Exact Match using Like query [Resolved]

Hi,
if we use
1. like '%1%' --> it checks if any part of string has 1 or not
2. like '1%' --> it checks if the string starts with 1 so here 11,2,1, also comes
3. like '%,1,%' --> ,1, anything.
So, better use like '%,1,%' --> this means only 1. since "," is included to separate.. it takes the whole part (,1,) and it is only 1.
Note: If the string is 12,2,3,1 --> Here ending with 1 so specify a or condition with like '%,1' that means string ending with 1.
In the same way, if starts with 1 like '1,%' or if the string is only 1 then '1'
Totally:

 SELECT proid FROM e2 WHERE catid LIKE '%,1,%' OR catid LIKE '1,%' OR catid LIKE '%,1' OR catid = '1' .

I think there might be a more simple query than this.

Regards,

Awesome Coding !! :)

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hi Alokssm

please try this.....query as i have tried and the output will come as you required....

select proid from e2 where catid like '1,%'

Output:

proid
-----------
1
3

Sourabh07

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select proid from e2 where catid like '1,%' it will execute suppose if catid contains 1 instead of 1,3 means it not execute that records

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hi Alokssm

i thought it is fixed that we have to find out after comma....but it is not so....

Now, try this query instead.

select proid from e2 where catid not like '11%' and catid like '1%'

Output:
proid
-----------
1
3

Sourabh07

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hi Alokssm,

the following query also helpfull in finding the exact result........

Query:

select proid from e2 where catid like '1[^1]%'

Output:

proid
-----------
1
3

Sourabh07

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Thanks Vasanthmvp,

You provide the best soluition and my query runs successfully..

Thanks & Regards,
Alokssm

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