该错误的解决方案是什么？

Question

if($placed != true){    
  $_SESSION["eventid"][]   = "$r[id]";
  $_SESSION["selection"][] = "$selection";
  $_SESSION["title"][] = "$r[hometeam] - $r[awayteam]";

And the error is:

Warning: Cannot use a scalar value as an array in /home2/**/bet/add_bet.php on line 54
Fatal error: [] operator not supported for strings in /home2/**/bet/add_bet.php on line 55

I know it has to do with an array; but which would be the solution here??? Im confused!

It's not like i can type $_SESSION["eventid"][] = array();

FULL CODE CAN BE SEEN HERE

Answer 1

If you var_dump your session variables, you will see that you have defined $_SESSION['eventid'] et al. as strings somewhere in your code. As such, treating them as arrays will fail.

You will need to initialise your session variables to arrays explicitly.

Answer 2

You shouldnt use double quotes, also you need youse single quotes in $r , also use concatenation with single quotes:

     $_SESSION["eventid"][]   = $r['id'];
     $_SESSION["selection"][] = $selection;
     $_SESSION["title"][] = $r['hometeam'] .' - ' . $r['awayteam'];

I think you want something like this:

    $_SESSION["eventid"] = $r['id'];
    $_SESSION["selection"] = $selection;
    $_SESSION["title"] = $r['hometeam'] .' - ' . $r['awayteam'];