Hi,

Friday, March 19, 2004, 12:44:40 AM, you wrote:
F> I would be very interested in a solution to this problem because I have a similar 
one.

F> I stored tables on a page.

F> examp:

F> $table_1 = "<table><tr><td>".$name."</td></tr></table>";
F> $table_2 = "<table><tr><td>".$message."</td></tr></table>";
F> etc...

F> and included the page and used the variables (echo $table_1;)

F> The table displayed as it should,
F> however the variables didn't have values( for $name, $message, etc..)

F> I began testing for the problem by defining the string variable
F> for the table on the same page as the php code

F> examp:

F> <?php
F> $table_1 = "<table><tr><td>".$name."</td></tr></table>";
F> $table_2 = "<table><tr><td>".$message."</td></tr></table>";

F> $name = "blah blah";

F> if(blah blah){
F>     echo $table_1;}
F> else{
F>     echo $table_2;}
?>>

F> Again, the tables printed fine, but the variables ($name, $message) had no values.


F> Most recently, I put the whole dam table in each conditional option.

F>     if (blah blah){  echo
F> "<table><tr><td>".$name."</td></tr></table>";}

F>     else{  echo  "<table><tr><td>".$message."</td></tr></table>";}


F> Voila, the variables have values.

F> It would seem that if a solution was found for Jake's problem
F> it would also solve, or provide a key for solving, this one.


You have to define variable $name before you try to use it.
The only other way would be to eval the string but that is not the way
to go :)

-- 
regards,
Tom

-- 
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Reply via email to