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|Using a variable for an "if "condition|
is this possible?
| 10:03 pm on Feb 16, 2006 (gmt 0)|
Is there any way to have a condition for an if statement be evaluated according to its contents and not whether the variable is defined or not? In other words:
$a = 'no';
$the_condition = '$a == "yes"';
echo "$a is equal to yes";
echo "$a is not equal to yes";
This will always evaluate as true and print yes because $the_condition variable is defined.
Is there any syntax so that I can have it actually evaluate the condition within the variable and whether that is true or not?
| 10:32 pm on Feb 16, 2006 (gmt 0)|
Sounds like a job for switch [php.net].
| 10:50 pm on Feb 16, 2006 (gmt 0)|
A blast from the past ...
Variable condition statements [webmasterworld.com]
| 2:27 am on Feb 17, 2006 (gmt 0)|
Thanks very much, those posts helped me solve it using the eval function.
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