It's my table rows:
++ id ---- text ++++++++++++++
-- 1 ---- '90','80,'50' -----
-- 2 ---- '30','2','1_2' --
-- 3 ---- '10_2','5_3' -----
as you see, text contains 2 types of numbers, one doesn't have underscore, and the other does.
I want to select rows which have at least one number without underscore (type 1). Something like this: (result-set)
++ id ---- text ++++++++++++++
-- 1 ---- '90','80,'50' -----
-- 2 ---- '30','2','1_2' --
(3 is ignored)
How to do that? (I think it's possible with NOT LIKE, but I don't know how to write)
How long may be your numbers? Try this:
SELECT t1.id,t1.txt FROM t t1, t t2 WHERE t1.txt LIKE "%'__'%" AND t2.txt NOT LIKE "%\__',%"
The below query counts the number of commas in the strings, number of distinct numbers can be calculated as 1 more than the number of commas as the numbers are separated by commas, number of underscores in the string:
select id,
len(text) - len(replace(text,',','')) as count_of_commas,
len(text) - len(replace(text,',','')) + 1 as count_of_number,
len(text) - len(replace(text,'_','')) as count_of_underscore,
len(text) - len(replace(text,',','')) + 1 - (len(text) - len(replace(text,'_',''))) as zero_if_no_number_without_underscore_exists
from t1
The above query gives the following results:
Now using the logic of above query, following query can be used to get the desired result:
select * from t1
where len(text) - len(replace(text,',','')) + 1 - (len(text) - len(replace(text,'_',''))) != 0
i.e. it returns the rows where atleast one number exists without the underscore.
You can't do it with LIKE, but you can with a RLIKE, which uses regular expressions:
select * from mytable
where `text` rlike "'\d+_\d+'"
Related
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
Hi: I have a table made with rows like this:
ID_CATEGORIA CATEGORIA_DRG
------------ ---------------------------------------------------------------
1 001-002-003-543 Craniotomia
2 004-531-532 Interventi midollo spinale
3 005-533-534 Interventi vasi extracranici
4 006 Decompressione tunnel carpale
I'd like to get something like this:
ID CATEGORIA DESCRIZIONE
------------ ------------------ --------------------------------------
1 001-002-003-543 Craniotomia
2 004-531-532 Interventi midollo spinale
3 005-533-534 Interventi vasi extracranici
4 006 Decompressione tunnel carpale
I don't need to alter the table, a 'formatted' query can be enough.
I Think SUBSTRING() is the right function for me, but I don't know how to mesaure the lenght of the first (numbers, dash-separated) field.
In Python I'll find that size with len("005-533-534 Interventi vasi extracranici".split(' ')[0])', but I don't have idea about how to write it in SQL
Something like this should do -
SELECT ID_CATEGORIA AS ID ,SUBSTRING(CATEGORIA_DRG,1,CHARINDEX(' ',CATEGORIA_DRG)) as CATEGORIA,SUBSTRING(CATEGORIA_DRG,CHARINDEX(' ',CATEGORIA_DRG),LEN(CATEGORIA_DRG)) AS DESCRIZIONE
FROM TABLENAME
Try this:
select id_categoria ID,
substring(categoria_drg, 1, idx) CATEGORIA,
substring(categoria_drg, idx + 1, 1000) DESCRIZIONE
from (
select id_categoria, categoria_drg, charindex(' ', categoria_drg) idx from my_table
) a
It uses charindex to detect when the code is finished, because it is followed by first space in the string, which the function finds :)
I have data in a table of length 9 where data is like
999999969
000000089
666666689
I want to delete only those data in which any number from 1-9 is repeating more than 5 times.
OK, so the logic here can be summed up as:
Find the longest series of the same consecutive digit in any given number; and
Return true if that longest value is > 5 digits
Right?
So, lets split it into series of consecutive digits:
regress=> SELECT regexp_matches('666666689', '(0+|1+|2+|3+|4+|5+|6+|7+|8+|9+)', 'g');
regexp_matches
----------------
{6666666}
{8}
{9}
(3 rows)
then filter for the longest:
regress=>
SELECT x[1]
FROM regexp_matches('6666666898', '(0+|1+|2+|3+|4+|5+|6+|7+|8+|9+)', 'g') x
ORDER BY length(x[1]) DESC
LIMIT 1;
x
---------
6666666
(1 row)
... but really, we don't actually care about that, just if any entry is longer than 5 digits, so:
SELECT x[1]
FROM regexp_matches('6666666898', '(0+|1+|2+|3+|4+|5+|6+|7+|8+|9+)', 'g') x
WHERE length(x[1]) > 5;
can be used as an EXISTS test, e.g.
WITH blah(n) AS (VALUES('999999969'),('000000089'),('666666689'),('15552555'))
SELECT n
FROM blah
WHERE EXISTS (
SELECT x[1]
FROM regexp_matches(n, '(0+|1+|2+|3+|4+|5+|6+|7+|8+|9+)', 'g') x
WHERE length(x[1]) > 5
)
which is actually pretty efficient and return the correct result (always nice). But it can be simplified a little more with:
WITH blah(n) AS (VALUES('999999969'),('000000089'),('666666689'),('15552555'))
SELECT n
FROM blah
WHERE EXISTS (
SELECT x[1]
FROM regexp_matches(n, '(0{6}|1{6}|2{6}|3{6}|4{6}|5{6}|6{6}|7{6}|8{6}|9{6})', 'g') x;
)
You can use the same WHERE clause in a DELETE.
This can be much simpler with a regular expression using a back reference.
DELETE FROM tbl
WHERE col ~ '([1-9])\1{5}';
That's all.
Explain
([1-9]) ... a character class with digits from 1 to 9, parenthesized for the following back reference.
\1 ... back reference to first (and only in this case) parenthesized subexpression.
{5} .. exactly (another) 5 times, making it "more than 5".
Per documentation:
A back reference (\n) matches the same string matched by the previous
parenthesized subexpression specified by the number n [...] For example, ([bc])\1 matches bb or cc but not bc or cb.
SQL Fiddle demo.
Horrible and terrible in terms of performance, but it should work:
DELETE FROM YOURTABLE
WHERE YOURDATA LIKE '%111111%'
OR YOURDATA LIKE '%222222%'
OR YOURDATA LIKE '%333333%'
OR YOURDATA LIKE '%444444%'
OR YOURDATA LIKE '%555555%'
OR YOURDATA LIKE '%666666%'
OR YOURDATA LIKE '%777777%'
OR YOURDATA LIKE '%888888%'
OR YOURDATA LIKE '%999999%'
I have a table in which I want to get the strings which are like ab aabb aaabbb ...... a n times followed by b n times as shown below.
Eg TABLE:
value
----------
ab
aabb
aaabbb
aaaabbbb
1
1a
abababa
I want the result TABLE to be:
value
----------
ab
aabb
aaabbb
aaaabbbb
I've tried like this
select * from [NumTest] where value LIKE '[a]+[b]+'
but it's returning zero rows.
Can anybody help me how to use special characters in SQL Server's LIKE ?
Here is something that can work:
(EDIT - after O/P comment, commented parts not needed)
--WITH CTE_GoodValues AS
--(
SELECT value
FROM Table1
WHERE LEFT(VALUE,LEN(VALUE)/2) = REPLICATE('a',LEN(VALUE)/2)
AND RIGHT(VALUE,LEN(VALUE)/2) = REPLICATE('b',LEN(VALUE)/2)
AND LEN(VALUE)%2=0
--)
--SELECT REPLICATE(' ', (SELECT MAX(LEN(VALUE))/2 FROM CTE_GoodValues)- LEN(VALUE)/2) + VALUE
--FROM CTE_GoodValues
In the CTE - select values that have left half all a-s and right half all b-s. Then find MAX length and use it to replicate needed empty spaces in front
DEMO (after edit)
How can I write a select statement to select only integers (and nothing more) from a char column in SQL Server. For example, my table name is POWDER with 2 columns, ID (int) and Name(char (5))
ID Name
-- ----------
1 AXF22
2 HYWWW
3 24680
4 8YUH8
5 96635
I want to be able to select only those rows that contain an integer and nothing more (ID 3 and ID 5 in this example)
If I try:
SELECT *
FROM POWDER
WHERE Name LIKE '[0-9]%'
...it will return:
ID Name
-- ----------
3 24680
4 8YUH8
5 96635
Any ideas how to get the rows containing just integers?
SELECT * FROM POWDER WHERE IsNumeric(Name) = 1
IsNumeric returns 1 for some other characters that are valid in numbers, such as + and - and $ but for your input you should be fine.
Try this:
SELECT * FROM Table WHERE Name LIKE '[0-9]%%'
To avoid issues with ISNUMERIC and all spaces, -, +, . etc, use the fact that the column is char(5)
SELECT *
FROM POWDER
WHERE Name LIKE '[0-9][0-9][0-9][0-9][0-9]'
Edit: for any number of characters. Double negative...
SELECT *
FROM POWDER
WHERE Name NOT LIKE '%[^0-9]%'
Use positive and negative checks to make sure we have an integer: It must contain a digit. Only digits and spaces are allowed. No spaces are allowed between digits.
SELECT *
FROM POWDER
WHERE Name LIKE '%[0-9]%'
AND Name NOT LIKE '%[^0-9 ]%'
AND Name NOT LIKE '%[0-9]% %[0-9]%'
Try:
SELECT *
FROM POWDER
WHERE Name patindex ('%[a-z]%',name) != 0
The last one is the best,kind of works really really well.
SELECT * FROM POWDER
WHERE Name LIKE '%[0-9]%'
AND Name NOT LIKE '%[^0-9 ]%'
AND Name NOT LIKE '%[0-9]% %[0-9]%'