12dstring Posted August 30, 2007 Share Posted August 30, 2007 I have a list of 12 variables, like so $d1=31; $d2-28; $d3=31; $d4=30 .... $d12=31; and another one called $month which is a numerical value from 1-12 (calculated seperately) What I need is a variable, $dmonth, which is the value of $d"$month" For example if $month=1 then $dmonth=$d1 (=31) and likewise if $month=12, then $dmonth=$d12 I can't for the life of me figure out how to code a variable as part of a defined variable, so any help would be much appreciated. Thanks in advance, Dave Quote Link to comment Share on other sites More sharing options...
Jessica Posted August 30, 2007 Share Posted August 30, 2007 $dmonth = $d.$month; Quote Link to comment Share on other sites More sharing options...
akitchin Posted August 30, 2007 Share Posted August 30, 2007 not sure jesirose understood your question: $varname = 'd'.$month; echo 'days in this month: '.$$varname; it might be smarter to use an array and simply define the indeces by month: $d[1] = 31; $d[2] = 28; ... then to access the number of days, all you have to do is: echo 'days in this month: '.$d[$month]; Quote Link to comment Share on other sites More sharing options...
Jessica Posted August 30, 2007 Share Posted August 30, 2007 When I read it the first time I read it the way you saw, then I read it again and thought it was different. Gah. *headdesk* I had it right Quote Link to comment Share on other sites More sharing options...
12dstring Posted August 30, 2007 Author Share Posted August 30, 2007 The array route sorted it a treat. Thanks for the super-quick help guys Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.