CSS & HTML In PHP Echo Statement

[jamesjmann]

Hey, I posted a thread before about a news cms I've been developing. And right now I'm trying to flesh out the css of the articles I post on the home page. Now the problem I having is everytime I place a <span class=""></span> or <p class=""></p> or even <div class=""></div>, my web page does NOT load, and instead gives me an error.

 

Parse error: syntax error, unexpected T_STRING in /home/content/d/j/s/djsmiley/html/index.php on line 405

 

Now, I know it's possible to add css to a PHP echo statement, but I'm clearly doing something wrong.

 

Here's the code for the php echo statement with the css included:

 

<? 
include("dbconnect.php"); //include the file to connect to the database 
$getnews = mysql_query("SELECT * FROM mynews ORDER BY id DESC"); //query the database for all of the news 

while($r=mysql_fetch_array($getnews)){ //while there are rows in the table 
extract($r); //remove the $r so its just $variable 

echo("<br><span class="NewsID">$type</span> <span class=h2>$title</span><br><br>
<em>posted by <strong>$user</strong> | added on $time</em><br><br> 
$message<br><br>
<label class="fltlft2"><img src="../../Websites/DJSmiley.Net/images/Icons/Arrows/Right.png" width="20" height="20"/></label><a href=$url>Read more - $url</a>
		<div class="newsLikeShareRate">
				  <table width="100%" border="0">
					<tr>
					  <td width="3%" height="21"><script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script>
						  <fb:like href="$url" show_faces="true" width="450" font="arial"></fb:like>
                              </td>
					<td width="65%"><a name="fb_share" id="fb_share4" type="icon_link" 
		   share_url="$url">Share</a>
						  <script src="http://static.ak.fbcdn.net/connect.php/js/FB.Share" 
				type="text/javascript"></script></td>
					  <td width="32%">Rate this article:
					   </td>
					</tr>
				  </table>
		</div>"); 

} 

?>

 

If you're wondering why there's a facebook script in there, I made it to where everytime a new post is added, a facebook like and share button are added, and are assigned the url I specify when the article is posted.

 

Well, I think that's it for the code. Can anyone tell me what I'm doing wrong?

[Pikachu2000]

When using double quotes within a double quoted string, or single quotes within a single quoted string, they must be escaped with a backslash.

 

echo "Dave said, \"Hello\" to everyone he saw that day.";
echo 'That\'s Jimmy\'s beach ball.';

[kenrbnsn]

You have surrounded your string with double quotes and have double quotes in the HTML. As soon as PHP sees the second double quote, that terminates the string and you get an error. Just surround your string with single quotes. Also you don't need to put parenthesis around the string when using echo.

 

Ken

[jamesjmann]

Ok. Well, I achieved a similar result when eliminating all of the quotes within the echo statement as a whole. I just googled it, and someone pretty much said what you guys said. Thanks!