I would like to remove all occurrences of a certain string pattern across all fields in a table.
For example, find all occurrences of the pattern "<XY>*<Xy>" where * represents any possible configuration of characters. I want to remove just those substrings and leave the remainder of the string intact.
This is an example of what I would like to use as my SQL command, but of course this doesn't work:
UPDATE Table SET Field = replace(Field, '<XY>*<Xy>', '');
What is the solution?
Here is an option which attempts to splice around the <XY>...</XY> tags:
UPDATE yourTable
SET Field = SUBSTR(Field, 1, INSTR(Field, '<XY>') - 1) ||
SUBSTR(Field, INSTR(Field, '</XY>') + 5)
WHERE Field LIKE '%<XY>%</XY>%'
It updates fields containing this pattern with the concatenation of everything coming before the first <XY> and everything coming after the second </XY>.
Note that I used <XY>...</XY> rather than what you had originally, because INSTR() is not case sensitive, and both tags would appear as being the same thing.
Demo
The demo is for MySQL but the sytnax is almost identical to SQLite.
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 have a column like:
Values
111.111.111-Dummy
111.111.111-Dummy2
111.111.111-X
222.222.222-Dummy
222.222.222-Dummy2
333.333.333-Dummy
333.333.333-Dummy2
333.333.333-X
I need to find the numbers that do not have an entry with "-X" in the end.
So in this scenario the query should show: 222.222.222.
My idea so far was to first trim the results to only have the numbers part (or everything before the '-')
But I don't know what to do next. How can I find entries that don't match in the same column and same table?
select substr(values_, 1, instr(values_, '-') - 1) as numbers
from {your-table}
group by substr(values_, 1, instr(values_, '-') - 1)
having count(case when values_ like '%-X' then 1 end) = 0;
values is a reserved keyword in Oracle, and therefore it can't be an identifier (such as a column name); I changed it by adding a trailing underscore.
Note that this assumes all "values" are followed by a dash and a (possibly empty) string. If you may also have values like 111.11.1111 (with no dash at the end) then the query must be modified slightly, but I assumed there aren't any - otherwise you should have included one or two in your sample.
Use not like in a having clause:
select substring_index(values, '-', 1)
from t
group by substring_index(values, '-', 1)
having sum(values like '%-x') = 0;
I am trying to figure out some commands/code in SQL.
I have database with names, addresses IDs etc, but I have to convert firstname values ending in “jnr” to “(Jnr)” and those ending in “snr” to “(Snr)”.
How do I do this?
update table TABLE_NAME set NAMES = '*xyz*Jnr' where NAMES like '%jnr'
Update or select:
PASTE(column, CHAR_LENGTH(column)-3, 1, UPPER(SUBSTRING(column FROM CHAR_LENGTH(column)-3 FOR 1)
WHERE column LIKE '%jnr' OR column LIKE '%snr'
PASTE is used to put in one character at position 3 from end,
CHAR_LENGTH to get length of column value,
UPPER converts character to upper case,
SUBSTRING is used to pick one character here (j or s),
LIKE is used to find values ending with jnr, or snr.
All ANSI SQL (no dbms specified!)
How can i query a column with Names of people to get only the names those contain exactly 2 “a” ?
I am familiar with % symbol that's used with LIKE but that finds all names even with 1 a , when i write %a , but i need to find only those have exactly 2 characters.
Please explain - Thanks in advance
Table Name: "People"
Column Names: "Names, Age, Gender"
Assuming you're asking for two a characters search for a string with two a's but not with three.
select *
from people
where names like '%a%a%'
and name not like '%a%a%a%'
Use '_a'. '_' is a single character wildcard where '%' matches 0 or more characters.
If you need more advanced matches, use regular expressions, using REGEXP_LIKE. See Using Regular Expressions With Oracle Database.
And of course you can use other tricks as well. For instance, you can compare the length of the string with the length of the same string but with 'a's removed from it. If the difference is 2 then the string contained two 'a's. But as you can see things get ugly real soon, since length returns 'null' when a string is empty, so you have to make an exception for that, if you want to check for names that are exactly 'aa'.
select * from People
where
length(Names) - 2 = nvl(length(replace(Names, 'a', '')), 0)
Another solution is to replace everything that is not an a with nothing and check if the resulting String is exactly two characters long:
select names
from people
where length(regexp_replace(names, '[^a]', '')) = 2;
This can also be extended to deal with uppercase As:
select names
from people
where length(regexp_replace(names, '[^aA]', '')) = 2;
SQLFiddle example: http://sqlfiddle.com/#!4/09bc6
select * from People where names like '__'; also ll work
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.