Hi, I have a database with table name as test in that i have 6 colums they are name varchar (20) address varchar (20) position varchar (20) nametype1 varchar (20) nametype2 varchar (20) nameval varchar(20) now in the nametype1 and nametype2 there are values like nametype1 nametype2 "AB" "BA" "BB" "BB" "AA" "AA" "BA" "AB" now depending upon the combination i want to assign value to the thrid field that is nameval like example below nametype1 nametype2 nameval "AB" "BA" 1 "BB" "BB" 2 "AA" "AA" 2 "BA" "AB" 1 please suggest query in sql which i can run to do this . Regards
Hi, You do not specify what is the logic for generating the values for the nameval column, but based on your example seems it is 2 when the values are equal and 1 when they are different. Based on that assumption, here is a query that will do the update: update test set nameval = (case when nametype1 = nametype2 then 2 else 1 end) Regards, Plamen Ratchev http://www.SQLStudio.com
[quoted text, click to view] On Jan 30, 6:47 pm, "Plamen Ratchev" <Pla...@SQLStudio.com> wrote: > Hi, > > You do not specify what is the logic for generating the values for the > nameval column, but based on your example seems it is 2 when the values are > equal and 1 when they are different. > > Based on that assumption, here is a query that will do the update: > > update test > set nameval = (case when nametype1 = nametype2 then 2 else 1 end) > > Regards, > > Plamen Ratchev http://www.SQLStudio.com hi i have an update to the query , please suggest if the table structure and results are below then what should i run for no of fileds name varchar (20) address varchar (20) position varchar (20) nametype1 varchar (20) nametype2 varchar (20) nametype3 varchar(20) nametype4 varchar(20) nameval varchar(20) nameval1 varchar(20) nameval2 varchar(20) nameval3 varchar(20) now in the nametype1 and nametype2 there are values like nametype1 nametype2 nametype3 nametype4 "AB" "BA" "BB" "BB" "AA" "AA" "BA" "AB" "AB" "BA" "BB" "BB" "AA" "AA" "BA" "AB" now depending upon the combination i want to assign value to the thrid field that is nameval like example below nametype1 nametype2 nameval "AB" "BA" 1 "AA" "AA" 2 "AB" "BA" 1 "AA" "AA" 2 nametype1 nametype3 nameval1 "AB" "BB" 1 "AA" "BA" 1 "AB" "BB" 1 "AA" "BA" 1 nametype1 nametype4 nameval2 "AB" "BB" 1 "AA" "AB" 1 "AB" "BB" 1 "AA" "AB" 1 please suggest query in sql which i can run to do this also i would like to know is it possible to have some kind of loop which can check each nametype with other like the combination above please suggest. Regards
[quoted text, click to view] On Jan 31, 11:42 am, "Plamen Ratchev" <Pla...@SQLStudio.com> wrote: > I am still confused about the logic and the purpose of this, but since it > seems to follow the same pattern (when equal then 2 else 1), here it is (you > just keep repeating the same for the other "nameval" columns): > > update test > set nameval = (case when nametype1 = nametype2 then 2 else 1 end), > nameval1 = (case when nametype1 = nametype3 then 2 else 1 end), > nameval2 = (case when nametype1 = nametype4 then 2 else 1 end) > > Also, you can create those "nameval" columns as computed columns and then > you do not have to run the update statements. Something like this: > > create table test( > name varchar (20), > address varchar (20), > position varchar (20), > nametype1 varchar (20), > nametype2 varchar (20), > nametype3 varchar(20), > nametype4 varchar(20), > nameval as (case when nametype1 = nametype2 then 2 else 1 end), > nameval1 as (case when nametype1 = nametype3 then 2 else 1 end), > nameval2 as (case when nametype1 = nametype4 then 2 else 1 end)) > > insert into test (nametype1, nametype2, nametype3, nametype4) values ('AA', > 'AA', 'AB', 'BA') > > select * from test > > drop table test > > Perhaps the table should be normalized too, but since no requirements are > given I do not want to guess... > > HTH, > > Plamen Ratchev http://www.SQLStudio.com is it possible to run the query in existing table , please suggest how and also if there more than 20 nametypes is there query which can loop through all the name types and do the job. please suggest any reference too if any. Regards
Please post DDL, so that people do not have to guess what the keys, constraints, Declarative Referential Integrity, data types, etc. in your schema are. Sample data is also a good idea, along with clear specifications. It is very hard to debug code when you do not let us see it. Even your narrative is wrong. Did you know that double quotes are not used for string values in SQL? Have you heard of ISO-11179 rules for data element names? Absurd things like "name_type_1" look like a repeating group in violationof 1NF which will lead to some really horrible kludges and a loss of data integrity. Please try again and pretend that you have to work from these specs without any prior knowledge -- we do SQL, not mind-reading :)
I am still confused about the logic and the purpose of this, but since it seems to follow the same pattern (when equal then 2 else 1), here it is (you just keep repeating the same for the other "nameval" columns): update test set nameval = (case when nametype1 = nametype2 then 2 else 1 end), nameval1 = (case when nametype1 = nametype3 then 2 else 1 end), nameval2 = (case when nametype1 = nametype4 then 2 else 1 end) Also, you can create those "nameval" columns as computed columns and then you do not have to run the update statements. Something like this: create table test( name varchar (20), address varchar (20), position varchar (20), nametype1 varchar (20), nametype2 varchar (20), nametype3 varchar(20), nametype4 varchar(20), nameval as (case when nametype1 = nametype2 then 2 else 1 end), nameval1 as (case when nametype1 = nametype3 then 2 else 1 end), nameval2 as (case when nametype1 = nametype4 then 2 else 1 end)) insert into test (nametype1, nametype2, nametype3, nametype4) values ('AA', 'AA', 'AB', 'BA') select * from test drop table test Perhaps the table should be normalized too, but since no requirements are given I do not want to guess... HTH, Plamen Ratchev http://www.SQLStudio.com
Yes, the query will update an existing table, just keep adding lines for each column, like this: update test set nameval = (case when nametype1 = nametype2 then 2 else 1 end), nameval1 = (case when nametype1 = nametype3 then 2 else 1 end), nameval2 = (case when nametype1 = nametype4 then 2 else 1 end), nameval3 = (case when nametype1 = nametype5 then 2 else 1 end), ... As for you reference to do the update in a loop, then this is doable via dynamic SQL. Here is a quick sketch of how it may look (by adjusting the number 10 you will get different number of columns, beware of the 4000 character limit on the SQL string): DECLARE @sql nvarchar(4000), @count int, @numcolumns int SELECT @sql = 'update test set ', @count = 1, @numcolumns = 10 WHILE @count <= @numcolumns BEGIN IF @count = 1 SELECT @sql = @sql + 'nameval = (case when nametype1 = nametype' + CAST(@count + 1 as nvarchar) + ' then 2 else 1 end)' ELSE SELECT @sql = @sql + ', nameval' + CAST(@count - 1 as nvarchar) + ' = (case when nametype1 = nametype' + CAST(@count + 1 as nvarchar) + ' then 2 else 1 end)' SELECT @count = @count + 1 END EXEC(@sql) Erland Sommarskog has an excellent guide on dynamic SQL at http://www.sommarskog.se/dynamic_sql.html. I would recommend reading it before jumping into using dynamic SQL. Also, please read the comment from Celko, he is correct that without posting DLL and specifications it is very difficult to get good answers. Regards, Plamen Ratchev http://www.SQLStudio.com
[quoted text, click to view] Tradeorganizer wrote: > I have a database with table name as test in that i have 6 colums > they are > > name varchar (20) > address varchar (20) > position varchar (20) > nametype1 varchar (20) > nametype2 varchar (20) > nameval varchar(20) > > now in the nametype1 and nametype2 there are values like > nametype1 nametype2 > "AB" "BA" > "BB" "BB" > "AA" "AA" > "BA" "AB" > > now depending upon the combination i want to assign value to the thrid > field that is nameval like example below > > nametype1 nametype2 nameval > "AB" "BA" 1 > "BB" "BB" 2 > "AA" "AA" 2 > "BA" "AB" 1 > > please suggest query in sql which i can run to do this .
A couple of things that others have touched on, but not made quite this explicit: /Why/ should nametype1 = "AB" and nametype2 = "BA" lead to nameval = 1? What is the general rule that you want to apply? Why do you have more than 20 nametypes? Please give serious consideration to splitting this table into two tables, e.g. [table1] person_id, name, address, position 1, 'John Doe', '123 Cherry Lane', 'Regional Manager' 2, 'Thomas Atkins', '987 Easy Street', 'President' [table2] person_id, nameindex, nametype, nameval 1, 1, 'AB', 1 1, 2, 'BA', 1 2, 1, 'BB', 2
Thanks for great help , yes its working for me. Thanks to all for taking time to explain me. Regards [quoted text, click to view] Ed Murphy wrote: > Tradeorganizer wrote: > > > I have a database with table name as test in that i have 6 colums > > they are > > > > name varchar (20) > > address varchar (20) > > position varchar (20) > > nametype1 varchar (20) > > nametype2 varchar (20) > > nameval varchar(20) > > > > now in the nametype1 and nametype2 there are values like > > nametype1 nametype2 > > "AB" "BA" > > "BB" "BB" > > "AA" "AA" > > "BA" "AB" > > > > now depending upon the combination i want to assign value to the thrid > > field that is nameval like example below > > > > nametype1 nametype2 nameval > > "AB" "BA" 1 > > "BB" "BB" 2 > > "AA" "AA" 2 > > "BA" "AB" 1 > > > > please suggest query in sql which i can run to do this . > > A couple of things that others have touched on, but not made > quite this explicit: > > /Why/ should nametype1 = "AB" and nametype2 = "BA" lead to > nameval = 1? What is the general rule that you want to apply? > > Why do you have more than 20 nametypes? Please give serious > consideration to splitting this table into two tables, e.g. > > [table1] > person_id, name, address, position > 1, 'John Doe', '123 Cherry Lane', 'Regional Manager' > 2, 'Thomas Atkins', '987 Easy Street', 'President' > > [table2] > person_id, nameindex, nametype, nameval > 1, 1, 'AB', 1 > 1, 2, 'BA', 1 > 2, 1, 'BB', 2 > 2, 2, 'BB', 2
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