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Question

Why does this PHP script interfere with my CSS layout?

Answer 1

+2 A:

You can't just die in the middle of the page - your page will be incomplete and invalid. What you need to do is to check your condition before opening table, output error message html in case of error and the table with content otherwise. Suicide is never a solution.

Michael Krelin - hacker 2010-05-19 18:05:11

I disagree, suicide is the FINAL solution.

Brent Baisley 2010-05-19 18:07:50

Brent, no, you forgot about respawning.

Michael Krelin - hacker 2010-05-19 19:38:45

That is why his K/D ratio is so terrible.

Dolph 2010-05-19 20:00:38

Answer 2

+4 A:

You are very limited when you integrate code and markup this way. The approach that you should take is to run your code, process your inputs and generate a suitable output which then is passed to a template that knows how to output this data.

You can then check for a mysql error before you output anything, set a variable that if present and output a user friendly error message. If no error, then you can loop through your data to display your results.

If you want to loop through the data only once (vs. loop through the dataset first to build the array then again in the template loop), you can assign the mysql result to a variable and process it that way(though you can't close the db until you are done). It doesn't really make that much difference unless you are trying to save milliseconds in an enterprise application, or have 10s of 1000s of rows of data.

Either way you choose to do it, then only way to handle cases like this is to "process first, then present"... this gives you the maximum flexibility for how to handle output.

FilmJ 2010-05-19 18:21:21

+1 for pointing out extreme messyness. :)

Arve Systad 2010-05-21 11:59:39

Answer 3

+3 A:

mmattax 2010-05-19 19:47:46

Answer 4

+4 A:

Just skimming through your code, you might want to fix this broken <tr>:

echo "<tr<td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td>Company:</td><td>";
         ^

Dolph 2010-05-19 19:58:43

Answer 5

+2 A:

Your layout breaks because you "die" right in the middle of outputting HTML. Your page will output the content up to the table just after "View Asset", then if the script dies, it will stop right there and not output a single character below. You'll end up with a table start tag being output but not closed, nor will any of the divs, the body or html tags be closed.

You'll also find that you cannot use the setcookie function once any HTML has started to be output.

I'd suggest completely separating your PHP logic and HTML output - preferably into separate files, or alternatively PHP code at the top of the file and HTML below. Do your logic and save your variables in PHP first, then output all your PHP normally, just including the appropriate variables where necessary. Something like this:

<?php
// connect to the database and run your query here
// save results into an array, something like this:
$rows = array();
while($row = mysql_fetch_assoc($result))
{
    // processing with $row here if needed
    $rows[] = $row;
}

//rest of PHP code

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"&gt;
<html xmlns="http://www.w3.org/1999/xhtml"&gt;
<!-- rest of the opening HTML code here -->

<table width="100%" border="0" cellpadding="2">
<?php foreach ( $rows as $row ) : ?>
<tr>
    <td>Addet ID:</td>
    <td><?php echo $row['id'] ?></td>
</tr>
<tr>
...etc
</tr>
<?php endforeach; ?>
</table>

<!-- rest of the closing HTML code here -->

DisgruntledGoat 2010-05-19 20:27:46

Thank you - I like this.

CT 2010-05-19 21:04:12

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Why does this PHP script interfere with my CSS layout?

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