PHP Code

Question

redous 0 Newbie Poster

13 Years Ago

Hi,

I have some PHP code in a project which if a user is logged in the in will display the username and other options and if the user is a guest it wont.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Home</title>
</head>
<body>

<!-- HTML Layout here -->

<?php
if($session->logged_in){
echo "Welcome $user"
  ."<a href="index.php">Home</a><br>"
  ."<a href="forums.php">Forums</a><br>"
  ."<a href="downloads.php">Downloads</a><br>"
  ."<a href="profile.php">Profile</a><br>"
  ."<a href="settings.php">Settings</a><br>";
}
else {
echo "Welcome Guest"
  ."<a href="index.php">Home</a><br>"
  ."<a href="forums.php">Forums</a><br>"
  ."<a href="downloads.php">Downloads</a><br>";
}
?>
</body>
</html>

now even though this works my project needs to be validated but the validator fails because of the php coding it says the following

#  Error  Line 1208, Column 83: character data is not allowed here

…strong>$session->username</strong>  <span style='font-size: 80%;'></span></li>'

You have used character data somewhere it is not permitted to appear. Mistakes that can cause this error include:

    * putting text directly in the body of the document without wrapping it in a container element (such as a <p>aragraph</p>), or
    * forgetting to quote an attribute value (where characters such as "%" and "/" are common, but cannot appear without surrounding quotes), or
    * using XHTML-style self-closing tags (such as <meta ... />) in HTML 4.01 or earlier. To fix, remove the extra slash ('/') character. For more information about the reasons for this, see Empty elements in SGML, HTML, XML, and XHTML.

is there a way i can bypass this part of the coding or can i change something in the code which the validator will accept.

php validator

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9 Replies
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Latest Post 13 Years Ago Latest Post by redous

All 9 Replies

edwinhermann 57 Junior Poster

13 Years Ago

strong>$session->username </li>'

The bit that's wrong is the use of single quotes instead of double quotes in HTML.

The "span" tags should read:

<span style="font-size: 80%;"></span>

edwinhermann 57 Junior Poster

13 Years Ago

ive tried all ways of putting this part of the code
.'$session->username </li>'
."$session->username </li>"
.'$session->username </li>'
none of these work the middle one actually throws up 5 more validator errors. Am i using the wrong doctype due to the page having php in.

Can you show us the code which includes that line? THe original code you posted does not contain the code.

edwinhermann 57 Junior Poster

13 Years Ago

That code looks good. Verify in your "View Source" (in the browser) that it comes out as:

<span style="font-size: 80%;"></span></li>

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redous 0 Newbie Poster · Answer 1 · 2010-10-10T08:37:14+00:00

The bit that's wrong is the use of single quotes instead of double quotes in HTML.
The "span" tags should read:

ive tried all ways of putting this part of the code

.'$session->username </li>'
."$session->username </li>"
.'$session->username </li>'

none of these work the middle one actually throws up 5 more validator errors. Am i using the wrong doctype due to the page having php in.

redous 0 Newbie Poster · Answer 2 · 2010-10-10T08:38:44+00:00

i originally thought it was because the </li> was there without the beginning <li> but that again throws more errors up and leaves my site all messed up.

redous 0 Newbie Poster · Answer 3 · 2010-10-10T21:24:26+00:00

<?php
if($session->logged_in){
echo "<li class='userbar_row_title'>"
	."<strong>$session->username</strong>  <span style=\"font-size: 80%;\"></span></li>"
	."<li class='userbar_row'><a class='userbar_row content' href='../userinfo.php?user=$session->username'>"
	."<img class='userbar_row_large_icon' src='modules/userbar/images/awards.png' alt=''>Profile</a></li>"
	."<li class='userbar_row'><a class='userbar_row content' href='#'>"
	."<img class='userbar_row_large_icon' src='modules/userbar/images/games.png' alt=''>Mail <span>(8)</span></a></li>"
	."<li class='userbar_row'><a class='userbar_row content' href='../useredit.php' title='Settings'>"
	."<img class='userbar_row_large_icon' src='modules/userbar/images/edit.png' alt=''>Edit Profile</a></li>"
	."<li class='userbar_row'><a class='userbar_row content' href='../process.php' title='Logout'>"			
	."<img class='userbar_row_large_icon' src='modules/userbar/images/logout.png' alt=''>Logout</a></li></ul>"
	."<a class='userbar_link' href='../userinfo.php?user=$session->username' title='My Profile'>"
	."<img src='images/users/thumbnails/$session->username.jpg' class='userbar_photo' style='position: absolute; top: 0px; left: 0px;' alt=''>"
	."<div class='userbar_caption' style='padding-left: 26px;'>$session->username</div></a></div>"
	."<div class='userbar_right_line'></div>"

redous 0 Newbie Poster · Answer 4 · 2010-10-11T02:22:23+00:00

yea it does but it still flags in xhtml validator has being wrong is it something to do with the validator itself not liking php code.

edwinhermann 57 Junior Poster · Answer 5 · 2010-10-11T02:27:29+00:00

The PHP code should not display to the browser. Your query as to why the XHTML validator is failing your code is not really a PHP question but more an HTML question.

I suggest you post the entire HTML code (copy/pasted from the browser View Source). Some people here may be able to help you work out why the XHTML validator doesn't like your code, but a better forum would be the HTML and CSS one here: http://www.daniweb.com/forums/forum143.html

redous 0 Newbie Poster · Answer 6 · 2010-10-11T05:20:35+00:00

okay for some wierd reason i gave up on solving the problem last night this morning i logged on looked at it again and still there was a problem. Again i had enough and logged out after changing a few things (nothing to do with the line in question) but was still getting a error i then retried just and the error messaged has gone and all is validated i have no idea what i did (or who did) but its all clear now.

Case closed

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