Re: reporting errors when $ sign is missing in front of a variable

Thanks... Well I just changed the <?php error_reporting (E_ALL ^ E_NOTICE); ?> to <?php error_reporting (E_ALL ); ?> and that does it for me. Notice: Use of undefined constant my_age - assumed 'my_age' in D:\Hosting\5291100\html\blueprint\bp_library.php on line 40 my_age Now back in business :) Notice: Use of undefined constant my_age - assumed 'my_age' in D:\Hosting\5291100\html\blueprint\bp_library.php on line 40my_age On Wed, Jan 11, 2012 at 9:12 PM, Tommy Pham <tommyhp2@gmail.com> wrote: > On Wed, Jan 11, 2012 at 8:43 PM, Haluk Karamete <halukkaramete@gmail.com> wrote: >> >> Hi, I'm coming from ASP background. >> There, there is a life saver option called "option explicit". It >> forces you to declare your variables using the "dim" statement. The >> good thing about that is that if you were to mis-spell one of your >> variables, asp.dll throws an error stating that on line so and so, >> variable so and so not declared. This allows you to immediately fix >> the error saving lots of time. If you did not use "option explicit", >> then that misspelled variable would not have caused any error and you >> woud have spent much more time debugging your app as to what went >> wrong where. >> >> Now, I undersand with PHP, that we do not have a variable declaration >> per se; you put a $ sign in front of a word, and that becomes a >> variable. Since in asp, we do not use $ much. I keep forgetting that. >> I first declare a var and set a value for it using the $. But then I >> refer to the darned thing, without the $. And there are no errors. Ths >> behaviour seems extremely odd to me. >> >> How do I achieve the functionality that if I forget to use $ sign for >> a previously declared variable, php throws me an error. >> >> example >> >> $my_var = 90; >> echo my_var; >> >> I want an error to be thrown in line 2. what do I need to do?" >> I was assuming  that since there is no function titled "my_var", PHP >> would have complain right there and then. But instead, it simply >> echoes "my_var". >> >> I would have expected "my_var" to be outputted only if I were to write >> echo "my_var";. This beats me. >> >> At the top of my page, I already have this <?php error_reporting >> (E_ALL ^ E_NOTICE); ?> >> >> Haluk >> >> > > This works for me in development environment without a debugger setup > using a web browser (note that I'm using 5.4RC2 so the default > behavior of error_reporting(E_ALL) is different [1]: > > Notice: Use of undefined constant my_var - assumed 'my_var' in > F:\dev\sites\wwwroot\php_apps\test.php on line 5 > my_var <?php > error_reporting(E_ALL); > ini_set('display_errors', 'on'); > $my_var = 90; > echo my_var; > > highlight_file(__FILE__); > > Good luck, > Tommy > > [1] http://php.net/function.error-reporting