output from mysql

[Anon]

Hello,

I have a mysql variable e.g. $sql[text] .
In this variable could be:

/-- start
hello friend,
your name is <?echo("$name");?>
yours boy.
/-- stop

but I need to print this variable out with

echo("$sql[text]");

It works ok, but all php-codes are gone.

Php have to think, that $sql[text] is a php-file, how do it that?

//Ulrik Q K
Hope you understand...

[stereo]

I think I understand.....

Your selecting info from a table, extracting it, and then displaying it, yes?

If that's the case, an example would be below......

$result = mysql_query("SELECT (blah) FROM (blah) WHERE (blah) = (blah);");

if(mysql_num_rows($result)) {

$r = mysql_fetch_array($result);

$Blah = $r[(blah)]; // Blah could be name (without the bracets)

echo $Blah;

Or I think you can echo it like this....

echo "$Blah = $r[(blah)];

Not too sure about that one.

Hope I helped,
~~Stereo

[dstockto]

If you do a view source on your page, I bet your PHP codes are still there, but you can't see them because they are not being run. You should probably use the eval() function to get your PHP code to run if you are grabbing it from a database. I asked a similar question a few days ago, but over in the database support area.

http://www.phpbuilder.com/forum/read.php3?num=5&id=10603&loc=0&thread=10603

The manual page for the command is at
http://www.php.net/manual/en/function.eval.php

[Anon]

Thanks!