UPDATE and REPLACE part of a string

Question

I've got a table with two columns, ID and Value. I want to change a part of some strings in the second column.

Example of Table:

ID            Value
---------------------------------
1             c:\temp\123\abc\111
2             c:\temp\123\abc\222
3             c:\temp\123\abc\333
4             c:\temp\123\abc\444

Now the 123\ in the Value string is not needed. I tried UPDATE and REPLACE:

UPDATE dbo.xxx
SET Value = REPLACE(Value, '%123%', '')
WHERE ID <= 4

When I execute the script SQL Server does not report an error, but it does not update anything either. Why is that?

Answer 1

You don't need wildcards in the REPLACE - it just finds the string you enter for the second argument, so the following should work:

UPDATE dbo.xxx
SET Value = REPLACE(Value, '123', '')
WHERE ID <=4

If the column to replace is type text or ntext you need to cast it to nvarchar

UPDATE dbo.xxx
SET Value = REPLACE(CAST(Value as nVarchar(4000)), '123', '')
WHERE ID <=4

Answer 2

Try to remove % chars as below

UPDATE dbo.xxx
SET Value = REPLACE(Value, '123', '')
WHERE ID <=4

Answer 3

To make the query run faster in big tables where not every line needs to be updated, you can also choose to only update rows that will be modified:

UPDATE dbo.xxx
SET Value = REPLACE(Value, '123', '')
WHERE ID <= 4
AND Value LIKE '%123%'

Answer 4

query:

UPDATE tablename 
SET field_name = REPLACE(field_name , 'oldstring', 'newstring') 
WHERE field_name LIKE ('oldstring%');

Answer 5

You have one table where you have date Code which is seven character something like

"32-1000"

Now you want to replace all

"32-"

With

"14-"

The SQL query you have to run is

Update Products Set Code = replace(Code, '32-', '14-') Where ...(Put your where statement in here)

Answer 6

For anyone want to replace your script.

update dbo.[TABLE_NAME] set COLUMN_NAME= replace(COLUMN_NAME, 'old_value', 'new_value') where COLUMN_NAME like %CONDITION%

Answer 7

CREATE TABLE tbl_PersonalDetail
(ID INT IDENTITY ,[Date] nvarchar(20), Name nvarchar(20), GenderID int);

INSERT INTO Tbl_PersonalDetail VALUES(N'18-4-2015', N'Monay', 2),
                                     (N'31-3-2015', N'Monay', 2),
                                     (N'28-12-2015', N'Monay', 2),
                                     (N'19-4-2015', N'Monay', 2)

DECLARE @Date Nvarchar(200)

SET @Date = (SELECT [Date] FROM Tbl_PersonalDetail WHERE ID = 2)

Update Tbl_PersonalDetail SET [Date] = (REPLACE(@Date , '-','/')) WHERE ID = 2 

Answer 8

you should use the below update query

UPDATE dbo.xxx SET Value=REPLACE(Value,'123\','') WHERE Id IN(1, 2, 3, 4)

UPDATE dbo.xxx SET Value=REPLACE(Value,'123\','') WHERE Id <= 4

Either of the above queries should work.

Answer 9

replace for persian word

UPDATE dbo.TblNews
SET keyWords = REPLACE(keyWords, '-', N'،')

help: dbo.TblNews -- table name

keyWords -- fild name