Insert radio button value to mysql

Question

I have created a test form just to try to send my radio button value to mysql. I am having problems with it at the moment. The code below is just a test, I want the radio button to submit the value but it isn't.

    <table width="300" border="0" align="center" cellpadding="0" cellspacing="1">
<tr>
<td><form name="form1" method="post" action="insert_ac.php">
<table width="100%" border="0" cellspacing="1" cellpadding="3">
<tr>
<td colspan="3"><strong>TEST </strong></td>
</tr>
<tr>
<td width="71">Name</td>
<td width="6">:</td>
<td width="301"><input name="name" type="text" id="name"></td>
</tr>
<tr>
<td>Case</td>
<td>:</td>
<td><input name="case" type="radio" id="case1"> <input name="case" type="radio" id="case2"> <input name="case" type="radio" id="case3"></td>
</tr>
<tr>
<td>Email</td>
<td>:</td>
<td><input name="email" type="text" id="email"></td>
</tr>
<tr>
<td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</form>
</td>
</tr>
</table>

And here is the connection part of the database

<?php

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="123"; // Mysql password 
$db_name="store"; // Database name 
$tbl_name="test_mysql"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// Get values from form 
$name=$_POST['name'];
$case=$_POST['case'];
$email=$_POST['email'];

// Insert data into mysql 
$sql="INSERT INTO $tbl_name(name, case, email)VALUES('$name', '$case', '$email')";
$result=mysql_query($sql);

// if successfully insert data into database, displays message "Successful". 
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='insert.php'>Back to main page</a>";
}

else {
echo "ERROR";
}
?> 

<?php 
// close connection 
mysql_close();
?>

Answer 1

This is done by adding values to radio button input. For instance:

<form method="post">
<input type="radio" name="sex" value="male">Male<br>
<input type="radio" name="sex" value="female">Female
<input type="submit">
</form> 

Answer 2

You should start with closing all of your <input> tags with </input> or at least a slash at the end (like <input name="case" type="radio" id="case1"></input>).

You should set values to your radios (like this they always return 'on'), whereas the submit button needs neither name nor value.

EDIT:

Define a default radio with selected in yout input tag! If none is selected, there's no case getting transmitted and PHP will throw Undefined index: case when accessing $_POST['case'].

A good way to prevent such errors is to check if all necessary indices are set. You can do the following:

if(isset($_POST['name']) and isset($_POST['case']) and isset($_POST['email'])) { ... }

Answer 3

   <!-- Once you have created Mysql connection and column in specified database table,-->
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "website";

// Create connection
$conn = mysqli_connect($servername, $username, $password,$dbname);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";}           
$sql="INSERT INTO Registration(Name,FatherName,CNIC,FCNIC,Email,Password,Contact,Gender) VALUES ('$_POST[Name]','$_POST[FatherName]','$_POST[CNIC]','$_POST[FCNIC]','$_POST[Email]','$_POST[Password]','$_POST[Contact]','$_POST[Gender]')"; 
if (!mysqli_query($conn,$sql))
{
  die('Error:'.mysqli_error($conn));
}
echo "         & 1 record added";
mysqli_close($conn);




?>
 <!--after that you just need to  write this code and make sure to adjust this code because i'm posting some portion of my code."-->



<h4>Gender</h4> <input type="radio" value="Male"  name="Gender"> Male 
     &nbsp;&nbsp;&nbsp; 
<input type="radio" value="Female"   name="Gender" >Female <br><br>
        <button type="Submit">Submit</button><br>
//This answer is for you.