Site Sift Templates

A great Site-Sift template from BlogAboutMySite.com. The template is called Sunny Blue and is really good looking bluish template.

It has a left sidebar, space for banner on top right and a great looking category box. Honestly, it’s one of the better templates i have seen lately. Specially when it comes to site-sift templates. It’s made for 1.3 version of site-sift, so just take care if you are having some other version of site-sift.

This template, “Sunny Blue”, is brought to you free by BlogAboutMySite Web Directory and Aviva Web Directory. Template is free to use as long as the footer links are intact. Please view the ReadMe.txt file inside the template zip for all details and support. Editable logo file is included in the template zip. You are not authorized to use the template removing its footer links, or changing the position of the links or adding “nofollow” to them.

Screenshot:

Live Demo: Directory Guru

Site-Sift Version: 1.3

Download Link: Click Here.

Enjoy the template! If you like it, then leave your comment!

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JojoMart, a member from Site Sift forums has made a modification which allows you to show up number of sub categories and number of links in each of the categories. It will show up something like:

Web Directories (24) (150) wherein “Web Directories” is the category, (24) is the number of sub categories and (150) is number of links.

It’s a mod alright, but i haven’t had any review so far on this mod from other members. So if you are installing this, please backup your files. And take care of it… Do it on your own risk. And please, please, don’t hold me liable if anything goes wrong… Thanks.

Alright, so here is how the thing goes:

1. Take your backup of the files.

2. Open subcat.php

3. Find the code:

while($subcategory = mysql_fetch_assoc($subcat)){
//fetch children
$sql = “select * from pages where active = ‘Yes’ and catID = ‘” . $subcategory['id'] . “‘”;
//echo $sql;
$R = mysql_query($sql,$myconn) or die(mysql_error());
//print children links and total subcategories of children
$sql = “select * from settings where id = ‘1′”;
$S = mysql_query($sql,$myconn);
$settings = mysql_fetch_assoc($S);
if($settings['stats'] == ‘1′)
$stats = ‘ (’.mysql_num_rows($R).’)';

4. Replace the code with:

while($subcategory = mysql_fetch_assoc($subcat)){
//fetch children
$sql = “select * from pages where active = ‘Yes’ and catID = ‘” . $subcategory['id'] . “‘”;
//echo $sql;
$R = mysql_query($sql,$myconn) or die(mysql_error());
//print children links and total subcategories of children
$sql = “select * from settings where id = ‘1′”;
$S = mysql_query($sql,$myconn);
$settings = mysql_fetch_assoc($S);
if($settings['stats'] == ‘1′)
$sql = “select * from categories where fatherID = ‘” . $subcategory['id'] . “‘”;
$Results = mysql_query($sql,$myconn) or die(mysql_error());
$subcats = mysql_query($sql,$myconn);
$stats = ‘ (’.mysql_num_rows($subcats).’) (’.mysql_num_rows($R).’)';

This will show the number of subcategories and number of links under each category.

Hope you will find it useful. And if you do, then write a review and leave a commment!

Thanks!

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