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[SOLVED] integrating a function within HTML


Foser

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This is my function: (works perfecly)

function news(){
$query = mysql_query("SELECT `post_subject`, post_text FROM `phpbb_posts` WHERE `forum_id` = 2 ORDER by `post_id` DESC LIMIT 2");

while ($fetch = mysql_fetch_assoc($query)) {
$article[] = $fetch;
}
$article[0]['post_subject']; 
$article[1]['post_subject']; 
$article[0]['post_text'];
$artcile[1]['post_text'];
}

 

Now it's not showing up in my html.

 

here what've done. I called news(); at  the beginning and I put the variables where I needed to into my html such as:

<h2><?php echo $article[0]['post_subject']; ?></h2>
		<?php echo smiley_text($article[0]['post_subject']); ?><br /><br />
		<a href="#">Message Footer information</a>
	</div>

 

No text is showing up when I execute it. Whats a similar way to do this correctly?

 

thanks

 

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Are you sure your information is coming out of the database correctly? Your syntax looks like it has some problems with it. Try this:

 

$query = mysql_query("SELECT post_subject, post_text FROM phpbb_posts WHERE forum_id='2' ORDER by post_id DESC LIMIT 2");

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It very well may execute it, but you need to return the array from the function, then use it where you like. eg;

 

<?php $article = news(); ?>
<h2><?php echo $article[0]['post_subject']; ?></h2>
<?php echo smiley_text($article[0]['post_subject']); ?><br /><br />
<a href="#">Message Footer information</a>
</div>

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This is the error I get:

Fatal error: Cannot use string offset as an array in C:\WAMP\www\CodersDesk\index.php on line 47

 

not sure if this is exacly how you do the returns... but...

 

return $article[0]['post_subject']; 
return $article[1]['post_subject']; 
return $article[0]['post_text'];
return $artcile[1]['post_text'];

 

 

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You can only return one thing from a function.

 

<?php

function news(){
  $query = mysql_query("SELECT `post_subject`, post_text FROM `phpbb_posts` WHERE `forum_id` = 2 ORDER by `post_id` DESC LIMIT 2");

  while ($fetch = mysql_fetch_assoc($query)) {
    $article[] = $fetch;
  }
  return $article;
}

?>

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