create TABLE #OutputEmails

(

Email1 nvarchar(1000),

Email2 nvarchar(1000),

Email3 nvarchar(1000),

Email4 nvarchar(1000),

Email5 nvarchar(1000),

Email6 nvarchar(1000),

Email7 nvarchar(1000),

Email8 nvarchar(1000),

Email9 nvarchar(1000),

Email10 nvarchar(1000)

);

INSERT INTO #OutputEmails SELECT Email1,Email2,Email3,Email4,Email5,Email6,Email7,Email8,Email9,Email10 FROM OutputResume3

where (contains (originalresume, '"livelink" and "java"'))

select count(DISTINCT (Email)) FROM

(

SELECT top...

Just the number '15' as a count. No email addresses necessary.

I have attached 20 sample records.

10 are duplicates.

Total # of unique emails are: 31

email1 - 10

email2- 8

email3 - 6

email4 - 4

email5 - 2

email6 - 1

I would like only the...

GilaMonster (1/10/2010)

---

It probably is,...

USE [microsoft525]

GO

/****** Object: Table [dbo].[OutputResume3] Script Date: 01/12/2010 22:58:31 ******/

SET ANSI_NULLS ON

GO

SET QUOTED_IDENTIFIER ON

GO

CREATE TABLE [dbo].[OutputResume3](

[ID3] [int] IDENTITY(1,1) NOT NULL,

[OriginalResume] [nvarchar](max) NULL,

[Email1] [nvarchar](1000) NULL,

[Email2] [nvarchar](1000) NULL,

[Email3]...

I did get a couple of suggestions.

Use

CREATE PARTITION FUNCTION

or

CREATE PARTITION SCHEMA

.....but as you all know i suck at SQL, so i really dont know exactly how to use this to...

Hello guys, i have not forgotten you 😀

I guess the problem is a little more difficult than expected!

We do have big time zone differences to deal with.

......Help you i will, oh Jedi master.

The force is strong in this one!

Lynn,

You need to chill!

I can see that my lack of compliance is irritating you. I will just ask my Indian developers tomorrow.

Problem solved 🙂

SELECT COUNT(DISTINCT(variablename))

Let's assume i have 1000 variable names, with duplicates, which is why i need DISTINCT.

I only want to display the TOP 50.

SELECT TOP 50 COUNT(DISTINCT(variablename)) does not...

Lynn Pettis (1/9/2010)

---

ifila (1/9/2010)

---

--SELECT DISTINCT TOP 1000 Email

SELECT COUNT(DISTINCT(Email))

FROM(SELECT Email1,Email2,Email3, Email4,Email5,Email6,Email7,Email8,Email9,Email10 FROM outputresume3

WHERE ((contains (originalresume, '"project" and "manager" and "php"'))

...

Definition of a Forum "a public meeting or assembly for open discussion"

Based on the strict rules of this site, it does not qualify as a Forum!

....and people wonder why...

I think 'the drill' is BS! By the time i go thru all the things that email asks for, i can get the response elsewhere, for a LOT less $$$...

Thanks!

Thanks !