How to convert this to an IF statement

Manat

Hello All,

I am having problems with this script, and I was wondering if anyone would know if converting this to an IF statement would have the same effect?

$sql= "SELECT

CASE WHEN $winners > 40 THEN '" . noPrize() . " 😢'
WHEN MOD(($contestantID -10), 98) = 0 THEN '" . Dothis() . "'
WHEN MOD(($contestantID -10), 49) = 0 THEN '" . DoThat() . "'
ELSE '" . noPrize() . "'('
END AS DECISION";

If I made this an if statement, would it have the same effect?
And if so, how would I go about doing that?

Thx
MK

Manat

<?php
while ($winners <40)
{
if (($contestantID -10) % 98 = 0)
{ Dothis(); }
elseif (($contestantID -10) % 49 = 0)
{ Dothat(); }
else { doThis2(); }
}

Hi all,

This is my version of the IF statement of the Select.
Anyhow, I am getting a parse error at the line where it says:
<?php

So just <?php is giving me an error, so i'm totally unsure on how to tackle this since the error is not pointing at code.

Regarding the while $winners, below is the code on how I retrieve the winners variable.

<?php
$sql="select entrantID, count(entrantID) as winners
from tableName
where isWinner=1
group by entrantID";
$result = mysql_query($sql) or die(mysql_error());
?>

Can anyone help me on this?
Thx

stolzyboy

is there anything before the <?php that will cause an error, also when checking to see if something equals something else, use ==, two equals for equality, one equals when setting something equal to something

Manat

Hi,

Right, well I put the == so,

<?php
while ($winners < 40)
{
if (($contestantID -10) % 98 == 0)
{ Dothis(); }
elseif (($contestantID -10) % 49 == 0)
{ DoThis(); }
else { Dothat(); }
}

But I get a really weird error, the error is:
Unknown column 'contestantID' in 'field list'

What does that mean?

stolzyboy

that's an error from your sql statement

Manat

Hi,

Well i've combed my sql and I don't see any syntax errors.
I have one question, if I do a $HTTP_POST_VARS['age']; on the php page, and this isn't inside a function, then the $age variable is now automatically global right? Since it wasn't declared within a function, in PHP it would be as so, am I correct? So I don't have to pass any of these variables to other functions in the function call?

Also when you say error in my SQL statement, is this a syntax error or something else?

Thx

stolzyboy

it means it can't find that column name in the field list of your table

Manat

Well I don't know why it said contestandID doesn't exist, but I changed the column name to the same name, contestantID.

Now I have an error: Duplicate entry '1' for key 2

So i've done some research on what this means, and ways people solved it.

So what I did is, since contestandID was the primary key, i made a dummy column called commentID. Made commentID the primary key and an autonumber, (auto-increment), and changed contestantID to unique in the DB.

But I still get this problem. And what is key 2?

Manat

Duplicate entry '0' for key 2

Can anyone tell me what Key 2 is?
What causes the above error?

What I did is I put an echo functionName in each function.

And it echoed it twice, which means that the above duplicate key 0 for key 2 means it tries to insert into the same values twice.

<?php
if ($winners < 25)
{
if (($contestantID -10) % 98 == 0)
{ functionName(); }
elseif (($contestantID -10) % 49 == 0)
{ functionName(); }
else { functionName(); }
}
else { functionName(); }
}
?>

I also did a echo on $contestantID, and it echoes the right value, however the number 0(zero) is what goes into the DB.

This how I get contestandID

$sql = "SELECT contestantID FROM tableName";
$result = mysql_query($sql) or die(mysql_error());

$userID = $contestantID++;
$sql = "UPDATE tableName SET contestantID = $userID";

Why would my script cause the functioned to be called twice within the same script? If I define a value outside a function, I don't have to pass the variable to the function as a parameter, correct?
I would apreciate any help since I've been stuck at the same problem for some time now.

Thanks

Thx