Undefined index: userid - php

Question

I'm creating an edit user profile for my project. I came across this error "
Notice: Undefined index: userid in C:\xampp\htdocs\HelloWorld\EditProfile.php on line 18
". I've spent an hour trying to find the cause of the error but I can't seem to find it. I followed this guide here php -'Edit' function for forum posts and such Here's my code:
EditProfile.php
<?php
// connect to SQL
$dbcnx = mysql_connect("localhost", "root", "2345fypj");
if (!$dbcnx)
{
echo( "<P>Unable to connect to the database server at this time.</P>" );
exit();
}
// connect to database
$dbcon = mysql_select_db("my_db", $dbcnx);
if (!$dbcon) {
echo( "<P>Unable to locate DB table at this time.</P>" );
exit();
}
//data preparation for the query
$id = intval($_GET['userid']);
// selects title and description fields from database
$sql = "SELECT * FROM user_profile WHERE userid=$id";
$result = mysql_query($sql) or die(mysql_error());
# retrieved by using $row['col_name']
$row = mysql_fetch_array($result);
?>
<h3>Edit</h3>
<form action="save_edit.php" enctype="multipart/form-data" method="post" name="myForm" />
<table>
<tr>
<td><b>Name</b></td>
<td><input type="text" size="70" maxlength="100" name="title" value="<?php echo $row['name'] ?>"></td>
</tr>
<tr>
<td><b>Age</b></td>
<td><input type="text" size="70" maxlength="100" name="title" value="<?php echo $row['age'] ?>"></td>
</tr>
</table>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input name="enter" type="submit" value="Edit">
</form>
<?php
mysql_close($dbcnx);
?>
save_edit.php
<?php
// connect to SQL
$con = mysql_connect("localhost", "root", "2345fypj");
if (!$con) {
echo( "<P>Unable to connect to the database server at this time.</P>" );
exit();
}
// connect to database
$dbcon = #mysql_select_db("user_profile", $con);
if (!$dbcon) {
echo( "<P>Unable to locate DB table at this time.</P>" );
exit();
}
#data preparation for the query
$id = intval($_POST["userid"]);
foreach ($_POST as $key => $value) $_POST[$key] = mysql_real_escape_string($value);
$sql = "UPDATE user_profile SET
name='$_POST[name]',
age='$_POST[age]',
WHERE userid=$id";
if (!mysql_query($sql,$con)) {
die('Error: ' . mysql_error());
}
mysql_close($con);
header ("location: http://www.domain.com/url_to_go_to_after_update");
?>
Thanks in advance.

Answer 1

$id = intval($_GET['userid']);
It means set the $id variable to the "userid" variable in your URL. For example, if your URL is mySite.com?userid=12, $id will be set to "12". if your URL doesn't have "username=aValue" at the end section of it, you'll get the error you're seeing. :)
You could change it to this to set a default value:
$id = (isset($_GET['userid']) ? intval($_GET['userid']) : -1);

Answer 2

The problem is you're trying to use variable $_GET['userid'] while it's undefined. This variable refers to GET parameter in URL (e.g. EditProfile.php?userid=42). If this parameter isn't passed in URL, you will get this warning. You should check existence of variable:
if (!isset($_GET['userid'])) {
die("Parameter is missing!");
}
$id = intval($_GET['userid']);

Answer 3

You should probably be sure that the value is correctly set before trying to access it. Try using the isset function beforehand.
http://php.net/manual/en/function.isset.php

Answer 4

The problem with your code is apart from using a variable before its used, your continuing todo the query when setting to a default value with intval (will always return atleast 0), which if a non numeric character is passed your always going to update the row 0 user in the table, what you should be doing is not updating anything and returning the user with an error. You also have a rouge , after you age column in the query.
<?php
if(isset($_POST["userid"]) && is_numeric($_POST["userid"])){
$_POST = array_walk($_POST,'mysql_real_escape_string');
$sql = "UPDATE `user_profile` SET `name`='{$_POST['name']}', `age`='{$_POST['age']}'
WHERE `userid` = {$_POST['userid']}";
mysql_query($sql);
}else{
header('Location: ./failed');
}
?>