First of all, it sounds like a boolean (based on the name of the variable), so either choose a less misleading name or just do
if (variable).
I think you're trying to check the value of a variable whose name you don't know beforehand? This should work:
PHP Code:
temp.varName = "foo";
echo(clientr.(@ temp.varName));
I've always used an @ there and thought it was unnecessary, but a quick test reveals that apparently it
is necessary.