If variable equals value php [duplicate]

Question

I am trying to do a check before the data inserts into the MySQL query. Here is the code;

$userid = ($vbulletin->userinfo['userid']);
$sql3 = mysql_query("SELECT * FROM table WHERE ID='$_POST[hiddenID]'");

while ($row = mysql_fetch_array($sql3)){

$toon = $row['toonname'];
$laff = $row['tlaff'];
$type = $row['ttype'];

if ($type == 1){
$type == "Bear";
} elseif ($type == 2){
$type == "Cat";
} elseif ($type == 3){
$type == "Dog";
}            

}

However, this isn't working. Basically, there are different values in the 'table' for each type. 1 means Bear, 2 means Cat, and 3 means Dog.

Thanks to whomever can help see a problem in my script!

Answer 1

You are comparing, not assigning:

if ($type == 1){
  $type = "Bear"; 
}

You compare values with == or ===.

You assign values with =.

You could write less code to achieve the same result too, with a switch statement, or just a bunch of ifs without the elseifs.

if ($type == 1) $type = "Bear";
if ($type == 2) $type = "Cat";
if ($type == 3) $type = "Dog";

I would make a function for it, like this:

function get_species($type) {
    switch ($type):
        case 1: return 'Bear';
        case 2: return 'Cat';
        case 3: return 'Dog';
       default: return 'Jeff Atwood';
    endswitch;
}

$type = get_species($row['ttype']);

Answer 2

You are using == instead of =. It compares the variable to the new value. Use = to set the value.

if ($type == 1){
$type = "Bear";
} elseif ($type == 2){
$type = "Cat";
} elseif ($type == 3){
$type = "Dog";
}