CODEWITHSUNDEEP
phpif-statement
Goulash
Goulash

Reputation: 3838

Confusing result with if / else statement

this is some of my script so far:

$check = ("SELECT username FROM users WHERE username = '$us3r'");
$check2 = ("SELECT password FROM users WHERE username = '$us3r'");
$check3 = ("SELECT userID FROM users WHERE username = '$us3r'");
$check4 = ("SELECT userrole FROM users WHERE username = '$us3r'");

//Role

$role = mysql_query($check4);
$arr5 = mysql_fetch_row($role);
$roles = ($arr5[0]);

echo $roles;

if($roles = 1) {

    //Username

    $results3 = mysql_query($check);
    $arr2 = mysql_fetch_row($results3);
    $results4 = ($arr2[0]);

    //Password

    $results5 = mysql_query($check2);
    $arr3 = mysql_fetch_row($results5);
    $results6 = ($arr3[0]);

    //UID

    $id1 = mysql_query($check3);
    $arr4 = mysql_fetch_row($id1);
    $id = ($arr4[0]);

    echo 1;

}

else if($roles = 2) {

    //Username

    $mresults3 = mysql_query($check);
    $marr2 = mysql_fetch_row($mresults3);
    $mresults4 = ($marr2[0]);

    //Password

    $mresults5 = mysql_query($check2);
    $marr3 = mysql_fetch_row($mresults5);
    $mresults6 = ($arr3[0]);

    //UID

    $mid1 = mysql_query($check3);
    $marr4 = mysql_fetch_row($mid1);
    $mid = ($marr4[0]);

    echo 2;

};

However there is something wrong with my if / else if for some reason the echo shows 21 when I use a user with a userrole of 2, I want it to be either 11 or 22 :/

Upvotes: 0

Views: 172

Answers (7)

Your Common Sense
Your Common Sense

Reputation: 157839

to let you know, your whole code makes absolutely no sense.
instead of 2 pages it have to be of just a few lines.

$us3r = mysql_real_escape_string($us3r);
$sql  = "SELECT * FROM users WHERE username = '$us3r'"; 
$res  = mysql_query($sql); 
$userdata = mysql_fetch_assoc($role); 
// now you have all user's data in the $userdata array.
// echo $userdata['username'];  for example will echo a username
// no need for the separate queries, no need to write the same code twice.

// and, finally
echo $userdata['userrole'];
//it seems the only thing your code does - echoing the actual userrole value:

Upvotes: 0

David W.
David W.

Reputation: 107040

Hmmm...

I'm not a big PHP programmer, but the if statement not working caught my attention:

$roles = ($arr5[0]);
if ($roles = 2) {

is not doing what you think. That's a set value equals sign. What you're doing is setting the value of $roles to 2.

So, why would you want to set something in your if statement?

You can then do things like this:

if (($roles = $arr5[0]) == 2) {

Here, I'm setting the value of $roles and checking the value at the same time.

In most modern programming languages, you use a doubled up equals sign to test for equality:

if ($roles == 2) {

PHP also has a triple equal sign which can test not only for equality, but sameness.

if ($a === $b) {

Not only are $a and $b equal to each other, but they're also the same type.

Upvotes: 0

Brendan Long
Brendan Long

Reputation: 54242

So is there any reason you can't just do this (no if/else)?

$check = ("SELECT username FROM users WHERE username = '$us3r'");
$check2 = ("SELECT password FROM users WHERE username = '$us3r'");
$check3 = ("SELECT userID FROM users WHERE username = '$us3r'");
$check4 = ("SELECT userrole FROM users WHERE username = '$us3r'");

//Role

$role = mysql_query($check4);
$arr5 = mysql_fetch_row($role);
$roles = ($arr5[0]);

echo $roles;

//Username

$results3 = mysql_query($check);
$arr2 = mysql_fetch_row($results3);
$results4 = ($arr2[0]);

//Password

$results5 = mysql_query($check2);
$arr3 = mysql_fetch_row($results5);
$results6 = ($arr3[0]);

//UID

$id1 = mysql_query($check3);
$arr4 = mysql_fetch_row($id1);
$id = ($arr4[0]);

echo $roles;

Upvotes: 0

Teofilo Israel Vizcaino Rodrig
Teofilo Israel Vizcaino Rodrig

Reputation: 1480

Change 

if($roles = 1)

by

if($roles == 1)

same for 

else if($roles = 2)

Upvotes: 0

Marc B
Marc B

Reputation: 360592

Typos everywhere:

if($roles = 1) {

should be

if($roles == 1) {

The first one is doing an assignment, so the if() is doing an asignment. The new version is doing a comparison instead, and can potentially evaluate to false.

Upvotes: 1

Clive
Clive

Reputation: 36957

You're setting the value of $roles instead of checking for equality. Try changing your code to:

if($roles == 1) {
  ...
}
else if ($roles == 2) {
  ...
}

Upvotes: 1

Daniel
Daniel

Reputation: 31569

You need to use == for comparison instead of =: Change

if($roles = 1)

to

if($roles == 1)

and

else if($roles = 2)

to

else if($roles == 2)

If you use assignment (=) instead of comparison (==) it will not only evaluate to true, but it also will change the variable.

Upvotes: 4

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