I want to update a column that contains a string like 12,43,433 I want to only replace 43 with another number say 54 so that the column value becomes 12,54,433.
How can I do that??
You can use REPLACE() function like this:
UPDATE YourTable a
SET a.StringColumn = REPLACE(a.StringColumn,',43,',',54,')
WHERE a.StringColumn like '%,43,%'
Storing lists as strings is a very bad idea. SQL has a great data structure for storing lists. It is called a table, not a string. The proper way to store lists is to use a junction table.
Sometimes we are stuck with other people's really bad design decisions. If so, you can do:
update t
set col = trim(replace(',' || col || ',', ',43,', ',54,'), ',')
where ',' || col || ',' like '%,43,%';
Notes:
This works regardless of where the "43" appears in the string (including at the beginning and end).
This maintains the format of the string with no commas at the beginning and end.
This only attempts to update rows that have the particular elements in the list.
Use of such a query should really be a stopgap while you figure out how to fix the data structure.
Related
I have a lot of character varying records in this format: {'address': 'New Mexico'}.
I would like to update all those columns to have it like this: New Mexico.
I've been investigating how to do it, and it could be with regexp, but I don't know how to make for all columns in a table, and I never used regex in PostgreSQL before.
I have an idea that is something like this:
SET location = regexp_replace(field, 'match pattern', 'replace string', 'g')
Valid JSON literals require double-quotes where your sample displays single quotes. Maybe you can fix that upstream?
To repair (assuming there are no other, unrelated single-quotes involved):
UPDATE web_scraping.iws_informacion_web_scraping
SET iws_localizacion = replace(iws_localizacion, '''', '"')::json ->> 'address'
WHERE iws_id = 3678
AND iws_localizacion IS DISTINCT FROM replace(iws_localizacion, '''', '"')::json ->> 'address';
The 2nd WHERE clause prevents updates to rows that wouldn't change. See:
How do I (or can I) SELECT DISTINCT on multiple columns?
Optional if such cases can be excluded.
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 know I'm close to figuring this out but need a little help. What I'm trying to do is all grab a column from a particular table, but chop off the first 4 characters. For example if in a column the value is "KPIT08L", the result I was is 08L. Here is what I have so far but not getting the desired results.
SELECT LEFT(FIELD_NAME, 4)
FROM TABLE_NAME
First up, left will give you the leftmost characters. If you want the characters starting at a specific location, you need to look into mid:
select mid (field_name,5) ...
Secondly, if you value performance,portability and scalability at all, this sort of "sub-column" manipulation should generally be avoided. It's usually far easier (and faster) to patch columns together than to split them apart.
In other words, keep the first four characters in their own column and the rest in a separate column, and do your selects on the relevant one. If you're using anything less than a full column, then it's technically not one attribute of the row.
Try with
SELECT MID(FIELD_NAME, 5) FROM TABLE_NAME
Mid is very powerfull, it let you select the starting point and all the remainder, or,
if specified, the length desidered as in
SELECT MID(FIELD_NAME, 5, 2) FROM TABLE_NAME ' gives 08 in your example text
SELECT RIGHT(FIELD_NAME,LEN(FIELD_NAME)-4)
FROM TABLE_NAME;
If it is for a generic string then the above one will work...
Don't have Access at my current location, but please try this.
SELECT RIGHT(FIELD_NAME, LEN(FIELD_NAME)-4)
FROM TABLE_NAME
The LEFT(FIELD_NAME, 4) will return the first 4 caracters of FIELD_NAME.
What you need to do is :
SELECT MID(FIELD_NAME, 5)
FROM TABLE_NAME
If you have a FIELD_NAME of 10 caracters, the function will return the 6 last caracters (chopping the first 4)!
I have a table in oracle with a column with comma separated values. What i need is when a user enters a value and if that value is present in any of the rows, it should be removed.
eg.
COL
123,234
56,123
If user enters 123, the 1st column should have only 234 and second row should have only 56.
How do we do this in oracle??
Please help
Thanks
delete from yourtable t
where
instr(','||t.col||',', '123') > 0
You can replace '123' with a parameter if you like.
But a better way would be not to store comma separated values, and create a detail table instead. If you need to look for a specific value within a comma separated list, you cannot make use of indices, amongst other limitations.
[edit]
Misunderstood the question. You meant this:
update YourTable t
set
t.col = substr(substr(replace(','||t.col||',', ',123,', ','), 2), -2)
where
instr(','||t.col||',', '123') > 0
Add ',' before and after to match items at the beginning or end of the value.
Replace using the value ',123,' (within comma's) to prevent accidentally matching 1234 too.
Use substr twice to remove the first and last character (the added commas)
Use instr in the where to prevent updating records that don't need to be updated (better performance).
try this :
UPDATE t
SET col = REPLACE(REPLACE(col, '&variable', ''), ',', '') FROM t ;
I have a table with a field that denotes whether the data in that row is valid or not. This field contains a string of undetermined length. I need a query that will only pull out rows where all the characters in this field are N. Some possible examples of this field.
NNNNNNNNNNNNNNNNNNN
NNNNNNNNNNNNNNNNNNNNNNN
NNNNNEEEENNNNNNNNNNNN
NNNNNOOOOOEEEENNNNNNNNNNNN
Any suggestions on a postcard please.
Many thanks
This should do the trick:
SELECT Field
FROM YourTable
WHERE Field NOT LIKE '%[^N]%' AND Field <> ''
What it's doing is a wildcard search, broken down:
The LIKE will find records where the field contains characters other than N in the field. So, we apply a NOT to that as we're only interested in records that do not contain characters other than N. Plus a condition to filter out blank values.
SELECT *
FROM mytable
WHERE field NOT LIKE '%[^N]%'
I don't know which SQL dialect you are using. For example Oracle has several functions you may use. With oracle you could use condition like :
WHERE LTRIM(field, 'N') = ''
The idea is to trim out all N's and see if the result is empty string. If you don't have LTRIM, check if you have some kind of TRANSLATE or REPLACE function to do the same thing.
Another way to do it could be to pick length of your field and then construct comparator value by padding empty string with N. Perhaps something like:
WHERE field = RPAD('', field, 'N)
Oracle pads that empty string with N's and picks number of pad characters from length of the second argument. Perhaps this works too:
WHERE field = RPAD('', LENGTH(field), 'N)
I haven't tested those, but hopefully that give you some ideas how to solve your problem. I guess that many of these solutions have bad performance if you have lot of rows and you don't have other WHERE conditions to select proper index.