Undefined variable when trying to print variable into HTML

Question

PHP

if($count == 1) {
     session_register("username");
     $_SESSION['username'] = $username;

     header("location: ../home");
  }else {
     $error = "Your Login Name or Password is invalid";
  }

HTML -- line 43

<?php echo "<p class='text-danger'>$error</p>"; ?>

Notice: Undefined variable: error in directory on line 43

`session_register` REALLY!!!!! How 1990's What version of PHP are you using?? — RiggsFolly, Jan 21 '17 at 12:19
in .html extension file you can't do anything related to php. i hope that it's not your case and also define `$error='';`before `if` — Alive to die - Anant, Jan 21 '17 at 12:19
`session_register()` **Warning** This function has been DEPRECATED as of PHP 5.3.0 and REMOVED as of PHP 5.4.0. And we are currently on PHP7.1 — RiggsFolly, Jan 21 '17 at 12:28
If you dont pass throught the ELSE `$error` will not exist, initialize it above the IF to a default value — RiggsFolly, Jan 21 '17 at 12:29
I still get the same error message even after declaring the $error variable before if statement — Mark, Jan 21 '17 at 12:50

score 2 · Answer 1 · answered Jan 21 '17 at 12:29

Error occurred because, when if statement is true then only under if statement block code would be executed not else block code. So $error variable did not initialized. Each variable should be initialized if you want to use it. $error variable did not initialize but you use it in HTML-- line 43. Thanks

  $error ="";
  if($count == 1) {
     session_register("username");
     $_SESSION['username'] = $username;

     header("location: ../home");
  }else {
     $error = "Your Login Name or Password is invalid";
  }

I defined the variable before if statement, same error – Mark Jan 21 '17 at 12:52 — Mark, Jan 21 '17 at 12:52

score 0 · Answer 2 · answered Jan 21 '17 at 12:29

0

$error = '';

if($count == 1) {
     session_register("username");
     $_SESSION['username'] = $username;

     header("location: ../home");
  }else {
     $error = "Your Login Name or Password is invalid";
  }

answered Jan 21 '17 at 12:29

Vinod Kirte

189
1
6

2

Why should the OP "use this code"? A **good answer** will always have an explanation of what was done and why it was done that way, not only for the OP but for future visitors to SO that may find this question and be reading your answer. – RiggsFolly Jan 21 '17 at 12:30
I defined the variable before if statement, same error – Mark Jan 21 '17 at 12:52

Undefined variable when trying to print variable into HTML

2 Answers2