I'm probably missing something really obvious here, but this has been a bear to search for on Google (Maybe I don't have the right terminology).
I want to replace an unknown value with another value from a temp table. I know the length of the value so my thought was to use underscores as you would in a LIKE statement. The following DOES NOT work however:
UPDATE MyTable
SET Name =
Replace(Name, '__SomeString', TempTable.value + ' SomeString')
FROM MyTable INNER JOIN TempTable
ON Name LIKE TempTable.Name
This is MS SQL 2000 FWIW.
EDIT: To try and clarify it looks like the underscore '_' wildcard that is used in a LIKE statement is taken literally inside of the replace function. Is there another way?
Any thoughts?
UPDATE MyTable
SET Name =
CASE WHEN (Name like '_SomeString')
THEN TempTable.value + SUBSTRING(Name,2,LEN(Name)-1)
ELSE Name END
FROM MyTable INNER JOIN TempTable
ON MyTable.Name = TempTable.Name
WHERE MyTable.Name = 'TheNameToReplace' -- I don't know if it will be for a specific name hence the where...
This will then replace 'SomeString' in the Name field, with the value from TempTable.value
Is this what you were looking for or something else?
Perhaps you can use stuff instead of replace. You need to know the start position in the string where you want to replace the characters and you need to know the length of the expression that is to be replaced. If you don't know that perhaps you can use charindex or patindex to figure that out.
select stuff('A123', 1, 1, 'B ')
Result:
(No column name)
B 123
Would somethi8ng like this work?
UPDATE mytable
SET field1 = 'A' + SUBSTRING(field1,2,LEN(field1))
WHERE LEFT(field1) IN (0,1,2,3,4,5,6,7,8,9)
Apparently it is not possible to use wild cards in the REPLACE function. This is the closest match on SO that I could find: MySQL Search & Replace with WILDCARDS - Query While the link is for MySQL I believe it is true for MS SQL as well.
The other answers here are all creative solutions to the problem, but I ended up going the brute force route.
Related
I am attempting to generate a new field based on an existing field where some of the entries contain different special characters. (ie. *, ') The special characters are at the end of the string, which is either second or third position.
I am NEW to SQL but not new to data. I am using a CASE WHEN statement. I have tried several approaches and several other commands within the CASE statement.
what I want is:
SELECT *
CASE WHEN grde_code_mid LIKE '[*]' THEN 'Remedial'
WHEN grde_code_mid LIKE '[']' THEN 'Continuing'
ELSE NULL
END AS class_type
FROM grde_tble
I keep getting the same error: "FROM keyword not found where expected". I expect to have all 3 returns in the new field.
If you're looking for ' character you should escape it.
Change
WHEN grde_code_mid LIKE '[']' THEN 'Continuing'
by
WHEN grde_code_mid LIKE '['']' THEN 'Continuing'
Have a look at this question: How do I escape a single quote in SQL Server?
There are several issues with your query:
you are missing a comma in the SELECT clause between * and the CASE expression (this is causing the error that you are currently getting)
the bracket notations is only supported in SQL-Server; if you want to match on strings that end with * in a portable manner, you need expression ... LIKE '%*', not LIKE '[*]'
single quotes embedded in a string need to be escaped
SELECT *, other_field FROM ... is not supported on all RDBMS (and, as commented by jarhl, actually isn't standard ANSI SQL notation); you usually need to prefix the * with the table name or alias
Consider:
SELECT
g.*,
CASE
WHEN grde_code_mid LIKE '%*' THEN 'Remedial'
WHEN grde_code_mid LIKE '%''' THEN 'Continuing'
ELSE NULL
END AS class_type
FROM grde_tble g
Thank you all. I am still getting the hang of the language and proper terms. I am using Oracle DB. Also, yes, I did need to reference the table in the select statement (SELECT tablename.*). Found a work around:
CASE WHEN regexp_like(grde_code_mid, '[*]') THEN 'Remedial'
WHEN regexp_like(grde_code_mid, '['']') THEN 'Continuing'
ELSE NULL END AS special_class
I try to create a query from an Oracle DB.
that is, SELECT FROM and WHERE.
the column "ORG" is centered and always has 4 letters. I would like to filter that on one specific Item/ value.
I already have WHERE ORG = 'HHAH'
or with SBSTRG (ORG ...:
somehow nothing works.
Does somebody has any idea?
I have values of ' HHAH ' instead of 'HHAH' in the column. There are blanks befor and after the value
You could remove the leading and trailing spaces with the trim() function:
WHERE TRIM(ORG) = 'HHAH'
Using a function on the column value will prevent any index on that column being used (as will like with a leading wildcard); unless you add a function-based index for the trimmed value there isn't much you can do about that.
Do you need the LIKE operator:
WHERE ORG LIKE '%HHAH%'
or
WHERE ORG LIKE '%' || 'HHAH' || '%'
to search for values conatining 'HHAH'?
I would recommend fixing the data:
update t
set org = trim(org);
I see no reason to be storing spaces in the name of an org. If you need spaces for reporting purposes, put them there.
I have imported a table containing roughly 100,000 records, some of which need to be removed. I'd like to identify and remove any and all records where a particular field (called MyQuery) contains words surrounded in quotation marks, but if there are only TWO quotation marks in the field.
For example I would like to remove
"This is a test"
--but not--
"This is "a" test"
Thank you kindly for any assistance
My apologies to all of you for the terrible structure and clarity of my question. I think I was able to get to the answer based on your suggestions though!
select * from <table> where LEN(myquery) - LEN(REPLACE(myquery,'"','')) = 2 and myquery like '"%"'
Thank you all so much - you're the best!
You can do it like this:
DELETE
-- SELECT * -- To test first!
FROM YourTable
WHERE LEN(MyQuery) - LEN(REPLACE(MyQuery, '"', '')) = 2;
AND MyQuery LIKE '"%"'
You are basically comparing the length if the field after removing all quotes. This is an easy way to determine the number of occurrences of a specific character.
After your comment, I added another the condition so the quotes are always surrounding the field.
You can use the following query to delete all records that contain '"', excluding those that start and end with '"':
DELETE From <TABLENAME> Where COL like '%"%' and (COL not like '"%' Or COL not like '%"')
Or, you can use: Here we delete all records having more than 2 occurrences of "
DELETE FROM <TABLENAME> WHERE (LEN(COL)-len(replace(COL,'"',''))) > 2
My wording may not be correct so the solution could be on this site. In which case, pointing me to the solution would be great. Anyway, I am doing a search and need to do compares against the string and value that is in a column. It has to go beyond LIKE. For instance, I want to take the column put it in lower case and take out all spaces before comparing it to the string (which is too put in lower case and all spaces are gone). I want to do this without modifying the contents of the column. So just for the compare. Then if the compare evaluates to true take the original contents out of the column (not in the lower case and no spaces form). This may be to specific. Is it possible to do this, if so, any code sample would help. So I have something like this now:
SELECT *
FROM [MY_TABLE]
WHERE (
[MY_COLUMN] LIKE My_Search_String
)
so the My_Search_String is already formatted. I just need to format the [MY_COLUMN] without permanently modifying its contents. Any help is appreciated. Thank you!
Upper and lower case only make a difference if your default collations or column collations are case sensitive. In any case, you seem to want:
SELECT *
FROM [MY_TABLE]
WHERE lower(replace([MY_COLUMN], ' ', '')) LIKE lower(replace(My_Search_String, ' ', ''))
I think below query will solve your problem.!
IF EXISTS(select * from [MY_TABLE] where LOWER(REPLACE([MY_COLUMN],' ','')) like 'My_Search_String')
INSERT INTO Another_Table
select * from [MY_TABLE] where LOWER(REPLACE([MY_COLUMN],' ','')) like 'My_Search_String'
Does anyone have a LIKE pattern that matches whole words only?
It needs to account for spaces, punctuation, and start/end of string as word boundaries.
I am not using SQL Full Text Search as that is not available. I don't think it would be necessary for a simple keyword search when LIKE should be able to do the trick. However if anyone has tested performance of Full Text Search against LIKE patterns, I would be interested to hear.
Edit:
I got it to this stage, but it does not match start/end of string as a word boundary.
where DealTitle like '%[^a-zA-Z]pit[^a-zA-Z]%'
I want this to match "pit" but not "spit" in a sentence or as a single word.
E.g. DealTitle might contain "a pit of despair" or "pit your wits" or "a pit" or "a pit." or "pit!" or just "pit".
Full text indexes is the answer.
The poor cousin alternative is
'.' + column + '.' LIKE '%[^a-z]pit[^a-z]%'
FYI unless you are using _CS collation, there is no need for a-zA-Z
you can just use below condition for whitespace delimiters:
(' '+YOUR_FIELD_NAME+' ') like '% doc %'
it works faster and better than other solutions. so in your case it works fine with "a pit of despair" or "pit your wits" or "a pit" or "a pit." or just "pit", but not works for "pit!".
I think the recommended patterns exclude words with do not have any character at the beginning or at the end. I would use the following additional criteria.
where DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like 'pit[^a-z]%' OR
DealTitle like '%[^a-z]pit'
I hope it helps you guys!
Surround your string with spaces and create a test column like this:
SELECT t.DealTitle
FROM yourtable t
CROSS APPLY (SELECT testDeal = ' ' + ISNULL(t.DealTitle,'') + ' ') fx1
WHERE fx1.testDeal LIKE '%[^a-z]pit[^a-z]%'
If you can use regexp operator in your SQL query..
For finding any combination of spaces, punctuation and start/end of string as word boundaries:
where DealTitle regexp '(^|[[:punct:]]|[[:space:]])pit([[:space:]]|[[:punct:]]|$)'
Another simple alternative:
WHERE DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like '[^a-z]pit[^a-z]%' OR
DealTitle like '%[^a-z]pit[^a-z]'
This is a good topic and I want to complement this to someone how needs to find some word in some string passing this as element of a query.
SELECT
ST.WORD, ND.TEXT_STRING
FROM
[ST_TABLE] ST
LEFT JOIN
[ND_TABLE] ND ON ND.TEXT_STRING LIKE '%[^a-z]' + ST.WORD + '[^a-z]%'
WHERE
ST.WORD = 'STACK_OVERFLOW' -- OPTIONAL
With this you can list all the incidences of the ST.WORD in the ND.TEXT_STRING and you can use the WHERE clausule to filter this using some word.
You could search for the entire string in SQL:
select * from YourTable where col1 like '%TheWord%'
Then you could filter the returned rows client site, adding the extra condition that it must be a whole word. For example, if it matches the regex:
\bTheWord\b
Another option is to use a CLR function, available in SQL Server 2005 and higher. That would allow you to search for the regex server-side. This MSDN artcile has the details of how to set up a dbo.RegexMatch function.
Try using charindex to find the match:
Select *
from table
where charindex( 'Whole word to be searched', columnname) > 0