PHP IF Else Function Not Working

Question

I have a problem, I can't get the result of my code. When I try to debug the code. I using IF statement in my code.

It's just simple, example : if kelas(based on user_id login) = 1, then redirect it to kelas1.php, if kelas = 2, the redirect to kelas 2, else no have a kelas.

Here it's my code :

<?php
session_start();

if(!isset($_SESSION['user_id'])) {      
    header('Location: form_login_siswa.html');
}
$user_id = $_SESSION['user_id'];
include ("config.php");
$query = "SELECT kelas FROM t_siswa WHERE user_id = '$user_id'";
$hasil = mysql_query($query);

while ($data = mysql_fetch_array($hasil)) {
    $kelas = $data['kelas'];

    if($kelas = 1) {
        include ("kelas1.php");
    }
    if($kelas = 2) {
        include ("kelas2.php");
    } else {
        echo "Tidak ada kelas";
    }
}
?>

Anyone, please help to solve the problem. Appreciated with your helps.

Thank you.

You should probably start by reading the language reference portion of the PHP manual. If you're not comfortable using conditionals yet, then it's very likely that there are other fundamentals of the PHP language you're also insufficiently versed in, and this will seriously impede any other efforts you make to advance your knowledge if you don't have your foundations set. Also, format your code neatly. Find a standard coding style and follow it. Otherwise, it's going to make debugging problems in your code very difficult and introduce errors. — Lèse majesté
– Lèse majesté, Commented Jul 6, 2012 at 14:58

sikander · Accepted Answer · 2012-07-06 14:46:03Z

4

You're missing an else and also not using the correct operator for comparison.

Also, switch around the comparison so that the first value is always a valid value.

if(1 == $kelas)
{
    include ("kelas1.php");
}
else if(2 == $kelas)
{
    include ("kelas2.php");
}
else
{
    echo "Tidak ada kelas";
}

answered Jul 6, 2012 at 14:46

sikander

2,28616 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

miszczu Over a year ago

or use switch and case, the same effect

Shawn Over a year ago

Switching the comparison is not necessary. It only prevents you from making an error because 1 = $kelas will throw an error.

miszczu Over a year ago

yea, but i mean if you have 1 variable and many options, you can use switch and case to check all values rather than elseif or else if. But both ways give the same effect.

user1473757 Over a year ago

I've tried the code, but still not working... when I debug the code, nothing change.

sikander Over a year ago

what does "not working" mean? Is your code printing "Tidak ada kelas" all the time? Check what is in the $data variable if the results are not what you're expecting them to be.

|

Mark Baker · Accepted Answer · 2012-07-06 14:45:35Z

2

$kelas = 1

is assignment

$kelas == 1

is comparison

answered Jul 6, 2012 at 14:45

Mark Baker

213k34 gold badges354 silver badges390 bronze badges

Comments

Matt Gibson · Accepted Answer · 2012-07-06 14:45:51Z

1

if($kelas = 1)

should be

if($kelas == 1)

answered Jul 6, 2012 at 14:45

Matt Gibson

15k8 gold badges51 silver badges80 bronze badges

Comments

Fluffeh · Accepted Answer · 2012-07-07 08:00:33Z

1

Try using == instead of = . The "=" is an assignment operator and the expression gets the value of the right operand.

if($kelas == 1)
{
    include ("kelas1.php");
}
if($kelas == 2)
{
    include ("kelas2.php");
}
else
{
    echo "Tidak ada kelas";
}

Also, the statement will always echo "Tidak ada kelas"; if ($kelas == 1). Was this intentional?

edited Jul 7, 2012 at 8:00

Fluffeh

33.6k16 gold badges69 silver badges80 bronze badges

answered Jul 6, 2012 at 14:46

Razvan

10.2k6 gold badges41 silver badges51 bronze badges

Comments

Bruce · Accepted Answer · 2012-07-06 14:45:46Z

0

In your if statements using == for testing comparison rather than single = which is for assignment

answered Jul 6, 2012 at 14:45

Bruce

1,53813 silver badges17 bronze badges

Comments

imnotquitejack · Accepted Answer · 2012-07-06 14:51:03Z

0

Take a look at the elsif structure.

while ($data = mysql_fetch_array($hasil)) {
  $kelas = $data['kelas'];

  if($kelas == 1) {
    include ("kelas1.php");
  } elseif($kelas == 2) {
    include ("kelas2.php");
  } else {
    echo "Tidak ada kelas";
  }
}

Or the switch structure:

while ($data = mysql_fetch_array($hasil)) {
  switch($data['kelas']){
    case 1:
      include ("kelas1.php");
      break;
    case 2:
      include ("kelas2.php");
      break;
    default:
    echo "Tidak ada kelas";
  }
}

answered Jul 6, 2012 at 14:51

imnotquitejack

916 bronze badges

1 Comment

Rakesh Shetty Over a year ago

can you check by echoing in each condition,just to check in which condition it is going ? @user1473757

Collectives™ on Stack Overflow

PHP IF Else Function Not Working

6 Answers 6

6 Comments

Comments

Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

6 Comments

Comments

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related