How can I select records where in the column Value the 5th character is letter A?
For example the following records:
ID Value
-------------------------
1 1234A5636A6363
2 1234A4343B6363
3 1234B5353A6363
if I run
select * from table
where Value like '%A%'
this will return all records
but all I want is the first 2 where the 5th character is A, regardless if there are more A characters in the text or not
select *
from your_table
where substring(Value, 5, 1) = 'A'
The LIKE operator, in addition to %, which matches any number of any character, can use _, which matches any one single character. You may try:
SELECT *
FROM yourTable
WHERE Value LIKE '____A%'; -- 4 underscores here
use like below by using _(underscore)
LIKE '____A%'
SQL Server
select *
from YourTableName
where CHARINDEX('A', ColumnName) = 5
Note:- This finds where string 'A' starts at position 5
AND specify Your ColumnName
I have a DB column that has entries like this:
"56/45/34"
"78/34/145"
"45"
"" (i.e. NULL)
I want to search for the rows that match a certain number - for example "45" would should return the first and third rows but not the second.
We can try using a regex approach here with word boundaries:
select col
from your_table
where col ~* '\y45\y';
Demo
You can convert the delimited string to an array and then test the array
select *
from the_table
where '45' = any(string_to_array(the_column, '/'))
I would like to perform a "special group by" on strings with SQL language, some ending with "*". I use postgresql.
I can not clearly formulate this problem, even if I have partially solved it, with select, union and nested queries which are not elegant.
For exemple :
1) INPUT : I have a list of strings :
thestrings
varchar(9)
--------------
1000
1000-0001
1000-0002
2000*
2000-0001
2000-0002
3000*
3000-00*
3000-0001
3000-0002
2) OUTPUT : That I would like my "special group by" return :
1000
1000-0001
1000-0002
2000*
3000*
Because 2000-0001 and 2000-0002 are include in 2000*,
and because 3000-00*, 3000-0001 and 3000-0002 are includes in 3000*
3) SQL query I do :
SELECT every strings ending with *
UNION
SELECT every string where the begining NOT IN (SELECT every string ending with *) <-- with multiple inelegant left functions and NOT IN subqueries
4) That what I'm doing return :
1000
1000-0001
1000-0002
2000*
3000*
3000-00* <-- the problem
The problem is : 3000-00* staying in my result.
So my question is :
How can I generalize my problem? to remove all string who have a same begining string in the list (ending with *) ?
I think of regular expressions, but how to pass a list from a select in a regex ?
Thanks for help.
Select only strings for which no master string exists in the table:
select str
from mytable
where not exists
(
select *
from mytable master
where master.str like '%*'
and master.str <> mytable.str
and rtrim(mytable.str, '*') like rtrim(master.str, '*') || '%'
);
Assuming that only one general pattern can match any given string, the following should do what you want:
select coalesce(tpat.thestring, t.thestring) as thestring
from t left join
t tpat
on t.thestring like replace(tpat.thestring, '*', '%') and
t.thestring <> tpat.thestring
group by coalesce(tpat.thestring, t.thestring);
However, that is not your case. However, you can adjust this with distinct on:
select distinct on (t.thestring) coalesce(tpat.thestring, t.thestring)
from t left join
t tpat
on t.thestring like replace(tpat.thestring, '*', '%') and
t.thestring <> tpat.thestring
order by t.thestring, length(tpat.thestring)
I have a table column that has data like
NA_PTR_51000_LAT_CO-BOGOTA_S_A
NA_PTR_51000_LAT_COL_M_A
NA_PTR_51000_LAT_COL_S_A
NA_PTR_51000_LAT_COL_S_B
NA_PTR_51000_LAT_MX-MC_L_A
NA_PTR_51000_LAT_MX-MTY_M_A
I want to parse each column value so that I get the values in column_B. Thank you.
COLUMN_A COLUMN_B
NA_PTR_51000_LAT_CO-BOGOTA_S_A CO-BOGOTA
NA_PTR_51000_LAT_COL_M_A COL
NA_PTR_51000_LAT_COL_S_A COL
NA_PTR_51000_LAT_COL_S_B COL
NA_PTR_51000_LAT_MX-MC_L_A MX-MC
NA_PTR_51000_LAT_MX-MTY_M_A MX-MTY
I'm not sure of the Postgresql and I can't get SQL fiddle to accept the schema build...
substring and length may vary...
Select Column_A, substr(columN_A,18,length(columN_A)-17-4) from tableName
Ok how about this then:
http://sqlfiddle.com/#!15/ad0dd/56/0
Select column_A, b
from (
Select Column_A, b, row_number() OVER (ORDER BY column_A) AS k
FROM (
SELECT Column_A
, regexp_split_to_table(Column_A, '_') b
FROM test
) I
) X
Where k%7=5
Inside out:
Inner most select simply splits the data into multiple rows on _
middle select adds a row number so that we can use the use the mod operator to find all occurances of a 5th remainder.
This ASSUMES that the section of data you're after is always the 5th segment AND that there are always 7 segments...
Use regexp_matches() with a search pattern like 'NA_PTR_51000_LAT_(.+)_'
This should return everything after NA_PTR_51000_LAT_ before the next underscore, which would match the pattern you are looking for.
I have (and don't own, so I can't change) a table with a layout similar to this.
ID | CATEGORIES
---------------
1 | c1
2 | c2,c3
3 | c3,c2
4 | c3
5 | c4,c8,c5,c100
I need to return the rows that contain a specific category id. I starting by writing the queries with LIKE statements, because the values can be anywhere in the string
SELECT id FROM table WHERE categories LIKE '%c2%';
Would return rows 2 and 3
SELECT id FROM table WHERE categories LIKE '%c3%' and categories LIKE '%c2%'; Would again get me rows 2 and 3, but not row 4
SELECT id FROM table WHERE categories LIKE '%c3%' or categories LIKE '%c2%'; Would again get me rows 2, 3, and 4
I don't like all the LIKE statements. I've found FIND_IN_SET() in the Oracle documentation but it doesn't seem to work in 10g. I get the following error:
ORA-00904: "FIND_IN_SET": invalid identifier
00904. 00000 - "%s: invalid identifier"
when running this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories); (example from the docs) or this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories) <> 0; (example from Google)
I would expect it to return rows 2 and 3.
Is there a better way to write these queries instead of using a ton of LIKE statements?
You can, using LIKE. You don't want to match for partial values, so you'll have to include the commas in your search. That also means that you'll have to provide an extra comma to search for values at the beginning or end of your text:
select
*
from
YourTable
where
',' || CommaSeparatedValueColumn || ',' LIKE '%,SearchValue,%'
But this query will be slow, as will all queries using LIKE, especially with a leading wildcard.
And there's always a risk. If there are spaces around the values, or values can contain commas themselves in which case they are surrounded by quotes (like in csv files), this query won't work and you'll have to add even more logic, slowing down your query even more.
A better solution would be to add a child table for these categories. Or rather even a separate table for the catagories, and a table that cross links them to YourTable.
You can write a PIPELINED table function which return a 1 column table. Each row is a value from the comma separated string. Use something like this to pop a string from the list and put it as a row into the table:
PIPE ROW(ltrim(rtrim(substr(l_list, 1, l_idx - 1),' '),' '));
Usage:
SELECT * FROM MyTable
WHERE 'c2' IN TABLE(Util_Pkg.split_string(categories));
See more here: Oracle docs
Yes and No...
"Yes":
Normalize the data (strongly recommended) - i.e. split the categorie column so that you have each categorie in a separate... then you can just query it in a normal faschion...
"No":
As long as you keep this "pseudo-structure" there will be several issues (performance and others) and you will have to do something similar to:
SELECT * FROM MyTable WHERE categories LIKE 'c2,%' OR categories = 'c2' OR categories LIKE '%,c2,%' OR categories LIKE '%,c2'
IF you absolutely must you could define a function which is named FIND_IN_SET like the following:
CREATE OR REPLACE Function FIND_IN_SET
( vSET IN varchar2, vToFind IN VARCHAR2 )
RETURN number
IS
rRESULT number;
BEGIN
rRESULT := -1;
SELECT COUNT(*) INTO rRESULT FROM DUAL WHERE vSET LIKE ( vToFine || ',%' ) OR vSET = vToFind OR vSET LIKE ('%,' || vToFind || ',%') OR vSET LIKE ('%,' || vToFind);
RETURN rRESULT;
END;
You can then use that function like:
SELECT * FROM MyTable WHERE FIND_IN_SET (categories, 'c2' ) > 0;
For the sake of future searchers, don't forget the regular expression way:
with tbl as (
select 1 ID, 'c1' CATEGORIES from dual
union
select 2 ID, 'c2,c3' CATEGORIES from dual
union
select 3 ID, 'c3,c2' CATEGORIES from dual
union
select 4 ID, 'c3' CATEGORIES from dual
union
select 5 ID, 'c4,c8,c5,c100' CATEGORIES from dual
)
select *
from tbl
where regexp_like(CATEGORIES, '(^|\W)c3(\W|$)');
ID CATEGORIES
---------- -------------
2 c2,c3
3 c3,c2
4 c3
This matches on a word boundary, so even if the comma was followed by a space it would still work. If you want to be more strict and match only where a comma separates values, replace the '\W' with a comma. At any rate, read the regular expression as:
match a group of either the beginning of the line or a word boundary, followed by the target search value, followed by a group of either a word boundary or the end of the line.
As long as the comma-delimited list is 512 characters or less, you can also use a regular expression in this instance (Oracle's regular expression functions, e.g., REGEXP_LIKE(), are limited to 512 characters):
SELECT id, categories
FROM mytable
WHERE REGEXP_LIKE('c2', '^(' || REPLACE(categories, ',', '|') || ')$', 'i');
In the above I'm replacing the commas with the regular expression alternation operator |. If your list of delimited values is already |-delimited, so much the better.