Using PHP $_POST to fetch input from another PHP script's echo form

Question

I understand the title is pretty ambiguous so let me explain in more detail. I am using PHP to output a form which will display certain user information, the information is that which the user had previously stored in the database. This part of the code is working fine and the data is being displayed as I wish. Once the data is displayed in the input boxes, I want the user to be able to edit the information and then once the user clicks the submit button the updateuserinfo.php script will be called:

    while($return = mysql_fetch_assoc($result)) {
    echo "<form action=UpdateUserInfo.php method=post>";
    echo "<p> First Name: <input type=text name=firstname value= ".$return['FirstName']."/></p>";
    echo "<p> Surname: <input type=text value= ".$return['Surname']." /></p>";
    echo "<p> House Number: <input type=number value= ".$return['HouseNumber']." /></p>";
    echo "<p> Address Line One: <input type=text value= ".$return['AddressLineOne']." /></p>";
    echo "<p> Address Line Two: <input type=text value= ".$return['AddressLineTwo']." /></p>";
    echo "<p> County: <input type=text value= ".$return['County']." /></p>";
    echo "<p> Phone Number: <input type=number value= " .$return['PhoneNumber']." /></p>";
    echo "<p> Email: <input type=email value= ".$return['Email']." /></p>";
    echo "<p> Username: <input type=text value= ".$return['Username']." /></p>";
    echo "<p> Password: <input type=password value= ".$return['Password']." /></p>";
    echo "<p> User Type: <input type=text value= ".$return['UserType']." /></p>";
    echo "<input id=submit type=submit value='Update Info' Info />";
    echo "</form>";

In the script found in updateuserinfo.php I was hoping I could firstly display the users info i.e what the user has changed the input box values to and then update the database if the user confirms the changes. I am confident I will be able to update the information stored on the database as this is a simple SQL query but the problem I am having is that I can not display the information found in the input boxes when the user has made changes. I thought I could apply a name attribute to each input tag, like I have done with 'firstname' and then use:

    echo "<p>First Name: " $_POST[firstname] "</p>";

in the updateuserinfo.php in order to display the information the user has entered in the input box. However this is not working. Any suggestions would be greatly appreciated as I'm extremely new to programming in PHP. Thanks in advance.

Answer 1

You should read $_REQUEST array, not $_POST...

UPDATE

Sorry, I did never use $_POST... It has been introduced in 4.1.0. So, only check your PHP version is >= 4.1.0... Otherwise please post

print_r($REQUEST);

results...

Answer 2

Try to show the complete $_POST array by using this code: var_export($_POST);

Also you are missing some quotes in your Form. For example the line

echo "<p> First Name: <input type=text name=firstname value= ".$return['FirstName']."/></p>";

should be

echo "<p> First Name: <input type=\"text\" name=\"firstname\" value=\" ".$return['FirstName']."\"/></p>";

or easier within single quotes (so you don't have to mask each double quote):

echo '<p> First Name: <input type="text" name="firstname" value= "'.$return['FirstName'].'"/></p>';

Read What is the difference between single-quoted and double-quoted strings in PHP? if you want to know what's the difference between single and double quotes in PHP

And in your update Script you are missing the dot operator and also the quotes:

echo "<p>First Name: " $_POST[firstname] "</p>";

should be

echo "<p>First Name: " . $_POST["firstname"] . "</p>";