Noobie issue. PHP else if statement

Question

I'm hitting my head to a wall... This is a fraction of my code:

$stmt = $db->prepare("SELECT vote,id FROM votes WHERE post_id = ? AND user_id = ? "); 
$stmt->bind_param('ii', $post_id, $user_id);
$stmt->execute(); 
$stmt->store_result();
$stmt->bind_result($result, $vote_id);
$stmt->fetch();
$num_rows = $stmt->num_rows;
$stmt->close();

if ($num_rows>=1)
{

if ($result!==$vote)   
{        
$stmt = $db->prepare("UPDATE votes SET vote = ? WHERE id=?"); 
$stmt->bind_param('ii', $vote, $vote_id);
$stmt->execute();
$stmt->close();
}

else if ($result==$vote) 
{
$stmt = $db->prepare("UPDATE votes SET vote = 0 WHERE id=?"); 
$stmt->bind_param('i', $vote_id);
$stmt->execute();
$stmt->close();
}    
}

No errors are shown, the script works just fine until it reaches ELSE IF part. It doesn't update although $vote value is the same as in a database.

you don't need an else **if** just an else, its either going to be match or not — user557846
– user557846, Commented Dec 4, 2013 at 23:23

scrowler · Accepted Answer · 2013-12-04 23:24:39Z

3

!== is a strict comparison operator, and this statement is saying that $result must not be identical (in type as well as value) to $vote. You don't need to be that specific for this.

if ($result!==$vote)

What you should be doing is this:

if ($result != $vote) // only one equal sign, not two
{
   ...
} else { // they are equal implicitly, no need to redefine the ==
    ...
}

answered Dec 4, 2013 at 23:24

scrowler

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6 Comments

Phil Over a year ago

+1 can only assume $vote is an integer whilst any result from bind_result / fetch will be a string

Saulius s Over a year ago

@Phil, yeah $vote is either 1 or -1... So the reason of this issue is different data types? And how should I make them both int?

scrowler Over a year ago

@SauliusSlavinskas Just use != instead, it automatically converts variable types and compares them.

Saulius s Over a year ago

@scrowler God bless ya! Just tried your suggestion and it works. Thank you so much!

Phil Over a year ago

@SauliusSlavinskas For completeness, you can also cast $result to an integer via (int) $result

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