RE: [PHP] No cookie is being set?

--- Larry Brown <larry.brown@dimensionnetworks.com> wrote: > when I find a function doesn't work right and I am using variables > in its execution I run an echo/die combination immediately before > the function to verify the data being fed to it is as it should be. This is a very good suggestion, although your example might be confusing to some people due to all of the quote escaping. Basically, Larry is suggesting that you make your code appear in the browser with each of the function parameters evaluated: myfunc(foo, 1 + 2); Your code may have: myfunc($var1, $var2 + $var3); So you might try this for debugging: echo "myfunc($var1, $var2 + $var3);"; If this looks right, you can take it a step further to reveal the arithmetic: echo "myfunc($var1, " . $var2 + $var3 . ');'; Hope that helps. Chris ===== Chris Shiflett - http://shiflett.org/ PHP Security Handbook Coming mid-2004 HTTP Developer's Handbook http://httphandbook.org/