Undefined index: [duplicate] - php

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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
I get this error:
Notice: Undefined index: gender \form.php on line 26.please help..how solve this error..?
how get from form employee gender male or female and insert MySQL database
**my html code**
<form action="form.php" method="post">
<table border="1">
<tr>
<td align="center">Form Input Employees Data</td>
</tr>
<tr>
<td>
<table>
<form method="post" action="form.php">
<tr>
<td>Name</td>
<td><input name="employe_name" type="text" id="employe_name" size="20">
</td>
</tr>
<tr>
<td>Address</td>
<td><input name="employe_add" type="text" id="employe_add" size="40">
</td>
</tr>
<tr>
<td height="23">Gender</td>
<td>Male<input type="radio" name="gender"value="m" />
Female<input type="radio" name="gender"value="f"/>
</td></tr>
<td>
<td align="right"><input type="submit"
name="submit" value="Submit"> add</td>
</tr>
</table>
</td>
</tr>
</table></form>
And
my PHP code is...
<?php
//Open a new connection to the MySQL server
$mysqli = new mysqli('localhost','root','','employe');
//Output any connection error
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}else{
echo"connection good";
}
if(isset($_POST['gender']))
{
$gender=$_POST['gender'];
echo("your are select" .$gender);
}
//get values from form
$employe_name=$_POST['employe_name'];
$employe_add=$_POST['employe_add'];
$gender=$_POST['gender'];
//insert data into mysql
$sql = "INSERT INTO employe_detail(employe_name,employe_add,gender) VALUES('$employe_name','$employe_add','$gender')";
print '<h3>Insert a record</h3>';
if($sql)
{
print 'Success! ID of last inserted record is : ' .$mysqli->insert_id .'<br />';
}else{
die('Error : ('. $mysqli->errno .') '. $mysqli->error);
}
$mysqli->close();
?>
please help me............

Answer 1

Change your code from
if(isset($_POST['gender']))
{
$gender=$_POST['gender'];
echo("your are select" .$gender);
}
//get values from form
$employe_name=$_POST['employe_name'];
$employe_add=$_POST['employe_add'];
$gender=$_POST['gender'];
To
$gender = "";
if(isset($_POST['gender']))
{
$gender=$_POST['gender'];
echo("your are select" .$gender);
}
//get values from form
$employe_name=$_POST['employe_name'];
$employe_add=$_POST['employe_add'];
gender will be assigned or will be left blank