Facing an error - Undefined variable Error in PHP

Question

I am fairly new in PHP and I am facing this issue while practising PHP from a book -

Here is the HTML Code

        <form method="post" action="handle_post.php">
          <p>First Name:  <input type="text" name="first_name" size="20"></p>
                <p>Last Name: <input type="text" name="last_name" size="20"></p>
                <p>Email Address: <input type="email" name="email" size="30"></p>
                <p>Posting: 
                      <textarea name="posting" rows="5" cols="40">
                                        </textarea>
                 </p>
                <input type="submit" name="submit" value="Send My Fucking Email">                 
        </form>

Here is the PHP code .. when I am trying to make it work it is saying that something is wrong in this line -

$posting

I am pasting the code here .. let me know what is wrong in it

<?php

        date_default_timezone_set('Africa/Lagos');
        $first_name = $_POST['first_name']; 
        $last_name = $_POST['last_name']; 
        $posting =  $_POST['posting']; 
        $email = $_POST['email']; 

        $fullname = $first_name . ' ' . $last_name; 

        print "<div>Thank you my lord - $name, for your kind posting in this thread. This is the excerpt of the post: 
                    <p>$posting</p>
                    </div>";


        $fullname = urlencode((binary) $fullname); 
        $email = urlencode((binary) $email); 

        print "<p>Click <a href=\"thanks.php?name=$name&email=$email\">Here</a> to continue. </p>"; 



?>

`print` doesn't understand new lines. – u_mulder Dec 05 '15 at 18:28 — u_mulder, Dec 05 '15 at 18:28

score 1 · Answer 1 · answered Dec 05 '15 at 18:30

1

You use the variable $name which not declared:

print "<div>Thank you my lord - $name, for your kind posting in this thread. This is the excerpt of the post: 
                <p>$posting</p>
                </div>";

answered Dec 05 '15 at 18:30

Gouda Elalfy

6,888
1
26
38

Elafy okay i was testing .. the problem was not with $name but with the term $posting – danish deb Dec 06 '15 at 04:56

score 0 · Answer 2 · answered Dec 05 '15 at 18:29

0

Turn your print into echo when using variables like that.

answered Dec 05 '15 at 18:29

VIDesignz

4,703
3
25
37

Interpolation isn't a feature of `echo`, it's a feature of strings. – castis Dec 05 '15 at 18:51
@castis Using `echo "My name is $name" ;` Works, not sure what you are talking about. As long as there is double quotes to make it a string... – VIDesignz Dec 05 '15 at 18:53
@castis Though a good habit would be to use brackets like `echo "My name is {$name}";` – VIDesignz Dec 05 '15 at 18:59

Facing an error - Undefined variable Error in PHP

2 Answers2