Undefined index with $_POST function - php

Question

So I Basically copied the following code, execpt renaming columns names.
The form is folowing:
<table align="left" width="30%" border="0" >
<form action="company_edit_php.php" method="post">
<tr><td align="center" bgcolor="#ECE6E6">Location: <td align="center"> <input type="text" value="<?php echo $row['Location']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">User: <td align="center"> <input type="text" value="<?php echo $row['User']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Telephone: <td align="center"> <input type="text" value="<?php echo $row['Telephone']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Email: <td align="center"> <input type="text" value="<?php echo $row['Email']?>"/><br></td></tr>
<td> <input type="submit" value="Change" /></td></tr>
And PHP code:
<?php
$con = mysql_connect("localhost","username","password");
$id = isset($_GET['id']) ? (int)$_GET['id']: 1;
mysql_select_db("Database", $con);
if (isset($_POST['Location'])) {
echo $_POST['Location'];
} else {
echo 'empty';
}
$Location = $_POST['Location'];
Result of this code is Undefined index: Location and from if statement i get "empty".
Why this same code work for another page? What should I do now?
Thank you for your efforts

Answer 1

To correct the index post, use this command at the beginning of the file and uses the variable $Location
$Location = (isset($_POST["Location"])?$_POST["Location"]:"");

Answer 2

You have to change your form because you miss name into your input and to retrieve the value with $_GET you have to use it like this:
<tr><td align="center" bgcolor="#ECE6E6">Location: <td align="center"> <input type="text" name="Location" value="<?php echo $row['Location']?>"/><br></td></tr>

Answer 3

To get the value of textboxes, you need to give them name
<table align="left" width="30%" border="0" >
<form action="company_edit_php.php" method="post">
<tr><td align="center" bgcolor="#ECE6E6">Location: <td align="center"> <input type="text" value="<?php echo $row['Location']?>" name='location'/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">User: <td align="center"> <input type="text" value="<?php echo $row['User']?>" name='user'/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Telephone: <td align="center"> <input type="text" value="<?php echo $row['Telephone']?>" name='phone'/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Email: <td align="center"> <input type="text" value="<?php echo $row['Email']?>" name='mail'/><br></td></tr>
<td> <input type="submit" value="Change" /></td></tr>

Answer 4

Location is not passed in the code that you show. for that reason $_POST['Location'] is undefined. put it inside your if statement to avoid the error.
like:
$Location = 'empty';
if (isset($_POST['Location'])) {
echo $_POST['Location'];
$Location = $_POST['Location'];
} else {
echo 'empty';
}

Answer 5

Complete form
<table align="left" width="30%" border="0" >
<form action="company_edit_php.php" method="post">
<tr><td align="center" bgcolor="#ECE6E6">Location: <td align="center"> <input type="text" value="<?php echo $row['Location']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">User: <td align="center"> <input type="text" value="<?php echo $row['User']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Telephone: <td align="center"> <input type="text" value="<?php echo $row['Telephone']?>"/><br></td></tr>
<tr><td align="center" bgcolor="#ECE6E6">Email: <td align="center"> <input type="text" value="<?php echo $row['Email']?>"/><br></td></tr>
<td> <input type="submit" value="Change" /></td></tr>

Answer 6

Try this:
<?php
$con = mysql_connect("localhost","username","password");
$id = isset($_GET['id']) ? (int)$_GET['id']: 1;
mysql_select_db("Database", $con);
$location= "";
if (isset($_POST['Location'])) {
$location= $_POST['Location']; // <------------- put this here!
//echo $_POST['Location'];
} else {
echo 'empty';
}
echo $location; // <------------- added this
//$Location = $_POST['Location']; // it was erroring here as I am guessing that you were sometimes accessing the page without posting the location. If the variable does not exist, you cannot echo it.. thus the changes above

Answer 7

From what you are writing it is so that either there is no field named "Location" on the page you made, or it is outside of the form (thus not inside ....).
As it works on the site where the original sourcecode has been taken from there are only 2 options there.
The site in question HAS a field named location inside the appropriate form
The site in question suppresses the error message that would come.
In order for your code to work you would need to 1.) put the field "Location" inside the form if it is not already and 2.) You need to "secure" the getting of the variable so that an error is not shown even if Location is empty.
I'm giving different examples for both variants:
1.) Putting Location inside the form
<form action="company_edit.php" method="post">
......
<input type="text" name="Location" id="Location>
......
</form>
2.) Making sure that the error does not occur even if Location does notee xist,....
I would even advice doing the following step for every $_POST and $_GET variable you want
to use in your application as it reduces the chance of the error message that a specific
$_GET/$_POST is not set considerably (although for debugging and finding writing errors
it is good to have the error messages).
if (isset($_POST['Location'])
$Location = $_POST['Location'];
else
$Location='';
On another note as you are giving parameters via POST and you don't put the id into the URL the $_GET of id will always result in an empty id field (thus the "1" is always used as id). For it to work you would either need to use id as post and put an appropriate field into the form or you would have to eidt the "action" part of the form so that id is given as "?id=" parameter there