Problem With [Left] In SQL Server 2005?
Aug 25, 2006
We have a table that contains a column named "Left". (Yea, I know, this was a bad idea - but it is not within my power to change it.)
Our stored procedures access it using [Left] without any problems in Sql Server 2000 and Sql Server Express Edition. HOWEVER, it does NOT work with SQL Server 2005. It generates the error: Column 'Left' is read-only.
Has anyone seen this? Is there a problem with SQL Server 2005 and reserved words using [ ] notation?
Thanks!
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Oct 28, 2015
I have a table PLACE with a character column CODE
[Code] [nchar](4) NULL
I need to left pad the column with 0 if it is less than 4 characters long and extract the first 2 characters on the left into a new column COUNTY_CODE.
How can I do that in transact SQL?
I tried:
SELECT RIGHT(RTRIM('0000'+ISNULL([Code],'')),4)
FROM [Place]
WHERE [Place Code]='B' and [Code]='627'
And I got 0627. And how do I extract the first 2 characters?
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Dec 5, 2007
Hi,
We are migrating from SQL server 2000 to SQL server 2005
I want ot convert the non-ansi standard query to ansi standard.
My Query is
Non-ANSI standard (has the left outer join symbol *= )
Select *
from vw_rest_rest vw,
prty pa,
adduser ad
where vw.rono *= pa.partyno
and vw.runo *= ad.bkorgid
and pa.ADDRCD = 'Business'
and
Ansi standard (*= repalced left outer join )
select *
FROM VW_REST_REST VW
LEFT OUTER JOIN PRTY PA
ON VW.RUNO = PA.PARTYNO
LEFT OUTER JOIN ADDUSER AD
ON VW.RUNO = AD.BKORGID
where pa.ADDRCD = 'Business'
when i run both the queries on the same database I get different output.
Non Ansi query gives 10029 Rows
Ansi query gives 5625 Rows.
Can somebody tell me where am i going wrong? y its giving different output
If possible can somebody convert the non ansi standard query to ansi standard so that, both the query gives same output
Thanks is advance.
Ranjan
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Oct 24, 2013
How to remove space left to right and right to left
If I give limit >60 for first 60 character; limit 60< second 60 character
Result would be check if space at 60 character if yes remove and go the 59 character check then space remove and 58 character check if there is charater then display
As well as after 60 character to till 120 for right space
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Apr 16, 2008
Anyone know why using
SELECT *
FROM a LEFT OUTER JOIN b
ON a.id = b.id
instead of
SELECT *
FROM a LEFT JOIN b
ON a.id = b.id
generates a different execution plan?
My query is more complex, but when I change "LEFT OUTER JOIN" to "LEFT JOIN" I get a different execution plan, which is absolutely baffling me! Especially considering everything I know and was able to research essentially said the "OUTER" is implied in "LEFT JOIN".
Any enlightenment is very appreciated.
Thanks
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Jan 25, 2015
-- Why is the left table in a LEFT JOIN limited by the where clause on the right table?eg
DECLARE @LeftTable TABLE (LeftID INT NOT NULL IDENTITY(1, 1), LeftValue INT NULL)
INSERT @LeftTable (LeftValue)
VALUES (111)
INSERT @LeftTable (LeftValue)
VALUES (222)
[code]....
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Dec 20, 2006
hello
for MS SQL 2000, i cannot get it
i am having names like AB_12 I want to get all rows with left part similar , AB im that case
SELECT id, name
FROM Users
WHERE LEFT(name, CHARINDEX('_', name) - 1) AS name IN
(
SELECT LEFT(name, CHARINDEX('_', name) - 1) AS ns
FROM Users
GROUP BY LEFT(name, CHARINDEX('_', name) - 1)
HAVING (COUNT(*) > 1)
)
does not work
is there any way to use a variable ?
declare @nm nvarchar
set @nm = SELECT LEFT(name, CHARINDEX('_', name) - 1) AS ns
FROM Users
thank you for helping
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Jul 1, 2015
Normally it is recommended to leave an empty partition on both the front and back ends of a table to avoid data movement when merging/splitting. But I have some questions based on my scenario, which is a table partitioned by a load date, so all records in a partition contain the same date, not a range of dates.
If I use range left, once I switch out the first partition it would become empty, so would there be data movement when I merge it into the next partition? The real issue though is that we will not just be removing the first partition, but "random" partitions throughout the table. Will this work?
If I use range right, when I split the last partition to create a new one it doesn't seem there would be any data movement there either. Am I missing something?
Basically I'm wondering if I should use range left or right. Most recommend using right, but then the boundary value is not the value in the partition. This could potentially result in someone deleting the wrong data if they are not careful. So is there any reason not to use left in my scenario?
I did a test of removing a partition in the middle and it worked just fine; this was using range right. I have about 6 million rows per partition. I also tested splitting at the end and it worked fine. I'll rebuild it with range left and test.
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Mar 30, 2015
SELECT LEFT(InstanceName,CHARINDEX(' ',InstanceName)-1)
FROM ServerDB
Returns the following error.
Invalid length parameter passed to the SUBSTRING function.
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Sep 8, 2015
I need to join 2 tables but the join needs to account for 2 seperate columns.
for example:
select
a. type
a. prod_code
a. prod_type
b. division
from table1 a
left join table2 b
on a. prod_code = b. prod_code
and a. prod_type = b. prod_type
The issue is that you may have only the prod_code or prod_type and null value for the other in table1.
Ideally I want it to check for both then if 1 isn't available then it draws the division of the available. having both or one or the other determines the division it falls under.
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Oct 20, 2015
I have imported a txtfile to access and have some problem making the data appear in my wanted way. I want to shift the cells in the first row one step to the left, i.e to overwrite the zeros with the left value. Then I want to assign the first row values as field names.
Current data
Field1 Field2 Field3 Field4 Field5
0 Name Major Sector Hi
karl
per
anna
Wanted outcome
Name Major Sector Hi
karl
per
anna
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Sep 1, 2005
Hi All,
Im having a problem with a statement i cannot seem to get 2 left joins working at the same time 1 works fine but when i try the second join i get this error:-
Microsoft OLE DB Provider for ODBC Drivers error '80040e14'
[Microsoft][ODBC Microsoft Access Driver] Syntax error (missing operator) in query expression 'children_tutorial.school_id=schools.idx LEFT JOIN regions ON children_tutorial.region_id=region.idx'.
My SQL statment is as follows :-
SELECT children_tutorial.*,schools.schoolname,regions.rname FROM children_tutorial LEFT JOIN schools ON children_tutorial.school_id=schools.idx LEFT JOIN regions ON children_tutorial.region_id=region.idx
I am using an Access database i have tried all sorts to get it working and its driving me mad!! any help would be really appreciated.
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Dec 6, 2004
Hello!
I am trying to run a query on an SQL-server database:
SELECT (Left(Comments, 20)) As shortComment FROM myTable
The Field Comments is a Text field (datatype is text), and I can't get the LEFT function to work with this datatype...
How can I bypass this problem? Can I convert the commentfield to varchar in the QUERY?
noccy
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Nov 1, 2004
I am getting different results with LEFT outer join operator and *= operator. With *= I am getting the expected results. Can anyone look at SQL and tell what I am doing wrong?
SQL with Left Outer join operator:
select CurrentWeekFinMetrics.[Hub+],
WeeklyMetricsFormat.line#,
WeeklyMetricsFormat.MetricsType,
WeeklyMetricsFormat.Metrics,
WeeklyMetricsFormat.Measure, WeeklyMetricsFormat.jobs, case when dataformatchar is not null then
case when IsPrefix = 'Y' then dataformatchar + convert (varchar, CurrentWeekFinMetrics.displayCol )
else convert (varchar, CurrentWeekFinMetrics.displayCol ) + dataformatchar end
else
convert (varchar, CurrentWeekFinMetrics.displayCol )
end
from WeeklyMetricsFormat LEFT JOIN CurrentWeekFinMetrics on (WeeklyMetricsFormat.Line# = CurrentWeekFinMetrics.Line#)
where CurrentWeekFinMetrics.WeekEndingDate = '10/09/04' and CurrentWeekFinMetrics.[Hub+] = 'Amstelveen'
order by CurrentWeekFinMetrics.[Hub+], WeeklyMetricsFormat.Line#
SQL with *= operator
select CurrentWeekFinMetrics.[Hub+],
WeeklyMetricsFormat.line#,
WeeklyMetricsFormat.MetricsType,
WeeklyMetricsFormat.Metrics,
WeeklyMetricsFormat.Measure, WeeklyMetricsFormat.jobs, case when dataformatchar is not null then
case when IsPrefix = 'Y' then dataformatchar + convert (varchar, CurrentWeekFinMetrics.displayCol )
else convert (varchar, CurrentWeekFinMetrics.displayCol ) + dataformatchar end
else
convert (varchar, CurrentWeekFinMetrics.displayCol )
end
from WeeklyMetricsFormat , CurrentWeekFinMetrics
where CurrentWeekFinMetrics.WeekEndingDate = '10/09/04' and CurrentWeekFinMetrics.[Hub+] = 'Amstelveen'
AND (WeeklyMetricsFormat.Line# *= CurrentWeekFinMetrics.Line#)
For Left outer join operator, I am getting 54 rows, *= I am getting 69 rows.
Thanks and Regards
mb
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Apr 15, 2015
I'm attempting to improve a vendor created query.
Here is the original -
SELECT m.lID
FROM Message m
inner join Message_Cc mCC on m.lID=mCC.lMessage and mCC.lOfficeRecipient = 200321
INNER JOIN UserRole d on mCC.szRecipient=d.szRoleName
inner Join Map_UserAtOfficeToRole a2 on a2.lUserRole = d.lid AND d.nRecordStatus = 1
[Code] ....
If I run this without the LEFT OUTER JOIN and the is null statement I get 648 rows. But If I include it I get 0 rows. I can't understand why I get 0 rows with the outer join.
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Aug 3, 2015
I have a query that has a left join with a large partitioned table. The partitioned table has 10s of millions of records, and each partition has about 100,000 records.
The left join is part of an insert that gets a column from the partitioned table, if the column exists. The query contains the partition ID and all other joined columns are part of a non-clustered index.
Through the profiler, I found that there were millions of reads and the execution plan was giving me a table scan on the partitioned table.
I changed the query to do the insert followed by an update with inner join. That did the trick, but it worries me that SQL Server 2014 behaves differently from 2012 or 2008R2, which can make upgrading very time consuming.
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Sep 1, 2014
Looking to improve performance of the following code.
It basically generates future days for each dog. So there is a dog table and a day table with every day.
These 2 table cross join and then fill in missing rows. As time moves i will fill in further future dates but will need the initial insert to be a reasonable query.
All columns are covered by index's but the queries at the end take quite a long time. I would hope for index scan to just point out the missing rows especially on the final query.
How to make the last query as fast as possible.
IF OBJECT_ID('dbo.[AllDates]', 'U') IS NOT NULL
DROP TABLE dbo.[AllDates]
CREATE TABLE dbo.[AllDates] (
[Date] date not null PRIMARY KEY
)
;WITH Dates AS
[Code] .....
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Mar 8, 2015
We have a view with many left joins. The original creators of this view might have been lazy or sloppy, I don't know. I have rewritten the query to proper inner joins where required and also nested left joins.
So rather then the following exemplary fragment
select <many items>
from A
left join B on B.id_A = A.id
left join C on C.id_B = B.idthis now looks like
select <many items>
from A
left join (B
join C on C.id_B = B.id
) on B.id_A = A.id
Compilation time of the original view was 18s, of the new rewritten view 4s. The performance of execution is also better (not counting the compile of course). The results of the query are identical. There are about 30 left joins in the original view.
I can imagine that the optimizer has difficulty with all these left joins. But 14s is quite a big difference. I haven't looked into detail in the execution plans yet. I noticed that in both cases the Reason for Early Termination of Statement Optimization was Time Out.
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Jun 16, 2015
I am trying to erase some erroneous bad data in my table. The description column has a lot of </div>oqwiroiweuru</a> weird data attached to it, i want to keep the data to the left of where the </div> erroneous data begins
update MyTable
set Description = LEFT(Description(CHARINDEX('<',Description)-1)) where myid = 1
that totally works.
update MyTable
set Description = LEFT(Description(CHARINDEX('<',Description)-1)) where myid >= 2
gives me a Invalid length parameter passed to the LEFT or SUBSTRING function. The statement has been terminated error.
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Jun 25, 2015
Is there a way to leave the graphical 'Include Execution Plan' on by default in SSMS? I don't know how many times I run a long-running query, say to myself, "wow, that took a while; I wonder what the execution plan looks like?" only to realize that I left it turned off. Now I have to turn it on, and wait for the query to run again. I'm guessing there's a setting in the options somewhere to always leave it on, but I'm not sure where
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Mar 2, 2015
I have 10 Gb index and disk space only left 5gb .
How can i rebuild index ?
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Oct 7, 2014
Following is the query that I'm running:
create table a (id int, name varchar(10));
create table b(id int, sal int);
insert into a values(1,'John'),(1,'ken'),(2,'paul');
insert into b values(1,400),(1,500);
select *
from a
cross apply( select max(sal) as sal from b where b.id = a.id)b;
Below is the result for the same:
idname sal
1John500
1ken500
2paulNULL
Now I'm not sure why the record with ID 2 is coming using CROSS APPLY, shouldn't it be avoided in case of CROSS APPLY and only displayed when using OUTER APPLY.
One thing that I noticed was that if you remove the Aggregate function MAX then the record with ID 2 is not shown in the output. I'm running this query on SQL Server 2012.
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Apr 7, 2008
Is there any difference between left join and left outer join in sql server 2000?please reply with example if any?
Thanks in advance
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May 14, 2008
Hi,
Whats the diference between a left join and a left outer Join
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Aug 8, 2006
When I installed VS 2005, it installed the default version of SQL Server 2005 Express that ships with Visual Studio 2005 installer media.
How can apply SQL Server 2005 Express SP1 to update this existing instance?
Currently, if I run this query:
SELECT @@version
I get the following:
Microsoft SQL Server 2005 - 9.00.1399.06 (Intel X86) Oct 14 2005 00:33:37 Copyright (c) 1988-2005 Microsoft Corporation Express Edition on Windows NT 5.1 (Build 2600: Service Pack 2)
After applying SP1, I should get 9.00.2047.00.
Should I just go to this link and download & install the SQL Server 2005 Express Edition SP1:
http://msdn.microsoft.com/vstudio/express/sql/download/
Thank you,
Bashman
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Dec 1, 2006
how can I can update a table :: I must get all the left part of a columns part1_part2I dont know how many characterssomething like :UPDATE Users SET Users.Info = left(Users.name, _ ??? or split('_'))in that case Users.name = part1i must update only the one with a '_' and do nothing for the othersthank you
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May 8, 2006
Hi,
I am trying to pad a field from the left with zeros that is defined in a table as "[char] (9)". The field can have from 1 to 9 chars populated. In this instance the value in the field is "133000", which I am trying to pad so that it will be "000133000". I have made several failed attempts to do this, the latest of which was this:
select
right(replicate('0',9)+ convert(varchar(9),a.GS06GRP_CTL_NBR),9)
fromTable1 a
Thanks,
Jeff
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Sep 3, 2007
I installed Visual Studio 2005 Professional then after that was installed and running, I tried to install the the SQL Server 2005 Developer's Edition which installed but I do not get the SQL Server 2005 Studio Manager. I have remove and reinstalled but it never gets installed. Any ideas?
Chuck
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Jan 21, 2008
We have been using Sql Server 2005 Compact Edition 3.1 RDA synchronization method successfully on Sql Server 2000 database. Recently we moved the database to Sql Server 2005, sync doesn't work anymore, it just hangs on one table. On further investigation, we found out that it's the index on that table that causes this. We removed the index, it works fine. We are wondering the root cause, removing the index is not a solution for us. Any thoughts?. Thanks.
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Feb 22, 2008
(1) I have already installed of SQL Server 2005 Developer Edition first.
(1) Can I install visual studio 2008 without the 2005 express edition of SQL server? Will be any problems because I don't have express edition of SQL server? Do I need to install the express edition of SQL server as well?
(3) How to use SQL Server 2005 Developer Edition instance on visual studio 2008?
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Jan 26, 2007
I'm trying to join 2 tables. I thought I was getting the correct results but it turns out I'm not.
My Query:
SELECT IVINVA, IVORDN, IVCSLN, IVRESR, IVCITM, CONVERT(varchar(12),CAST(IVIAMT as money),1) AS ExtPrice, CONVERT(varchar(12),CAST(IVPIVC as money),1) AS DistPrice, IVCSUM, IVQYCS, IVDESC, OIRESR, OIDPCT, CONVERT(varchar(12),CAST(IVPIVC - (OIDPCT / 100 * IVPIVC) as money),1) AS NetPrice FROM INVDET1_TBL LEFT JOIN ORDDIS_TBL ON ORDDIS_TBL.OIORDN = INVDET1_TBL.IVORDN AND ORDDIS_TBL.OIRESR = INVDET1_TBL.IVRESR WHERE IVORDN = '0859919' AND IVINVA = '00324024'
Basically, my problem lies in the seonc condition of the LEFT JOIN. I needed to set the two tables equal my item number, because in some situations I need that logic to get the correct result. It most other cases, that item column in the ORDDIS_TBL is NULL, thus giving me the wrong results. In that case, I would want the JOIN to only be ORDDIS_TBL.OIORDN = INVDET1_TBL.IVORDN, and not include the second part. Is there a way I can condition this with an If statement, If ORDDIS_TBL.OIRESR is Null then do this join, if not, then do this? I'm confused how to get the proper result here.
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Jun 28, 2000
What is the syntax of left triming characters seperated by space.
e.g elsie reed
and i want elsie to go to one field and reed in another field i do not want to use numbers because they don't all have the same numbers but are all seperated by spaces
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Jul 7, 2005
I need to make a left join from the freezefile f, to sped s, instead of having f.studentid = s.id in the where clause. Any help??
select f.studentid, f.studentname, f.sex, fs.mealstatus, s.except, s.lre, r.description, g.testid, g.scale_la, g.scale_ma, t.test_name, t.year
from freezefile f, fsapps fs, sped s, regtb_exception r, gqe_scores g, test_info t
where
f.type = 'ADM'
and
s.except = r.code
and
t.test_name = 'ISTEP'
and
t.year = 2004
and
g.testid = t.testid
and
f.studentid = fs.id
and
f.studentid = s.id
and
f.studentid = g.studentid
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