sql query order output
I have table called Property
and the data is like
pid city state state_abb address1 address2
x1 NewCity NHANy NH xxxx gfg
x2 Gloucester Manchestar MA newAde xxxx
x3 OtherC NewYork NY yyyy
I want a query with search keywo开发者_JS百科rd to display order like (state or state_abb), city, address1, adress2
For ex:
If I search keyword with New The result output should
pid city state state_abb address1 address2
x3 OtherC NewYork NY yyyy
x1 NewCity NHANy NH xxxx gfg
x2 Gloucester Manchester MA newAde xxxx
I don't want unmatched rows.I want display only matched rows.
Thanks in advance
It think the question is how to order the results, not how to perform the search. Try this as your ORDER BY clause
order by
case when charindex('NEW',state) >0 then 1000 else 0 end desc,
case when charindex('NEW',city) >0 then 100 else 0 end desc,
case when charindex('NEW',address) >0 then 10 else 0 end desc
Where clause to only get matching rows
select * from Property where
city like '%New%' or
state like '%New%' or
address1 like '%New%' or
address2 like '%New%'
You could try something like this:
SELECT pid, city, state, stat_abb, address1, address2
from property
where city like '%New%'
OR state like '%New%'
OR address1 like '%New%'
This should get you started:
select * from Property where
city like '%New%' or
state like '%New%' or
address1 like '%New%' or
address2 like '%New%'
I assume you don't want to look at the pid
column.
select *
from Property
order by case
when state like 'New%' then 1
when state_abb like 'New%' then 1
when city like 'New%' then 2
when address1 like 'New%' then 3
when address2 like 'New%' then 4
end
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