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Monday, December 31, 2012

List of New year parties and events

For a complete list of new year parties and event:
Click here

Sunday, December 30, 2012

Top 5 IT certifications for 2013


This is the right time to set your 2013 goals because 2012 is approaching to its end. Achieving a professional certification is a way to improve your skills and enhance your chances of getting hired. Now a days there are so many certifications that are in the market so if you are confused about which certification to achieve in 2013, following are the top 5 IT certifications for 2013 based on expert opinion, research and Google trends.
If you would like to see a list of retired Microsoft Certifications, please Click Here
If you would like to see a list of retired Microsoft Exams, please Click Here
1. MCSA (Microsoft Certified Solutions Associate)
Following is the salary range and average salary for professionals having this certification:
  • Salary Range: $52,000 to $115,000
  • Average Salary: $59,000
Following are the most popular certifications for MCSA:
  • MCSA: Windows 2012 Server
  • MCSA: SQL Server 2012
  • MCSA: Windows 8
  • MCSA: SharePoint 2013
Remember that Microsoft Certified Solutions Associate (MCSA) certification is a prerequisite to the MCSE expert-level certifications. You can refer to Microsoft Learning website for further details of these certifications.
2. MCSE: Private Cloud
Private Cloud gives you expertise in managing and implementing Microsoft private cloud computing technologies. By this certification you can build your Microsoft private cloud solution to optimize IT services as well as gain the automation flexibility for your IT infrastructure.
Following is the salary range and average salary for professionals having this certification
  • Salary Range: $52,000 to $102,000
  • Average Salary: $61,000
This private cloud certification is as popular as the rival CompTIA and CCP cloud certifications. Microsoft is discontinuing its MCITP program in 2014 and promoting MCSE’s. This way you can get jump start for 2014 certifications.
 3. PMP (Project Management Professional)
Following is the salary range and average salary for professionals having this certification:
  • Salary Range: $65,000 to $93,000
  • Average Salary: $86,000
PMP is kind of certification which will never be outdated. This is a certification which fits in every field. PMP certification allows IT professional to plan, budget and manage IT projects. It will be a very good addition in your resume.
4. VCP (VMware Certified Professional)
Following is the salary range and average salary for professionals having this certification:
  • Salary Range: $59,000 to $80,000
  • Average Salary: $69,000
The main purpose of virtualization or creating virtual platforms for computing is to increase scalability and reduce infrastructure and hardware cost. There is a lot of potential in the virtualization industry and it is rapidly growing. VMware is the leader in virtualization software as well as certifications. They are becoming popular and their certifications are very much valuable.
5. CISSP (Certified Information Systems Security Professional)
Following is the salary range and average salary for professionals having this certification:
  • Salary Range: $65,000 to $111,000
  • Average Salary : $80,000
In order to protect organization’s systems, data and networks, information security architects are becoming very much popular these days. Nowadays companies are shifting towards storing data in cloud systems so security experts have a very important role to secure these systems just like they are secured in-house. That is why the CISSP certification is in high demand and normally the CISSP professionals receive highest average salary.

Tuesday, December 18, 2012

Sunday, December 16, 2012

Top 18 pictures from Rainy day in Dubai...

Check out some kool pics of rainy day in Dubai taken today morning...



















Check out my facebook page

Guys!!!


Check out my facebook page to spend sometime in fun....

http://www.facebook.com/FunFuzon

Enjoyyyyyyyy

Monday, December 3, 2012

McAfee Anti-virus pioneer John McAfee caught

Have a look at the link below for the detailed information

http://en-maktoob.news.yahoo.com/anti-virus-software-pioneer-mcafee-captured-report-011224665.html

Interns required for iOS and Android platforms

Interns required for iOS and Android platforms. Optimtec Inc. is looking for fresh graduates for internship. Get trained for the future of smart devices. 

Requirements:

Good in Object orientation programming. 
Good analytical skills.

Should have keen interest in mobile development and game programing.

Interested candidates, please send your resumes to jobs@optimtec.com

Sunday, December 2, 2012

How to get comma separated values in a single select statement


Sometimes you want to get comma separated values from another table in the same select statement, for this you can use the following approaches


--Approach 1
SELECT SUBSTRING(((SELECT (', ' + Aud_Type)
                   FROM   Audience_Type
                   WHERE  Id_Aud_Type IN (SELECT value
                                          FROM   dbo.fn_Split ('1,2,3', ','))
                   FOR    XML PATH (''))), 3, 1000)
FROM   Request;

GO
--Approach 2
SELECT SUBSTRING(((SELECT (', ' + Aud_Type)
                   FROM   Audience_Type AS t2
                   WHERE  t2.RequestID = req.RequestID
                   FOR    XML PATH (''))), 3, 1000) AS Aud_Type
FROM   SOP_Request AS req;

GO

You just have to replace the table/column names (Aud_Type, Audience_Type, Id_Aud_Type, SOP_Request) with your ones

--Split Function  which is being used in approach 1
CREATE FUNCTION [dbo].[fn_Split]
(@sText VARCHAR (8000), @sDelim VARCHAR (20)=' ')
RETURNS
    @retArray TABLE (
        idx   SMALLINT       PRIMARY KEY,
        value VARCHAR (8000))
AS
BEGIN
    DECLARE @idx AS SMALLINT,
@value AS VARCHAR (8000),
@bcontinue AS BIT,
@iStrike AS SMALLINT,
@iDelimlength AS TINYINT;
    IF @sDelim = 'Space'
        BEGIN
            SET @sDelim = ' ';
        END
    SET @idx = 0;
    SET @sText = LTrim(RTrim(@sText));
    SET @iDelimlength = DATALENGTH(@sDelim);
    SET @bcontinue = 1;
    IF NOT ((@iDelimlength = 0)
            OR (@sDelim = 'Empty'))
        BEGIN
            WHILE @bcontinue = 1
                BEGIN
                    --If you can find the delimiter in the text, retrieve the first element and
                    --insert it with its index into the return table.
                    IF CHARINDEX(@sDelim, @sText) > 0
                        BEGIN
                            SET @value = SUBSTRING(@sText, 1, CHARINDEX(@sDelim, @sText) - 1);
                            BEGIN
                                INSERT  @retArray (idx, value)
                                VALUES           (@idx, @value);
                            END
                            --Trim the element and its delimiter from the front of the string.
                            --Increment the index and loop.
                            SET @iStrike = DATALENGTH(@value) + @iDelimlength;
                            SET @idx = @idx + 1;
                            SET @sText = LTrim(RIGHT(@sText, DATALENGTH(@sText) - @iStrike));
                        END
                    ELSE
                        BEGIN
                            --If you can’t find the delimiter in the text, @sText is the last value in
                            --@retArray.
                            SET @value = @sText;
                            BEGIN
                                INSERT  @retArray (idx, value)
                                VALUES           (@idx, @value);
                            END
                            --Exit the WHILE loop.
                            SET @bcontinue = 0;
                        END
                END
        END
    ELSE
        BEGIN
            WHILE @bcontinue = 1
                BEGIN
                    --If the delimiter is an empty string, check for remaining text
                    --instead of a delimiter. Insert the first character into the
                    --retArray table. Trim the character from the front of the string.
                    --Increment the index and loop.
                    IF DATALENGTH(@sText) > 1
                        BEGIN
                            SET @value = SUBSTRING(@sText, 1, 1);
                            BEGIN
                                INSERT  @retArray (idx, value)
                                VALUES           (@idx, @value);
                            END
                            SET @idx = @idx + 1;
                            SET @sText = SUBSTRING(@sText, 2, DATALENGTH(@sText) - 1);
                        END
                    ELSE
                        BEGIN
                            --One character remains.
                            --Insert the character, and exit the WHILE loop.
                            INSERT  @retArray (idx, value)
                            VALUES           (@idx, @sText);
                            SET @bcontinue = 0;
                        END
                END
        END
    RETURN;
END

--ATLTUsers.eoEmployingorganization_ID in (SELECT value from dbo.fn_Split('''+@EOID+ ''', '',''))'

Enjoy