Sql ошибка 4104

Even MienEven Mien

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It’s probably a typo. Look for the places in your code where you call [schema].[TableName] (basically anywhere you reference a field) and make sure everything is spelled correctly.

Personally, I try to avoid this by using aliases for all my tables. It helps tremendously when you can shorten a long table name to an acronym of it’s description (i.e. WorkOrderParts -> WOP), and also makes your query more readable.

Edit: As an added bonus, you’ll save TONS of keystrokes when all you have to type is a three or four-letter alias vs. the schema, table, and field names all together.

answered Oct 15, 2008 at 21:57

You probably have a typo. For instance, if you have a table named Customer in a database named Sales, you could refer to it as Sales..Customer (although it is better to refer to it including the owner name (dbo is the default owner) like Sales.dbo.Customer.

If you typed Sales…Customer, you might have gotten the message you got.

answered Oct 15, 2008 at 21:51

If you are sure that it is not a typo spelling-wise, perhaps it is a typo case-wise.

What collation are you using? Check it.

answered Oct 15, 2008 at 22:02

When updating tables make sure you do not reference the field your updating via the alias.

I just had the error with the following code

update [page] 
set p.pagestatusid = 1
from [page] p
join seed s on s.seedid = p.seedid
where s.providercode = 'agd'
and p.pagestatusid = 0

I had to remove the alias reference in the set statement so it reads like this

update [page] 
set pagestatusid = 1
from [page] p
join seed s on s.seedid = p.seedid
where s.providercode = 'agd'
and p.pagestatusid = 0

answered Jun 20, 2012 at 0:48

Adding table alias in front Set field causes this problem in my case.

Right

Update Table1
Set SomeField = t2.SomeFieldValue 
From Table1 t1 
Inner Join Table2 as t2
    On t1.ID = t2.ID

Wrong

Update Table1
Set t1.SomeField = t2.SomeFieldValue 
From Table1 t1 
Inner Join Table2 as t2
    On t1.ID = t2.ID

answered Dec 15, 2011 at 14:35

Mine was putting the schema on the table Alias by mistake:

SELECT * FROM schema.CustomerOrders co
WHERE schema.co.ID = 1  -- oops!

answered Feb 15, 2013 at 20:14

answered Jun 26, 2017 at 17:19

I actually forgot to join the table to the others that’s why i got the error

Supposed to be this way:

  CREATE VIEW reserved_passangers AS
  SELECT dbo.Passenger.PassName, dbo.Passenger.Address1, dbo.Passenger.Phone
  FROM dbo.Passenger, dbo.Reservation, dbo.Flight
  WHERE (dbo.Passenger.PassNum = dbo.Reservation.PassNum) and
  (dbo.Reservation.Flightdate = 'January 15 2004' and Flight.FlightNum =562)

And not this way:

  CREATE VIEW reserved_passangers AS
  SELECT dbo.Passenger.PassName, dbo.Passenger.Address1, dbo.Passenger.Phone
  FROM dbo.Passenger, dbo.Reservation
  WHERE (dbo.Passenger.PassNum = dbo.Reservation.PassNum) and
  (dbo.Reservation.Flightdate = 'January 15 2004' and Flight.FlightNum = 562)

answered Sep 4, 2017 at 23:17

Error Code

FROM                
    dbo.Category C LEFT OUTER JOIN           
    dbo.SubCategory SC ON C.categoryID = SC.CategoryID AND C.IsActive = 'True' LEFT OUTER JOIN          
    dbo.Module M ON SC.subCategoryID = M.subCategoryID AND SC.IsActive = 'True' LEFT OUTER JOIN          
    dbo.SubModule SM ON M.ModuleID = SM.ModuleID AND M.IsActive = 'True' AND SM.IsActive = 'True' LEFT OUTER JOIN 
    dbo.trainer ON dbo.trainer.TopicID =dbo.SubModule.subModuleID 

Solution Code

 FROM                
    dbo.Category C LEFT OUTER JOIN           
    dbo.SubCategory SC ON C.categoryID = SC.CategoryID AND C.IsActive = 'True' LEFT OUTER JOIN          
    dbo.Module M ON SC.subCategoryID = M.subCategoryID AND SC.IsActive = 'True' LEFT OUTER JOIN          
    dbo.SubModule SM ON M.ModuleID = SM.ModuleID AND M.IsActive = 'True' AND SM.IsActive = 'True' LEFT OUTER JOIN 
    dbo.trainer ON dbo.trainer.TopicID = SM.subModuleID 

as you can see, in error code, dbo.SubModule is already defined as SM, but I am using dbo.SubModule in next line, hence there was an error.
use declared name instead of actual name. Problem solved.

answered Jan 17, 2018 at 6:33

answered Mar 9, 2018 at 19:40

I was getting this error and just could not see where the problem was. I double checked all of my aliases and syntax and nothing looked out of place. The query was similar to ones I write all the time.

I decided to just re-write the query (I originally had copied it from a report .rdl file) below, over again, and it ran fine. Looking at the queries now, they look the same to me, but my re-written one works.

Just wanted to say that it might be worth a shot if nothing else works.

answered Feb 5, 2019 at 14:52

I faced this problem and solved it but there is a difference between your and mine code. In spite of I think you can understand what is «the multi-part identifier could not be bound»

When I used this code

 select * from tbTest where email = sakira@gmail.com

I faced Multi-part identifier problem

but when I use single quotation for email address It solved

 select * from tbTest where email = 'sakira@gmail.com'

answered Feb 11, 2020 at 5:34

answered Jun 9, 2021 at 9:19

Mine worked after removing square brackets in a SUBSTRING method. I changed from

SUBSTRING([dbo.table].[column],15,2)

to

SUBSTRING(dbo.table.column,15,2)

answered Jul 7, 2021 at 19:46

answered Jun 21, 2022 at 14:36

I was using an alias but that alias I was not using in select

It is important to use an alias when data is coming from another database and use that alias with your select statement with fields

Example
Database_Name.dbo.Table_Name as Alias_Name with(NOLOCK)
ON
Join Condition

in Select Statement Alias_Name.Columns_Name

answered Apr 28 at 7:58