Cms "pagetype" Problem.

[White_Lily]

Hi, I am currently developing a cms where people can edit their site however they please.

 

I have just a small problem, within the database the table "pages" has a column named "pageType" the types range from 1-4. in my CMS i have a while that is supposed to check if the pageType is equal to 1 or 4, however my problem is, is its displaying pages that have a type that equals 2, which isnt supposed to happen.

 

<?php

echo "<h2>Default Pages</h2>";
$newVar = select("*", "pages", "pageType = 1 OR 4");
$num = mysql_num_rows($newVar);

if($num != 0){
while($fetch = mysql_fetch_assoc($newVar)){
if($fetch["pageType"] == 1 OR 4){
if($fetch["active"] == 1){
echo "<div class='active'><img src='images/tick.png' /><a href='edit/edit-default-pages.php?name=".$fetch["name"]."'>".$fetch["name"]."</a></div>";
}else{
echo "<div class='non-active'><img src='images/cross.png' /><a href='edit/edit-default-pages.php?name=".$fetch["name"]."'>".$fetch["name"]."</a></div>";
}
}
}
}else{
echo "<p>No Default Pages found.</p>";
}

?>

 

The page can be seen at: http://www.janedealsart.co.uk/template/cms/

Edited September 28, 2012 by White_Lily

[darkfreaks]

what happens when you do

 

elseif($fetch['pageType'] == 2 OR 3)
{ echo "NO default page selected";}

Edited September 28, 2012 by darkfreaks

[White_Lily]

Nothing changes... it then displaying all types, same as "1 or 4" displays 2 and 3 as well. "2 or 3" displays 1 and 4 as well as 2 and 3

[darkfreaks]

have you tried the identical operator

===

 

if($fetch['pageType'] === 1 OR 4) { //***DO Stuff**//
}else { echo 'no default pages found.'; }

Edited September 28, 2012 by darkfreaks

[White_Lily]

Stays the same even with ===

[White_Lily]

i changed one of the pages to type 11, and it was still being displayed

[darkfreaks]

is select a custom function can we see the code for that???

[White_Lily]

<?php
/***************************
*Select data from database
***************************/
function select($amount = "*", $table, $where = NULL, $order = NULL, $limit = NULL){
$query = "SELECT ".$amount." FROM ".$table;
if($where != NULL){
$query .= " WHERE ".$where;
}
if($order != NULL){
$query .= " ORDER BY ".$order;
}
if($limit != NULL){
$query .= " LIMIT ".$limit;
}

$result = mysql_query($query);

if($result){
return $result;
}else{
return false;
}
}
?>

 

Not sure the error is in this function as all the other pages using this function are working properly.

Edited September 28, 2012 by White_Lily

[darkfreaks]

cleaned up the function abit not sure if it will help or not.

<?php
function select($amount = "*", $table, $where = NULL, $order = NULL, $limit = NULL){
$query = "SELECT ".$amount." FROM ".$table;
if($where !== NULL){
$query .= " WHERE ".$where;
}
if($order !== NULL){
$query .= " ORDER BY ".$order;
}
if($limit !== NULL){
$query .= " LIMIT ".$limit;
}

$result = mysql_query($query);

if($result!==FALSE){
return $result;
}else{
return false;
}
}
?>

Edited September 28, 2012 by darkfreaks

[White_Lily]

No difference. Im not sure where the problem is :/

[darkfreaks]

print_r($newvar); what does it output?

[White_Lily]

print_r($newVar); outputs: Resource id #7

[Barand]

if($fetch["pageType"] == 1 OR 4)

NO!

 

if( ($fetch["pageType"] == 1) OR ($fetch["pagetype"] == 4) )

Yes

[White_Lily]

O_O

 

Didn't know that sort of if statment was possible. However it did work Barand, thank you. *Notes that one down*

[darkfreaks]

after researching you need to add mysql_fetch_array inside your select function.

 

<?php
function select($amount = "*", $table, $where = NULL, $order = NULL, $limit = NULL){
$query = "SELECT ".$amount." FROM ".$table;
if($where !== NULL){
$query .= " WHERE ".$where;
}
if($order !== NULL){
$query .= " ORDER BY ".$order;
}
if($limit !== NULL){
$query .= " LIMIT ".$limit;
}

$result = mysql_query($query);
$result_two =mysql_fetch_array($result);

if($result_two!==FALSE){
return $result;
}else{
return false;
}
}
?>

[Pikachu2000]

what happens when you do

 

elseif($fetch['pageType'] == 2 OR 3)
{ echo "NO default page selected";}

 

That syntax isn't right. It should be:

 

elseif( $fetch['pageType'] == 2 || $fetch['pageType'] == 3 ) {