SQL select count of all rows when using IF ELSE - sql

Question

I have a situation when I use IF ELSE statement in SQL.
#searchString nvarchar(50),
#languageId int,
#count int,
#id int
IF(#count IS NOT NULL)
SELECT TOP (#count) Id, value
FROM TABLE1
WHERE (Id IN
(SELECT Id
FROM TABLE2
WHERE (Id = #id))) and languageId = #languageId AND value LIKE '%' + #searchString + '%'
ORDER BY value
ELSE
SELECT Id, value
FROM TABLE1
WHERE (Id IN
(SELECT Id
FROM TABLE2
WHERE (Id = #id))) and languageId = #languageId AND value LIKE '%' + #searchString + '%'
ORDER BY value
I would like to return number of all rows using
count(*) over() (or something similar)
(as I return only TOP count records for now), like it is answered here:
How to return total number of records with TOP * select
BUT: I wouldn't return this value for every instance, but I would like to return count just once.
Is there a way to do this with one query, or I have to write a separate query for this?
Any hint would be greatly appreciated!
EDIT: using SQL server 2008 r2.

Answer 1

I suggest you to use INNER JOIN, for Exemple with scripts:
{
SCRIPT
SELECT
l.localite_id,
p.patient_id,
p.patient_date_naissance,
p.patient_gsm, p.patient_tel
FROM
patient p
INNER JOIN localite l
ON l.localite_id = p.localite_id
INNER JOIN ville v
ON v.ville_id = l.ville_id
INNER JOIN gouvernorat g
ON v.gouv_id = g.gouv_id
INNER JOIN pays pays
ON pays.pays_id = g.pays_id
IF NEQ #patient_gsm# -1
WHERE p.patient_gsm like #patient_gsm#
AND p.patient_code like #patient_code#
ENDIF
END
}