Solved Search last word in a string (1 Viewer)

nirmal

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Today, 22:13
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Feb 27, 2020
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SQL:
SELECT
    SecondLastValue([Place]) AS SLV,
    Count(1) AS LCountRecords
FROM Place_tbl
GROUP BY SecondLastValue([Place]);
Thank You Sir
A last doubt
There are total of 33996 records in Place_tbl
How to get the aggregate of other places in the table with a parameter of <> "Jalna"
 

Eugene-LS

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How to get the aggregate of other places in the table with a parameter of <> "Jalna"
SQL:
SELECT 
        SecondLastValue([Place]) AS SLV, 
        Count(1) AS LCountRecords
FROM Place_tbl
GROUP BY SecondLastValue([Place])
HAVING Not SecondLastValue([Place])="Jalna";
 

nirmal

Member
Local time
Today, 22:13
Joined
Feb 27, 2020
Messages
82
SQL:
SELECT
        SecondLastValue([Place]) AS SLV,
        Count(1) AS LCountRecords
FROM Place_tbl
GROUP BY SecondLastValue([Place])
HAVING Not SecondLastValue([Place])="Jalna";
Sir actually I require a total count of rest other than Jalna
 

Eugene-LS

Registered User.
Local time
Today, 19:43
Joined
Dec 7, 2018
Messages
481
Sir actually I require a total count of rest other than Jalna
SQL:
SELECT Count(1) AS LCountRecords
FROM Place_tbl
WHERE (((SecondLastValue([Place]))<>"Jalna"));
 

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