PHP. undefined index [duplicate] - php

Question

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
Here is the php connectivity code:
<?php
$con=mysql_connect("127.0.0.1", "root", "");
mysql_select_db("society");
if(!$con)
{
echo "Failed to Connect to MySql".mysqli_connect_error();
}
else
{
echo"success";
}
$query="INSERT INTO users (fname, lname, email, password)
VALUES('".$_POST['fname']."','".$_POST['lname']."','".$_POST['email']."','".$_POST['password']."')";
mysql_query($query,$con);
?>
This is the error. Tried, but I am not able to solve the error:
Notice: Undefined index: fname in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC9\data\localweb\scripts\html\connect.php on line 13
Notice: Undefined index: lname in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC9\data\localweb\scripts\html\connect.php on line 13
Notice: Undefined index: email in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC9\data\localweb\scripts\html\connect.php on line 13
Notice: Undefined index: password in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC9\data\localweb\scripts\html\connect.php on line 13

Answer 1

Try this
<?php
$con=mysql_connect("127.0.0.1", "root", "");
mysql_select_db("society");
if(!$con)
{
echo "Failed to Connect to MySql".mysql_connect_error();
}
else
{
echo"success";
}
if(isset($_POST['fname']) && isset($_POST['lname']) && isset($_POST['email'])) {
$query="INSERT INTO users (fname, lname, email, password)
VALUES('".$_POST['fname']."','".$_POST['lname']."','".$_POST['email']."','".$_POST['password']."')";
mysql_query($query,$con);
}
?>

Answer 2

Check if there is values for the POST your using in the query. If it is returning undefined_index is because the variables of the post that you are using isn't set. They don't even exist at the moment.
For instance:
If you debug $_POST['fname'] it will return false or empty. Try it.
Resolution:
Right now I think you're running this code without validating if the form has been submited or not.
So, to start you should do this:
if(isset($_POST) && count($_POST) > 0){
$con=mysql_connect("127.0.0.1", "root", "");
mysql_select_db("society");
if(!$con)
{
echo "Failed to Connect to MySql".mysqli_connect_error();
}
else
{
echo"success";
}
$query="INSERT INTO users (fname, lname, email, password) VALUES('".$_POST['fname']."','".$_POST['lname']."','".$_POST['email']."','".$_POST['password']."')";
mysql_query($query,$con);
}