num"/><br>
        Gender <br>  <input type="radio" name="g1" value="Male"/>Male

        <input type="radio" name="g1" value="Female"/>Female
        <br>
        <input type="submit" style="min-width:100%" align="center" name='submit' value="SUBMIT">
        <br>
</form>

你的PHP端应该是这样的：

if(!isset($_POST['submit']))
{
//This page should not be accessed directly. Need to submit the form.
    echo "Submitted";
}
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$g1 = $_POST['g1'];   
               $sql = "INSERT INTO etrack.test SET
                    fname = '".$fname."',
                    lname = '".$lname."',
                    g1 = '".$g1."'";

    if ($conn->query($sql) === TRUE) {
        echo "";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
        echo("Error description: " . mysqli_error($con));
        echo "Error 1";
    }

我得到所要求的结果