Not Working

Sharif

I want to request a row from a sql database with this code:
<?php
if (isset($pageid) && is_numeric($pageid))
{
$display_page = mysql_query("SELECT html FROM community WHERE pageid=$pageid")
or die ("MySQL Error");
if (mysql_num_rows($display_page) > 0)
{
while ($row = mysql_fetch_array($display_page))
{
$page = $row["html"];
echo $page;
}
}
}
else
{
echo("OOX ERROR");
}
?>

The field that it outputs contains html mixed with php, such as this:
<table width="600" border="0" align="center" cellpadding="0" cellspacing="5">
<tr>
<td width="150" height="150" align="center" valign="middle">
<?php
$display_avatar = mysql_query("SELECT avatar FROM staff WHERE id=1")
or die ("MySQL Error");
while ($row = mysql_fetch_array($display_avatar))
{
$avatar = $row["avatar"];
echo $avatar;
}
?></td>
<td valign="top"><table width="100%" height="100%" border="0" cellpadding="2" cellspacing="0">
<tr>
<td><p>Name:
<?php
$display_name = mysql_query("SELECT name FROM staff WHERE id=1")
or die ("MySQL Error");
while ($row = mysql_fetch_array($display_name))
{
$name = $row["name"];
echo $name;
}
?><br>
Alias:<br>
Email:<br>
Position:<br>
Quote: <br>
</p>
</td>
</tr>
</table></td>
</tr>
</table>

When I test the page that is supposed to output (community.php?pageid=1), all I get is a page with none of the sql stuff showing up. Just the plain html (Name:, Alias:, etc.) What am I doing wrong?

tomhath

Echoing a string and running the code are different. Try the eval function, it should do what you want:

http://www.phpbuilder.com/manual/function.eval.php

Sharif

What else can I do?

tomhath

Relax for a minute, try not to think about what you expect your code to do (because it's not doing what you expect).

When you echo a string, php sends the string to the browser as part of the response; it makes no attempt to run any php code that's inside the string being echo'd.

So you need to force php to interpret the string as php code. The two most common ways to do this are to either "include" a file with some code in it, or put the code in a string and tell php to run what's in the string with the "eval" function.

Take a look at the manual entry for "include", it gives an example that would work for you; instead of storing the html in the database, you would have to store it in a file and store the name of the file in the database. Then your code would be almost the same as the example given here; instead of a "for" loop on the array of file names, you would have the "while" loop fetching them from the database.

http://www.phpbuilder.com/manual/function.include.php

The other way to do it is with the "eval" function, but I've always found code with lots of eval'd strings hard to get working right.

Try out "include" first.

Sharif

Okay, Using the eval() function, I made my code that is supposed to show the php code from the database field like this:
<?php
if (isset($pageid) && is_numeric($pageid))
{
$display_page = mysql_query("SELECT html FROM community WHERE pageid=$pageid")
or die ("MySQL Error");
while ($row = mysql_fetch_array($display_page))
{
$page = $row["html"];
eval ("\$page = \"$page\";");
echo $page;
}
}
else
{
echo("OOX ERROR");
}
?>

I get this error: Parse error: parse error, unexpected T_LNUMBER in community.php(21) : eval()'d code on line 1

tomhath

I assume the problem is that eval wants only php code, not a mix of php and HTML.

If you want to use eval, you'll probably have to use echo statements inside the string and also remove the <?php ?> tags.

One thing after another, you can't use echo, so you use eval, but to use eval you need to use echo. That's one reason I recommended include instead of eval... :mad: