PHP undefined index or variable [closed] - php

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I have searched the whole internet saw unlimited stack posts followed by letter some tutorials and still i cant get it working but why? it keeps telling me undefined index when i use isset becomes undefined variable. Sorry my coding level is low...
Thanks in advance.
<form name="ghostform" action="insert.php" method="post">
<input type="hidden" name="id" value="1" id="id" />
<input type="hidden" name="eidos" value="2" id="eidos" />
<input type="hidden" name="seidos" value="3" id="seidos" />
<input type="hidden" name="idiotites" value="4" id="idiotites" />
</form>
<?php
mysql_connect("localhost","root","")or die("cant connect to server");
mysql_set_charset('utf8');
#mysql_select_db("cafemanager")or die( "Unable to select database");
$id = $_POST['id'];
$eidos = $_POST['eidos'];
$seidos = $_POST['seidos'];
$idiotites = $_POST['idiotites'];
mysql_query("INSERT INTO Current(id, eidos, specificeidos, idiothtes) VALUES('$id','$eidos','$seidos','$idiotites')");
?>

Answer 1

Use something like this:
<?php
if($_POST) {
mysql_connect("localhost","root","")or die("cant connect to server");
mysql_set_charset('utf8');
mysql_select_db("cafemanager")or die( "Unable to select database");
$id = mysql_real_escape_string(strip_tags($_POST['id']));
$eidos = mysql_real_escape_string(strip_tags($_POST['eidos']));
$seidos =mysql_real_escape_string(strip_tags( $_POST['seidos']));
$idiotites = mysql_real_escape_string(strip_tags($_POST['idiotites']));
mysql_query("INSERT INTO Current(id, eidos, specificeidos, idiothtes) VALUES('$id','$eidos','$seidos','$idiotites')");
}
?>
<form action="" method="post">
<input type="hidden" name="id" value="1" id="id" />
<input type="hidden" name="eidos" value="2" id="eidos" />
<input type="hidden" name="seidos" value="3" id="seidos" />
<input type="hidden" name="idiotites" value="4" id="idiotites" />
</form>
But please start using mysqli not depracted mysql

Answer 2

Every time you need a GET or a POST, I'd recommend you to especify default values in case that you don't get all values, so the function will not throw any warning, or extrange things inside the SQL sentence.
Example:
In this case i define 0 as default.
$id = isset($_POST["id"])?$_POST["id"]:0;
And also, if you are waiting for a number as a value.
$id = isset($_POST["id"])?intval($_POST["id"]):0;
If 0 is an impossible value and don't makes any sense, then you know that something has gone wrong, and you can react to it.

Answer 3

how do you submit this?
and please use if(isset($_POST['name']){ // do stuffs }

Answer 4

The reason it shows "undefined index or variable" is because PHP cannot find the variables in the $_POST array.
If this page is handling inputs from a form, make sure that the action of the form points to this page and the method of the form is set as POST.
<form action="urlofthispage" method="POST">
If this page is also used for handling other scenarios, you can simply check if variables are set in $_POST. The isset function simply checks if the variable has been set. Like this:
if(isset($_POST['id']))
{
//then do what do u want if the variable is present
}
else
{
//then do what do u want if the variable is not present
}
For the official documentation of the isset function, you can check this for reference:
http://in2.php.net/isset

Answer 5

Your actual problem
You assume that PHP can get the values from your fields, which you can't..
Because you need to submit the values somehow to the server before PHP realizes that they are there, either do this by a submit button or send your values via javascript.
PHP
A server side language that is not executed until you request the content from the server. Meaning that if you want these values (that are located in the clients browser) you need to send them (POST) to the server for processing everything between <?php and ?>
Try this code
<form name="ghostform" action="insert.php" method="post">
<input type="hidden" name="id" value="1" id="id" />
<input type="hidden" name="eidos" value="2" id="eidos" />
<input type="hidden" name="seidos" value="3" id="seidos" />
<input type="hidden" name="idiotites" value="4" id="idiotites" />
<input type="submit" value="kablamo">
</form>
Tested the code, and it works.
As long as you have a submit button...
On a side note, here's how isset works
function getVal($var) {
if(!isset($_POST[$var]))
return 'default';
return $_POST[$var];
}
$id = getVal('id');
$eidos = getVal('eidos');
$seidos = getVal('seidos');
$idiotites = getVal('idiotites');
Use isset to catch if the value is set or not.
And you could use a function with default values (this is just a primal example) if the value is not set. Or you could raise some form of exception before connecting to the database.
#MiQUEL's solution is neater, only reason why I'd go with a function is because it's more readable and you can perform operations before setting the value.. like this:
function getVal($var) {
if(!isset($_POST[$var])) {
mysql_connect("localhost","root","")or die("cant connect to server");
mysql_select_db("mydb");
$result = mysql_fetch_array(mysql_query("SELECT defaultValue FROM mytable WHERE name=" . $var), MYSQL_NUM);
return $result[0];
}
return $_POST[$var];
}