CODEWITHSUNDEEP
phpstringvariablesauthenticationoutput
Jordy144
Jordy144

Reputation: 91

PHP Output is a string instead of the string from another variable

I am making for myself a script that checks how many mH/hash of my miners on mining pool websites are. Therefor I have a bunch of variables with $user and $pass for each site. Example:

   $user1 = 'A';
   $pass1 = 'B'

   $user2 = 'C';
   $pass2 = 'D';

So I made a PHP curl that login the form and checks the currently mH/hash. Also I've made a counter, to rise the variable user1 and pass1 up by +1.

   $count = 0;
   $count = $count + 1;
   $user = "$user$count";   
   echo $user;

But the problem is my output at the login form:

  $user2

Should be:

  C

I hope it is clear what I am looking for and thanks in advance.

Upvotes: 1

Views: 69

Answers (4)

dtengeri
dtengeri

Reputation: 937

I think it would be a better approach to specify your user and password pairs in an array:

$login_info = array();
$login_info[] = array(
  'name' => 'A',
  'pass' => 'B',
);
$login_info[] = array(
  'name' => 'C',
  'pass' => 'D',
);

And you can travers the array by foreach:

foreach ($login_info as $info) {
  $name = $info['name'];
  $pass = $info['pass'];
  ...
}

You can easily add new elements to the array without creating new variables.

Upvotes: 0

Snowburnt
Snowburnt

Reputation: 6932

Use an Array:

$user = [];
$pass = [];

$user[1] = 'A';
$pass[1] = 'B'

$user[2] = 'C';
$pass[2] = 'D';

$count = 0;
$count = $count + 1;
echo $user[$count];

Upvotes: 0

rjmunro
rjmunro

Reputation: 28076

Use an array. Start with something like:

$users = array(
    array('A', 'B'),
    array('C', 'D')
);

foreach ($users as $user) {
    echo "Username: ".$user[0]. "\n";
    echo "Password: ".$user[1]. "\n";
}

Upvotes: 3

MTranchant
MTranchant

Reputation: 485

Var concatenation is not very "recommended", but...

$user = ${'user'.$count}

You should consider using an array instead.

Upvotes: 3

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