can you try ?

update table set field=REPLACE(field, '@', '\')...

@sean, are you sure 5th May is a Saturday ? in this case i want to May 6th 2014.

Its allright to assume 4 weeks each month ?

Cool, thank you. Meanwhile i was trying to solve using my old attempt, got it also ..finally )

declare @Phrases TABLE

(

phrase varchar(100)

)

DECLARE @Searchs TABLE

(

word varchar(100)

)

declare @NumberWords tinyint

insert into @Searchs (word) select...

Sean Lange (4/19/2013)

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Even though that is the best approach for this situation you should still look at replacing your split function with the one I pointed you too. Post back...

reading time for full text search then )

much obliged Sean.

Yes, the example is very simplified, what i want to do is to find results that match all words of the search.

If i have the text "i have lunched with...

this is the Split function

FUNCTION [dbo].[Split](@String nvarchar(4000), @Delimiter char(1)) returns @Results TABLE (Items nvarchar(4000))

as

begin

declare @index int

declare @slice nvarchar(4000)

select @index = 1

if @String is null...

Obrigado Carlos ) ..funcionou muito bem.

ScottPletcher (12/18/2012)

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If you're 100% sure that no group_id will ever be more than one type, then that looks OK.

You probably want to consider using an IDENTITY column to automatically assign...

Jason-299789 (2/6/2012)

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It is possbible but needs to use a regular expresion, so you could code the the select as follows

select #temp.alunos_id, #groups.ID

from #temp

left...

The function Split works great 😛

Still, i did

update #tgroups set alunos_ids = replace(alunos_ids, ' ','')

and my previous query worked well, so i think i have to make sure...

Jason-299789 (2/6/2012)

---

Looking at the Data you have a Leading Space infront of all numbers but that isnt replicated on the Select.

so when looking for 565, you are trying to match...

but it if i do

select #groups.ID from #groups where alunos_ids like '%477%'

it does not answer NULL.

Done, as requested.