PHP - conditional statement not working properly

Question

I'm working in PHP and JSON at API for my mobile application. I tried to write a registration module, but a part of my conditional statements don't work as expected.

If statement 1 :

if(!isset($_GET['username']) || !isset($_GET['password']) || !isset($_GET['imei']) || !isset($_GET['imie']) || !isset($_GET['nazwisko']) || !isset($_GET['email']) || !isset($_GET['zgoda']) || !isset($_GET['telefon']) || !isset($_GET['zgoda2']) || !isset($_GET['kraj']));
 {
  $returning = array('error' => 'Invalid query');
  echo json_encode($returning);
  break;
 }

It should give an error, when there is an argument missing, but it is giving an error always.

My query :

username=konrad12&password=xxx&imei=000000000000000&nazwisko=Potter&imie=Ronald&[email protected]&zgoda=1&telefon=000&zgoda2=1&kraj=Poland

If statement 2 :

 if(strlen($c) != 15 || !validEmail($f) || strlen($g) != 1 || strlen($i) != 1 || wez_id_kraju($j) == 0)
 {
  $returning = array('error' => 'Invalid query');
  echo json_encode($returning);
  break;
 }

It should give an error, when var values are incorrect, but it is giving an error always.

My variables :

 $z = mysql_real_escape_string($_GET['username']);
 $b = mysql_real_escape_string($_GET['password']);
 $c = mysql_real_escape_string($_GET['imei']);
 $d = mysql_real_escape_string($_GET['nazwisko']);
 $e = mysql_real_escape_string($_GET['imie']);
 $f = mysql_real_escape_string($_GET['email']);
 $g = mysql_real_escape_string($_GET['zgoda']);
 $h = mysql_real_escape_string($_GET['telefon']);
 $i = mysql_real_escape_string($_GET['zgoda2']);
 $j = mysql_real_escape_string($_GET['kraj']);

If statement 3 :

 if($g != 0 or 1 || $i != 0 or 1)
 {
  $returning = array('error' => 'Invalid query');
  echo json_encode($returning);
  break;
 }

It should give an error, when value of $g or $i isn't 1 or 0, but it is giving an error always.

Please help me, I tried a lot of things, but I can't find a solution

@Edit :

My valid email function :

    function validEmail($email)
    {
       $isValid = true;
       $atIndex = strrpos($email, "@");
       if (is_bool($atIndex) && !$atIndex)
       {
          $isValid = false;
       }
       else
       {
          $domain = substr($email, $atIndex+1);
          $local = substr($email, 0, $atIndex);
          $localLen = strlen($local);
          $domainLen = strlen($domain);
          if ($localLen < 1 || $localLen > 64)
          {
             // local part length exceeded
             $isValid = false;
          }
          else if ($domainLen < 1 || $domainLen > 255)
          {
             // domain part length exceeded
             $isValid = false;
          }
          else if ($local[0] == '.' || $local[$localLen-1] == '.')
          {
             // local part starts or ends with '.'
             $isValid = false;
          }
          else if (preg_match('/\\.\\./', $local))
          {
             // local part has two consecutive dots
             $isValid = false;
          }
          else if (!preg_match('/^[A-Za-z0-9\\-\\.]+$/', $domain))
          {
             // character not valid in domain part
             $isValid = false;
          }
          else if (preg_match('/\\.\\./', $domain))
          {
             // domain part has two consecutive dots
             $isValid = false;
          }
          else if
    (!preg_match('/^(\\\\.|[A-Za-z0-9!#%&`_=\\/$\'*+?^{}|~.-])+$/',
                     str_replace("\\\\","",$local)))
          {
             // character not valid in local part unless 
             // local part is quoted
             if (!preg_match('/^"(\\\\"|[^"])+"$/',
                 str_replace("\\\\","",$local)))
             {
                $isValid = false;
             }
          }
          if ($isValid && !(checkdnsrr($domain,"MX") || 
     ↪checkdnsrr($domain,"A")))
          {
             // domain not found in DNS
             $isValid = false;
          }
// I add that text ...
      mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD);
      mysql_select_db(DB_BASE);
      $q = "SELECT * FROM `system_domeny`";
      $a = mysql_query($q);
      while($wynik = mysql_fetch_array($a))
      {
       if($domena == $wynik[1]) $isValid = false;
      }
// ...
       }
       return $isValid;
    }

Why not make your first statement more informative by splitting them up and making the error message announce which value it claims isn't set? 'Invalid query' isn't too helpful, I'd far rather have 'No username detected!' etc... — Tro
– Tro, Commented Jul 19, 2013 at 11:57
The last condition seems to me like if ($g != 0 or true || $i != 0 or true) which is unconditionally true — Ivan Yonkov
– Ivan Yonkov, Commented Jul 19, 2013 at 11:57
you use isset($_GET['imei']) but send parameter imie=Ronald; spelling error? — Ygg
– Ygg, Commented Jul 19, 2013 at 11:58
To make that first query a bit more readable, just create an array of values to check for inside of $_GET and use !in_array() instead of repeating !isset() — billyonecan
– billyonecan, Commented Jul 19, 2013 at 12:00

Egg Vans · Accepted Answer · 2013-08-12 15:43:31Z

3

remove the semicolon after the last bracket of the if.

first statement ends

 || !isset($_GET['kraj']));

I think the second statement looks ok, is it an issue with the email validation function or the other function in the last check.

the last statement should be something like

if(($g != 0 && $g != 1) || ($i != 0 && $i != 1))

edited Aug 12, 2013 at 15:43

answered Jul 19, 2013 at 11:57

Egg Vans

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1 Comment

Egg Vans Over a year ago

Try removing !validEmail($f) then wez_id_kraju($j) in turn to see if either of them is the issue

user1646111 · Accepted Answer · 2013-07-19 12:22:23Z

3

first change first if to:

 $gets = array('username', 'password', 'imei', 'imie', 'nazwisko', 'email', 'zgoda','telefon','zgoda2','kraj');

 $er = 0;
 foreach($gets as $get){
  if(!isset($_GET[$get])){
   $er++;
   $error[] = $get;
  }
 }

 if($er > 0){
  $returning = array('error' => 'Invalid query, please fill these parameters: ['.implode(", ", $error).']');
  echo json_encode($returning);
  //break is not like exit, there is no function of break inside if!
  exit;
 }

if three, as answered by others also, change it to:

 if( ($g != 0 && $g != 1) || ($i != 0 && $i != 1))
 {
  $returning = array('error' => 'Invalid query');
  echo json_encode($returning);
  exit;
 }

answered Jul 19, 2013 at 12:22

user1646111

Comments

Community · Accepted Answer · 2017-05-23 11:49:13Z

2

This statement

if($g != 0 or 1 || $i != 0 or 1)

is definitely not when value of $g or $i isn't 1 or 0. There are various solutions, this is what you could do just using logical operators (split out over several lines and decorated with a generous helping of parentheses for readability):

if ( 
     ( ( $g != 0) && ( $g != 1) ) 
     || 
     ( ( $i != 0) && ( $i != 1) ) 
   )

Also note that or and || have different precedence which can lead to quite puzzling situations. For simplicity sake it's better to stick to || (and &&). Read this SO question for more info about the difference between || and or

edited May 23, 2017 at 11:49

CommunityBot

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answered Jul 19, 2013 at 12:09

fvu

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Comments

Hamender · Accepted Answer · 2013-07-19 12:11:13Z

1

make your json array like this:

 $returning=array();
$str = array('error' => 'Invalid query'); 
array_push($returning,$str); 
echo "{\"response\":".json_encode($returning)."}";

and remove terminator(;) after while condition from first statement, and try this for third statement:

if($g != 0 ||$g != 1 || $i != 0 ||$i != 1)
 {
  $returning = array('error' => 'Invalid query');
  echo json_encode($returning);
  break;
 }

answered Jul 19, 2013 at 12:11

Hamender

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1 Comment

fvu Over a year ago

Your third statement is not correct, it will always return true as neither $g nor $i can be 0 or 1 at the same time.

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