Hi all,

I am trying this all from scratch again...Talk about persistency...Well that's what I am all about.

Ok, Here's where I am struggling again...I have a categories.php page and a subcategories.php page. The categories page has a very simple query on it to display the categories, and it works just fine.. The subcategories page on the other hand does not work fine. What I want the subcategories.php to display is the subcategories for the id associated with the categories. Here's what my database looks like:

catiid type int(11) auto_increment (primary key)
parentid type int(11)
name type (varchar(255)
description type text
cat_image_url type texr

Note: I have included both categories and subcategories.php files in a zip attachement as well a a screenshot expalining how the link to the subcategories should work.

Thank you in advance!
-Brian

~ bump ~

Maybe a Switch statement for subcategories?

PHP Code:

switch(categoryID) {
    case 1: //display this subcategories
    break;
    case 2: //display this subcategories
    break;
    //...
} 


I didn't read what you wrote on previous pages, but if i understood, this is one way to do it.

Originally Posted by miki86

Maybe a Switch statement for subcategories?

PHP Code:

switch(categoryID) {
    case 1: //display this subcategories
    break;
    case 2: //display this subcategories
    break;
    //...
} 


I didn't read what you wrote on previous pages, but if i understood, this is one way to do it.

Hey miki,

I haven't tried the switch statement before, but after looking at some examples I don't think that is what I should be using in this case because of the large amount of subcategories.

Thanks for that option though
-Brian

You have relative ID's stored in your db as i understand.
Use that ID to load different subcategories.

Example:

categories.php

PHP Code:

// You can do a loop here if you have to many categories
<a href="subcategories.php?catid=1">Category 1</a>
<a href="subcategories.php?catid=2">Category 2</a>
<a href="subcategories.php?catid=3">Category 3</a>
<a href="subcategories.php?catid=4">Category 4</a> 


subcategories.php

PHP Code:

if(isset($_GET['catid']) && ($_GET['catid'] != NULL)) $categoryID = $_GET['catid'];
else $categoryID = 1;
$query = "SELECT * FROM subcategories WHERE parentID = {$categoryID}";
$result = mysql_result($query);

while($row = mysql_fetch_assoc($result)) {
    echo $row[];
} 


Something like that should work.

EDIT:
btw im looking at your files and i noticed this in your subcategories.php:

PHP Code:

$sub_cats = $_GET['parentid'];

$query = "SELECT * FROM categories WHERE parentid != 0"; 


You are passing a parentid value, but you are not using it in your sql

Your query should look like this:

PHP Code:

$sub_cats = mysql_real_escape_string($_GET['parentid']);

$query = "SELECT * FROM categories WHERE parentid = {$sub_cats}"; 


Originally Posted by miki86

You have relative ID's stored in your db as i understand.
Use that ID to load different subcategories.

Example:

categories.php

PHP Code:

// You can do a loop here if you have to many categories
<a href="subcategories.php?catid=1">Category 1</a>
<a href="subcategories.php?catid=2">Category 2</a>
<a href="subcategories.php?catid=3">Category 3</a>
<a href="subcategories.php?catid=4">Category 4</a> 


subcategories.php

PHP Code:

if(isset($_GET['catid']) && ($_GET['catid'] != NULL)) $categoryID = $_GET['catid'];
else $categoryID = 1;
$query = "SELECT * FROM subcategories WHERE parentID = {$categoryID}";
$result = mysql_result($query);

while($row = mysql_fetch_assoc($result)) {
    echo $row[];
} 


Something like that should work.

EDIT:
btw im looking at your files and i noticed this in your subcategories.php:

PHP Code:

$sub_cats = $_GET['parentid'];

$query = "SELECT * FROM categories WHERE parentid != 0"; 


You are passing a parentid value, but you are not using it in your sql

Your query should look like this:

PHP Code:

$sub_cats = mysql_real_escape_string($_GET['parentid']);

$query = "SELECT * FROM categories WHERE parentid = {$sub_cats}"; 


Hi Miki,

I've tried to use your code posted above, and I am getting this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /var/www/subcategories.php on line 17

Line 17:

PHP Code:

while($row=mysql_fetch_assoc($result)) { 


I do have the closing curly brace btw. I just posted the line with the error, and I did put 'name' in the $row[]; part in the while loop, assuming that will give me the names of the subcategories for each of the categories in the database. Yes, I also do have subcategories in the subcategories table.

It might be better if you could offer an sql file that I can import to get the database structure you offering me. Otherwise, I am having a total brain freeze trying to figure this out...I am pretty good with navigation, but until I have the coding part understood, I am a total n00b. Give me credit, I am still trying though.

Thanks again for the help in advance!
-Brian

This should work.

I've done no checking to see if the id is passed across.

You'll need to add you connection file again.

Categories

<html>
<head><title>Reunite My Site - Category Navigation Page</title>
<style type="text/css">

#main_content {

background-color: #ccc;
border: 1px solid #000;
color: #000;
padding: 5px;
}
</style>

<?php 

// Turn on error reporting
error_reporting(E_ALL);
ini_set('display_errors','on');

/* Connect String */
$dbHost = "localhost";
    $dbUser = "root";
    $dbPass = "root";
    
    $con = mysql_connect($dbHost,$dbUser,$dbPass);
    if (!$con)
        {
        die('could not connect: ' .mysql_error());
        }
        
    mysql_select_db("TESTDB", $con);

/* Output Main Categories */

$query = "SELECT * from categories where parentid = 0";
$result = mysql_query($query);

echo " <div id=\"main_content\" ";

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

echo "<p>";
echo "<a href=\"subcategories.php?id=$row[id]\">$row[name]</a>";
echo "</p>";
}
echo "</div>";


?>

Sub cat

<H3> Subcategories Page: </H3>

<?php 

// Turn on error reporting
error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors','on');

$dbHost = "localhost";
    $dbUser = "root";
    $dbPass = "root";
    
    $con = mysql_connect($dbHost,$dbUser,$dbPass);
    if (!$con)
        {
        die('could not connect: ' .mysql_error());
        }
        
    mysql_select_db("TESTDB", $con);

$sub_cats = $_GET['id'];

$query = "SELECT * FROM categories WHERE parentid = $sub_cats";
$result = mysql_query($query);

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

echo "<p>";

echo $row['name'];
echo "</p>";

}
?>

I am getting this error:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /var/www/subcategories.php on line 17

That means that your query is not ok.
Try this and see what it comes out:

PHP Code:

$query = "SELECT * FROM categories WHERE parentid = {$sub_cats}";
$result = mysql_query($query);
if(!$result) {echo mysql_error(); die();} 


It might be better if you could offer an sql file that I can import to get the database structure you offering me.

I can't give you any sql file but i can explain you how it works.
You have a db table with the names of parent categories, each of them has a unique ID, you also have a db table with subcategories which looks like this:

ID---parentid---subcategory_name
1----1----------Subcategory 1
2----1----------Subcategory 2
3----2----------Subcategory 3
4----3----------Subcategory 4
5----3----------Subcategory 5

So basically parentid in subcategory table is the ID of the category table.
You just need to select all rows with the proper ID.

OK, I just got in from work not too long ago..Had to work a double today so I am quite tired, so I will try tonight and if all fails, I will double / triple check it tomorrow. Thank you both!

-brian

Originally Posted by HullBorn

This should work.

I've done no checking to see if the id is passed across.

You'll need to add you connection file again.

Categories

<html>
<head><title>Reunite My Site - Category Navigation Page</title>
<style type="text/css">

#main_content {

background-color: #ccc;
border: 1px solid #000;
color: #000;
padding: 5px;
}
</style>

<?php 

// Turn on error reporting
error_reporting(E_ALL);
ini_set('display_errors','on');

/* Connect String */
$dbHost = "localhost";
    $dbUser = "root";
    $dbPass = "root";
    
    $con = mysql_connect($dbHost,$dbUser,$dbPass);
    if (!$con)
        {
        die('could not connect: ' .mysql_error());
        }
        
    mysql_select_db("TESTDB", $con);

/* Output Main Categories */

$query = "SELECT * from categories where parentid = 0";
$result = mysql_query($query);

echo " <div id=\"main_content\" ";

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

echo "<p>";
echo "<a href=\"subcategories.php?id=$row[id]\">$row[name]</a>";
echo "</p>";
}
echo "</div>";


?>

Sub cat

<H3> Subcategories Page: </H3>

<?php 

// Turn on error reporting
error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors','on');

$dbHost = "localhost";
    $dbUser = "root";
    $dbPass = "root";
    
    $con = mysql_connect($dbHost,$dbUser,$dbPass);
    if (!$con)
        {
        die('could not connect: ' .mysql_error());
        }
        
    mysql_select_db("TESTDB", $con);

$sub_cats = $_GET['id'];

$query = "SELECT * FROM categories WHERE parentid = $sub_cats";
$result = mysql_query($query);

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

echo "<p>";

echo $row['name'];
echo "</p>";

}
?>

Thank you Hullborn.

That was *exactly* what I was trying to accomplish! You have done a super job! I appreciate your work. It's always the small things that make a big difference!

:: Edit ::

Miki86, I appreciate your effort as well, only I am not the best when it comes to finding syntax mistakes, so even though the code you posted may have been on the right track, I could not figure it out...Seems like I will be studying up more on my syntax errors.

Thanks again Hullborn!
-Brian