Variable content variables?

vr4lyfe

Ok. First of all, I am not even sure the thread name is an accurate description of my problem, but here goes:

I have a field within my MySQL db that contains a php variables.

The field is "text text text $phpvariable text text text";

I have a PHP scripts that query for that field.

What I want is when the script query for the field, the script can set, or change the value that $phpvariable is equal to.

so far, I can set it if i set $phpvariable = "something" BEFORE i query the database, but I cant set it AFTER.

Am I explainign the problem clearly?

Well, I hope so, any help is appreciated.

Thanks.

vr4lyfe

I decided to post an example just to help clarify:

Mysql field = "text text text $phpvar text text text";

if script is :

<?PHP
//connect to db
//build query string and query for mysql field
// set $textstring as query result

   //then set $phpvar;
   $phpvar = "something";

?>
[/I]
then the output of $textstring would be "text text text text text text"

if script is :

<?PHP
// first set $phpvar;
$phpvar = "something";

   // then connect to db
   //build query string and query for mysql field
   // set $textstring as query result

?>[/I]

then the output of $textstring would be "text text text something text text text"

What i want to achieve is:

<?PHP
//connect to db
//build query string and query for mysql field
// set $textstring as query result

   //set $phpvar as 'something'
   //print $textstring would output 'text text text something text text text'

   //then set $phpvar as 'something else'
   //print $textstring would output 'text text text something else  text text text'

?>[/I]

swr

You'll need to use [man]eval[/man] for this.

You kind of glossed over the details, but what you've described shouldn't work at all. Unless you are using eval, pulling a string from a database and echoing it will NOT cause variables to be evaluated. I think you are misunderstanding what is happening.

vr4lyfe

thanks for the reply.

Eval seems like my best bet, but I have been getting an error I cant debug.

my code is:

$left_stylesheet = '
<div style=\'padding:5\'>
echo $content
</div>
';

$content = 'akjdflkjsadfasdf';

eval("$left_stylesheet;");

[/i]
error is:

Parse error: parse error, unexpected '<' in C:\Program Files\Apache Group\Apache2\htdocs\homepage.php(54) : eval()'d code on line 2

Do you see a problem with my code?

Please let me know.

Thanks.

Simon_Hartley

$left_stylesheet = '<div style=\'padding:5\'>
echo $content
</div>';

I see a problem with this part of your code. You need to use the concantenation operator (.) as demonstrated below:

$left_stylesheet = '<div style=\'padding:5\'>.$content.</div>';

swr

Eval starts out in PHP mode. The HTML tags aren't valid PHP code, so you get an error.

Try this instead:

eval("?>$left_stylesheet<?php");

Also, sticking this:

<div style='padding:5'>
echo $content
</div>

Into a web page will display "echo $content" on the page, and doing it with eval is no different. You need PHP tags around the echo statement.

vr4lyfe

thanks alot guys .. you guys are always a great help to when I need.

sorry it took me so long to reply. I was working another problem also.

Who knew even function calls can be "dynamic"?

If you have

functionNamedBlah()

you can call it with this code:

$function_name = "functionNamedBlah";

$function_name();

and it will call your functionNamedBlah();

LOL.