Shantanu's Blog

Corporate Consultant

May 31, 2005

 

Articles

MySQL uses indexes in several ways:

As just described, indexes are used to speed up searches for rows matching terms of a WHERE clause or rows that match rows in other tables when performing joins.

For queries that use the MIN() or MAX() functions, the smallest or largest value in an indexed column can be found quickly without examining every row.

MySQL can often use indexes to perform sorting and grouping operations quickly for ORDER BY and GROUP BY clauses.

Sometimes MySQL can use an index to reading all the information required for a query. Suppose that you're selecting values from an indexed numeric column in a MyISAM table, and you're not selecting other columns from the table. In this case, when MySQL reads an index value from the index file, it obtains the same value that it would get by reading the data file. There's no reason to read values twice, so the data file need not even be consulted.


May 30, 2005

 

Excel array Formula

Columns B:C contains numbers representing hours and minutes accordingly.
How could we sum up the numbers in both columns to a single time value?

Solution:

To get a time value representing the sum of hours and minutes in columns B:C
use the following Array Formula:
{=SUM(TIME(B12:B14,C12:C14,0))}
Example:

Hours___Minutes
5_______20
6_______50
3_______10

Result: 15:20
Formula: {=SUM(TIME(B12:B14,C12:C14,0))}

Notes:
To perform an Array Formula: Insert the formula, press F2 and then press Ctrl+Shift+Enter simultaneously.
The format in the cell contains formula should be:[HH:MM] You can choose the format by clicking on Format - Cells - Time - HH:MM
_____

The other method is to type the following formula in the D column and change the format to General (From Format - Cells or using the Ctrl + Shift + ~)
=TIME(B12,C12,0)
=TIME(B13,C13,0)
=TIME(B14,C14,0)
Now add all the serial numbers and change the format back to HH:MM

May 29, 2005

 

Installing Mysql on Desktop

You can download and install the Apache, PHP + Mysql (Around 25 MB) pre compiled installation from...

http://umn.dl.sourceforge.net/sourceforge/xampp/xampp-win32-1.4.13-installer.exe
Once installed goto Start - Programs - apachefriends - xampp control panel.
Click on Start Apache and Start Mysql.
Now you can access phpMyAdmin by going to ...
http://localhost/phpmyadmin/
You can use this version for test purpose. You will have to change the root password if the data is important.

May 28, 2005

 

INDIRECT Function

Another useful feature of the INDIRECT function is that since it takes string argument, you can use it to work with cell references that you don't want Excel to automatically change when you insert or delete rows. Normally, Excel will change cell references when you insert or delete rows or columns, even when you use absolute referencing. If you have the formula =SUM($A$1:$A$10), and then insert a row at row 5, Excel will convert the formula to =SUM($A$1:$A$11). If you don't want this to happen, use the INDIRECT function to change a text string to a reference:

=SUM(INDIRECT('A1:A10'))

Since Excel sees 'A1:A10' as a text string rather than a range reference, it will not change it when rows or columns are deleted or inserted.
http://www.cpearson.com/excel/indirect.htm


 

scribe extension to save form information

IF you need to save the information typed in forms in a html file, "scribe" extension for Mozilla will help you a lot.
Install the extension from here...
http://prismicspiral.com/scribe/index.html

  1. Now go to any website where you need to fill in the form information.
  2. Type all the information that you will need to reuse later.
  3. Click on file and choose "Save Entry As..."
  4. Next time when you visit the same web form, simple click on File - "Open Entry..."
  5. All the information that you had saved at the first time will be loaded! Yes. It's that simple. You don't need to change the HTML code.

Note: You can save multiple forms with multiple data filled in. So the same form can be saved twice for X as well as Y. The Scribe html file is a normal XHTML document - You can edit long posts in your favorite HTML or text editor then use Scribe to load it back into the web form.

May 26, 2005

 

Excel Trouble shooting !>>

Does your file opens in hidden mode?

First thing I'd do is check to see if the workbook is actually opened.

On the worksheet menu bar, click on Window. Do you see the name of the workbook at the bottom of the dropdown? If no, then the workbook isn't opening or it's
opening hidden (unlikely if others can open the file on their computer!).

But just to make sure, click on Window and then if Unhide is greyed out, it's not hidden.
If Unhide is available, click on it to see if the workbook name is in there (and unhide it if it is).

If the workbook appears at the bottom of the Window dropdown, try Window|Arrange and make it tiled. Then resize by hand.
(The workbook's window could just be off the visible window.)

_____

Does File|open work ok, but double clicking on the filename in windows explorer cause trouble?

If yes, then...

Sometimes one of these works:

Tools|Options|General|Ignore other applications (uncheck it)

_____

Excel Start-up problems?

Try opening excel in safe mode:
close excel
windows start button|Run
excel /safe

OR

Close Excel and
Windows Start Button|Run
excel /unregserver
then
Windows Start Button|Run
excel /regserver

The /unregserver & /regserver stuff resets the windows registry to excel's factory defaults.
_____

Does it still cause a problem?

then maybe it's your *.xlb (where excel keeps your customized toolbar changes) that's causing the trouble.

Close excel
windows start button|find|*.xlb

rename all you find to *.xlbOLD

Then restart excel (normally) and open your workbook.

If that fixes the problem, delete all those *.xlbOLD files and recreate any changes to your toolbar.

 

Point-in-Time Data Recovery

Point-in-Time Data Recovery: Binary logging will record in a binary file all SQL transactions executed and attempted on the server. By using the mysqlbinlog utility, the contents of the binary log file can be extracted so that the SQL statements may easily be rerun. To enable binary logging, add the following line to your server's options file (i.e., /etc/my.cnf or c:\my.ini, depending on your system) in the [mysqld] group:

log-bin = /var/log/mysql/bin.log

The exact path to use will depend on your filesystem and your preferences.
http://dev.mysql.com/doc/mysql/en/binary-log.html

 

Replace Old Name in Comments

Excel -- Comments -- Programming: "If a previous user inserted comments, their name may appear at the top of the comment. Their name may also appear in the Status Bar, when you hover over the cell that contains a comment. The following macro will replace the old name with a new name.

Sub ChangeCommentName()
'replaces old names in comments
'deletes and reinserts comments
' so new name appears in status bar
Dim ws As Worksheet
Dim cmt As Comment
Dim strOld As String
Dim strNew As String
Dim strComment As String
strNew = 'New Name'
strOld = 'Old Name'
Application.UserName = strNew
For Each ws In ActiveWorkbook.Worksheets
For Each cmt In ws.Comments
strComment = Replace(cmt.text, strOld, strNew)
cmt.Delete
cmt.Parent.AddComment text:=strComment
Next cmt
Next ws

End Sub"

May 23, 2005

 

phpMyBackupPro - the MySQL backup tool

phpMyBackup Pro is a very easy to use, free web-based MySQL backup application, licensed under the GNU GPL. You can create scheduled backups, manage and restore them, download or email them and a lot more!
http://www.phpmybackuppro.net/index.php

May 18, 2005

 

MySQL Case Study - 25

Sort by
I have a table with a primary key field (idField) declared as varchar(64). I did so because most of the time the key stored is going to be a numeric but sometimes it can be a some kind of small Text.
But now I'm facing some issues trying to order by "idField" asc.

I would like the resultset to appear like this :

1
2
3
10256
14597
24573
32564

With all the text values on one end. And of course since it's sorting text it appears like that :

1
10256
14597
2
24573
3
32564

Any suggestions?

May 13, 2005

 

MySQL Case Study - 24

I am making a PHP-MySQL forum. I have a table "forum" that holds all the posts.
In my main forum view where I list the last 10 threads, I want returned in each
result set the ID, title, user, and the # of comments posted in that thread.

May 09, 2005

 

librarian.net

librarian.net: "The debate we’ve seen happening over the authority, or lack thereof, of collaborative information systems such as Wikipedia is just scratching the surface of the debates we’ll be seeing in the years to come. Librarians ignore Wikipedia, and by extension the new face of information, at their peril. Keep in mind I’m not saying that we all have to run to the Internet to answer our questions, just that if we fail to see the impact these systems are having, and the openness and transparency they bring with them, then we fail to learn something crucial about the downsides to the inflexible authority of print. Downsides that people have been living with and taking as a given all these years, and now may no longer have to."

 

The Entrepreneur's Secret is No

Learn to say no and accept no : "Learn how to handle no. Learn to accept it as an answer, but never let it be the final answer. Always try to turn rejection into opportunity. Always stay optimistic. Always believe that yes is around the corner."

 

Duplicate And Unique Items In Lists

If you need to determine whether a list in Excel has duplicate entries, you can use the following formula. It will display 'Duplicates' if the list in Range1 has duplicate entries, or 'No Duplicates' if the range does not have any duplicates.

=IF(MAX(COUNTIF(Range1,Range1))>1,'Duplicates','No Duplicates')

This is an array formula, so you must press Ctrl+Shift+Enter rather than just Enter when you first enter the formula, and when ever you edit it later. This formula requires that the complete range contain data.

http://www.cpearson.com/excel/duplicat.htm


May 08, 2005

 

MySQL Case Study - 23

Crosstab query

Dynamic crosstab queries let you produce pivoted, denormalized output from normalized data.

CREATE TABLE category(
category_id INT AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(20) NOT NULL,
parent INT DEFAULT NULL);


INSERT INTO category VALUES(1,'ELECTRONICS',NULL),
(2,'TELEVISIONS',1), (3,'TUBE',2), (4,'LCD',2),
(5,'PLASMA',2), (6,'PORTABLE ELECTRONICS',1),
(7,'MP3 PLAYERS',6), (8,'FLASH',7),
(9,'CD PLAYERS',6), (10,'2 WAY RADIOS',6);

SELECT * FROM category ORDER BY category_id;

+-------------+----------------------+--------+
| category_id | name | parent |
+-------------+----------------------+--------+
| 1 | ELECTRONICS | NULL |
| 2 | TELEVISIONS | 1 |
| 3 | TUBE | 2 |
| 4 | LCD | 2 |
| 5 | PLASMA | 2 |
| 6 | PORTABLE ELECTRONICS | 1 |
| 7 | MP3 PLAYERS | 6 |
| 8 | FLASH | 7 |
| 9 | CD PLAYERS | 6 |
| 10 | 2 WAY RADIOS | 6 |
+-------------+----------------------+--------+
10 rows in set (0.00 sec)



SELECT t1.name AS lev1, t2.name as lev2, t3.name as lev3, t4.name as lev4
FROM category AS t1
LEFT JOIN category AS t2 ON t2.parent = t1.category_id
LEFT JOIN category AS t3 ON t3.parent = t2.category_id
LEFT JOIN category AS t4 ON t4.parent = t3.category_id
WHERE t1.name = 'ELECTRONICS';

+-------------+----------------------+--------------+-------+
| lev1 | lev2 | lev3 | lev4 |
+-------------+----------------------+--------------+-------+
| ELECTRONICS | TELEVISIONS | TUBE | NULL |
| ELECTRONICS | TELEVISIONS | LCD | NULL |
| ELECTRONICS | TELEVISIONS | PLASMA | NULL |
| ELECTRONICS | PORTABLE ELECTRONICS | MP3 PLAYERS | FLASH |
| ELECTRONICS | PORTABLE ELECTRONICS | CD PLAYERS | NULL |
| ELECTRONICS | PORTABLE ELECTRONICS | 2 WAY RADIOS | NULL |
+-------------+----------------------+--------------+-------+
6 rows in set (0.00 sec)



We can use the depth value to indent our category names
with the CONCAT and REPEAT string functions:

SELECT CONCAT( REPEAT(' ', COUNT(parent.name) - 1), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;

+-----------------------+
| name |
+-----------------------+
| ELECTRONICS |
| TELEVISIONS |
| TUBE |
| LCD |
| PLASMA |
| PORTABLE ELECTRONICS |
| MP3 PLAYERS |
| FLASH |
| CD PLAYERS |
| 2 WAY RADIOS |
+-----------------------+

May 06, 2005

 

MySQL Case Study - 22

Different use of count

I am attempting to select (identify) records from a table where a specific field must equal certain values.
The problem I am having is that all the values must exist before any are to be selected.

For example

SELECT *
FROM `employees`
where Employee_ID = '1035'
and (Skill_Training_Type = 'SKILL1' or Skill_Training_Type = 'SKILL2')

Will select records with either of the values.

SELECT *
FROM `employees`
where Employee_ID = '1035'
and (Skill_Training_Type = 'SKILL1' and Skill_Training_Type = 'SKILL2')

Will select no records - obviously.

I want the select to select the records only if both records exist for the specific Employee_ID.

The statement is being used by a report generator so generating temporary tables and reselecting is not feasible.

Suggestions?

_____

If the id and skill combination are unique (unique key (id, skill)), you can just count the number of rows using an aggregate function and add a HAVING clause...

create table tmp(id int, skill char(1));
insert into tmp values (1,'a'), (1,'b'), (1,'c'), (2,'a'), (2,'c'), (3,'b');

NOTE: (skill='a' or skill='b') is the same as skill in ('a', 'b')

select id from tmp where id=1 and skill in ('a', 'b') group by id having count(*)=2;
+------+
| id |
+------+
| 1 |
+------+

select id from tmp where id=2 and skill in ('a', 'b') group by id having count(*)=2;

Empty set (0.00 sec)

or search ALL rows

select id from tmp where skill in ('a', 'b') group by id having count(*)=2;

---
You can also use a subselect or self-join ...

select DISTINCT id from tmp where id in (select id from tmp where skill='a') and id in (select id from tmp where skill='b');
+------+
| id |
+------+
| 1 |
+------+

or self join (one join needed for each skill)

select t1.id, t1.skill, t2.skill from tmp t1, tmp t2
where t1.id=t2.id and t1.skill='a' and t2.skill= 'b';
+------+-------+-------+
| id | skill | skill |
+------+-------+-------+
| 1 | a | b |
+------+-------+-------+

 

MySQL Case Study - 21

Managing Hierarchical Data in MySQL
I have a table with all my categories and subcategories, with the following schema:

id_category | id_parent | name

1 | 0 | category#1
2 | 1 | subcategory#1
3 | 1 | subcategory#2
4 | 2 | subsubcategory#1
5 | 0 | category#2
5 | 5 | subcategory#3
...

All categories have id_parent = '0' and subdirectories have id_parent = the category or subcategory they belong to.

I want to do a simple query that displays categories and all it's nested subcategories in order.

Is it possible to do this with one query or must I make a loop to get them all?

_____

Check this recent article on trees in mysql.

http://www.vbmysql.com/articles/database-design/managing-hierarchical-data-in-mysql.html

If you know the height of your tree (in your example there are three levels), you can use self joins to display paths using one join for each level in the tree. Note this displays one column for each level in the tree.

Something like this.

SELECT t1.name as t1, t2.name as t2, t3.name as t3
FROM cat t1
LEFT JOIN cat t2 ON t1.id_category=t2.id_parent
LEFT JOIN cat t3 ON t2.id_category=t3.id_parent
WHERE t1.id_parent=0;

If you want all subcategories, subsubcategories, etc listed in a single column, you should use recursion (via a dbi script), nested set models, or you can convert your tree to a path enumerated list below.
I will use the table from the web site link, but now I'm searching up the tree instead of down.

SELECT t4.name as t4, t3.name as t3, t2.name as t2, t1.name as t1
FROM category t1
LEFT JOIN category t2 ON t1.parent=t2.category_id
LEFT JOIN category t3 ON t2.parent=t3.category_id
LEFT JOIN category t4 ON t3.parent=t4.category_id

+-------------+----------------------+----------------------+----------------------+
| t4 | t3 | t2 | t1 |
+-------------+----------------------+----------------------+----------------------+
| NULL | NULL | NULL | ELECTRONICS |
| NULL | NULL | ELECTRONICS | TELEVISIONS |
| NULL | ELECTRONICS | TELEVISIONS | TUBE |
| NULL | ELECTRONICS | TELEVISIONS | LCD |
| NULL | ELECTRONICS | TELEVISIONS | PLASMA |
| NULL | NULL | ELECTRONICS | PORTABLE ELECTRONICS |
| NULL | ELECTRONICS | PORTABLE ELECTRONICS | MP3 PLAYERS |
| ELECTRONICS | PORTABLE ELECTRONICS | MP3 PLAYERS | FLASH |
| NULL | ELECTRONICS | PORTABLE ELECTRONICS | CD PLAYERS |
| NULL | ELECTRONICS | PORTABLE ELECTRONICS | 2 WAY RADIOS |
+-------------+----------------------+----------------------+----------------------+

Basically, just count up the number of non-null columns to get the level for each name in the last column, and concatenate non-null columns to get the path.


create table category_path as
SELECT t1.category_id, t1.name,
( if(t4.category_id is null, 0,1) + if(t3.category_id is null, 0,1) + if(t2.category_id is null, 0,1) + if(t1.category_id is null, 0,1)) as level,
trim(leading '; ' from concat(ifnull(t4.name, ''), '; ', ifnull(t3.name, ''), '; ', ifnull(t2.name, ''), '; ', ifnull(t1.name, ''))) as path
FROM category t1
LEFT JOIN category t2 ON t1.parent=t2.category_id
LEFT JOIN category t3 ON t2.parent=t3.category_id
LEFT JOIN category t4 ON t3.parent=t4.category_id;


+-------------+----------------------+-------+-------------------------------------------------------+
| category_id | name | level | path |
+-------------+----------------------+-------+-------------------------------------------------------+
| 1 | ELECTRONICS | 1 | ELECTRONICS |
| 2 | TELEVISIONS | 2 | ELECTRONICS; TELEVISIONS |
| 3 | TUBE | 3 | ELECTRONICS; TELEVISIONS; TUBE |
| 4 | LCD | 3 | ELECTRONICS; TELEVISIONS; LCD |
| 5 | PLASMA | 3 | ELECTRONICS; TELEVISIONS; PLASMA |
| 6 | PORTABLE ELECTRONICS | 2 | ELECTRONICS; PORTABLE ELECTRONICS |
| 7 | MP3 PLAYERS | 3 | ELECTRONICS; PORTABLE ELECTRONICS; MP3 PLAYERS |
| 8 | FLASH | 4 | ELECTRONICS; PORTABLE ELECTRONICS; MP3 PLAYERS; FLASH |
| 9 | CD PLAYERS | 3 | ELECTRONICS; PORTABLE ELECTRONICS; CD PLAYERS |
| 10 | 2 WAY RADIOS | 3 | ELECTRONICS; PORTABLE ELECTRONICS; 2 WAY RADIOS |
+-------------+----------------------+-------+-------------------------------------------------------+


Now you can print an indented tree

select concat(space(3*(level-1)), name) as tree from category_path order by path;

or display all subcategories under any node by searching the path string using LIKE (not very fast for large tables though).

select name from category_path where path like '%TELEVISIONS;%';

 

MySQL Case Study - 20

sample record x
id field1 field2
1 a q
2 b x
3 a y
4 b z

I want to set a "y" filter on field2 but want to have all records with the same field1 value as result.
so, in the field2 = "y" filter, I want record 1 and 3 as result because field1 is in both records "a"
(Show me all records with the same Field1 value when field2 is y)
Is this possible and if so, how?

 

MySQL Case Study - 19

Values repeats in joins
I have 3 tables connected via foreign keys.

Employees - PK EmployeeID
Holidays - FK EmployeeID
SickDays - FK EmployeeID

As you can probably guess, the holidays and sickdays tables have an EMployeeID field that links the holiday or sick day to the employee who took it.

SO right now, my tables look like

Employees (IDs only)
1
7
8

Holidays (EmployeeID and HolidayStartDate)
1 04/15/2005
1 05/01/2005
1 05/14/2005
7 03/15/2005
7 04/15/2005

Sickdays (EmployeeID and SickDate)
1 03/15/2005
1 04/02/2005
_____

DROP TABLE IF EXISTS emp;
CREATE TABLE emp (
id INT NOT NULL PRIMARY KEY,
name VARCHAR(100) NOT NULL
);
INSERT INTO emp (id, name) VALUES
(1, 'John'),
(2, 'Pete'),
(3, 'Workaholic');

DROP TABLE IF EXISTS holiday;
CREATE TABLE holiday (
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
emp_id INT NOT NULL,
dt_start DATE NOT NULL
);
INSERT INTO holiday (id, emp_id, dt_start) VALUES
(NULL, 1, '2005-01-01'),
(NULL, 1, '2005-02-01'),
(NULL, 2, '2005-01-01'),
(NULL, 2, '2005-03-01'),
(NULL, 2, '2005-03-10');

DROP TABLE IF EXISTS sick;
CREATE TABLE sick (
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
emp_id INT NOT NULL,
dt_start DATE NOT NULL
);
INSERT INTO sick (id, emp_id, dt_start) VALUES
(NULL, 1, '2005-04-01'),
(NULL, 1, '2005-04-03'),
(NULL, 1, '2005-04-05'),
(NULL, 1, '2005-05-01'),
(NULL, 2, '2005-03-10'),
(NULL, 2, '2005-05-01');

_____

I want to return a query that will have 3 columns a count of the sickdays for each employee and count of the holiday start dates for each employee, but all employees must be returned. I have tried about every combination of LEFT JOINS I can think of, and nothing seems to output the desired results. I keep either getting holidays repeated, or sick days repeated...or both.

I usually end up with something like this (I have been ignoring the aggregate functions for now, just to see the results)

EmployeeID, HolidayStartDate, SickDate
1, 04/15/2005, 03/15/2005
1, 04/15/2005, 04/02/2005
1, 05/01/2005, 03/15/2005
1, 05/01/2005, 04/02/2005
etc, etc

As you can see the holidays repeat for each unique sick day. I need to know unique entries in both columns.

 

is error?

If you've ever tried to set up a formula to divide in Excel before you've entered the data, you've probably seen this error:

#DIV/0!

Not only is it fairly cryptic, its also an extremely ugly thing to see in a worksheet. The ISERROR function can remedy that. The ISERROR function works by essentially allowing you to write your own custom error message in the ISERROR function. In the case of the divide by zero error, there could be any number of legitimate reasons that cause the error.

=IF(ISERROR(A1/B1), "Insufficient Data", (A1/B1))

What this is saying is, if dividing cell A1 by cell B1 causes an error (any error at all), then put the phrase "Insufficient Data" into this cell, otherwise put the result of the function (SUM A1/B1) into this cell.

ISERROR works to modify all errors including #DIV/0!, #VALUE!, etc. ISERROR can be used in conjunction with any Excel function to modify error messages so that they look more "user friendly".

May 05, 2005

 

MySQL Optimization

You can read the excerpts from chapter six of the book MySQL Administrator's Guide, by MySQL AB (Sams, 2004; ISBN: 0672326345)

http://tinyurl.com/c4o8x
http://tinyurl.com/a52p6

 

Backing up and restoring MySQL databases: a 3-step method

Export the database to a text file
The MySQL program to back up a database or table is called mysqldump, and it's included as a standard part of the MySQL distribution. It works by reading the selected database and reproducing its contents as a series of SQL commands, which serve as a snapshot and can be used to recreate the data at a later date.

Using mysqldump is simplicity itself, just run the program with the name of the database to be exported, as below:
$ mysqldump -h localhost -u us2049 -p secret db3 > backup.db3

This will create the file backup.db3 and populate it with the SQL commands needed to regenerate the database.

Note that you can back up all the databases on the system (assuming you have the credentials to read them) by replacing the database name in the call to mysqldump with the special keyword --all-databases. Here's an example:
$ mysqldump -h localhost -u us2049 -p secret --all-databases > backup.all

http://builder.com.com/5100-6388_14-5669527.html

May 03, 2005

 

My Web BETA - FAQ - Yahoo!

My Web BETA - FAQ - Yahoo!: "My Web is the newest way to save, share, and organize any page you want on the Web. You can save all the sites that interest you then re-find them, organize them, and share them with whomever you want. My Web is your personal collection of web knowledge to use as you please."

May 02, 2005

 

MySQL Case Study - 18

I am looking for the best way to model a "many to many" relationship. For example, say I have a table of images and a table of categories. An image can belong to any number of categories, and a category can be assoicated with any number of images.

Image table
ImageID | ImageURL

Category table
CategoryID | CategoryName

So, would the best way to associate images with categories to create another table, ImageCategoryRelationship which takes an ImageID and CategoryID? Is that the standard way?

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