Simple Question
I have the following type of results in a string field
'Number=123456'
'Number=1234567'
'Number=12345678'
How do I extract the value from the string with regard that the value can change between 5-8 figures
So far I did this but I doubt that fits my requirement
SELECT substring('Size' from 8 for ....
If I can tell it to start from the = sign till the end that would help!
The likely simplest solution is to trim 7 leading characters with right():
right(str, -7)
Demo:
SELECT str, right(str, -7)
FROM (
VALUES ('Number=123456')
, ('Number=1234567')
, ('Number=12345678')
) t(str);
str | right
-----------------+----------
Number=123456 | 123456
Number=1234567 | 1234567
Number=12345678 | 12345678
You could use REPLACE:
SELECT col, REPLACE(col, 'Number=', '')
FROM tab;
DBFiddle Demo
Based on this question:
Split comma separated column data into additional columns
You could probably do the following:
SELECT *, split_part(col, '=', 2)
FROM table;
You may use regexp_matches :
with t(str) as
(
select 'Number=123456' union all
select 'Number=1234567' union all
select 'Number=12345678' union all
select 'Number=12345678x9'
)
select t.str as "String",
regexp_matches(t.str, '=([A-Za-z0-9]+)', 'g') as "Number"
from t;
String Number
-------------- ---------
Number=123456 123456
Number=1234567 1234567
Number=12345678 12345678
Number=12345678x9 12345678x9
--> the last line shows only we look chars after equal sign even if non-digit
Rextester Demo
Related
I have a column of data that looks like this:
58,0:102,56.00
52,0:58,68
58,110
57,440.00
52,0:58,0:106,6105.95
I need to extract the character before the last delimiter (',').
Using the data above, I want to get:
102
58
58
57
106
Might be done with a regular expression in substring(). If you want:
the longest string of only digits before the last comma:
substring(data, '(\d+)\,[^,]*$')
Or you may want:
the string before the last comma (',') that's delimited at the start either by a colon (':') or the start of the string.
Could be another regexp:
substring(data, '([^:]*)\,[^,]*$')
Or this:
reverse(split_part(split_part(reverse(data), ',', 2), ':', 1))
More verbose but typically much faster than a (expensive) regular expression.
db<>fiddle here
Can't promise this is the best way to do it, but it is a way to do it:
with splits as (
select string_to_array(bar, ',') as bar_array
from foo
),
second_to_last as (
select
bar_array[cardinality(bar_array)-1] as field
from splits
)
select
field,
case
when field like '%:%' then split_part (field, ':', 2)
else field
end as last_item
from second_to_last
I went a little overkill on the CTEs, but that was to expose the logic a little better.
With a CTE that removes everything after the last comma and then splits the rest into an array:
with cte as (
select
regexp_split_to_array(
replace(left(col, length(col) - position(',' in reverse(col))), ':', ','),
','
) arr
from tablename
)
select arr[array_upper(arr, 1)] from cte
See the demo.
Results:
| result |
| ------ |
| 102 |
| 58 |
| 58 |
| 57 |
| 106 |
The following treats the source string as an "array of arrays". It seems each data element can be defined as S(x,y) and the overall string as S1:S2:...Sn.
The task then becomes to extract x from Sn.
with as_array as
( select string_to_array(S[n], ',') Sn
from (select string_to_array(col,':') S
, length(regexp_replace(col, '[^:]','','g'))+1 n
from tablename
) t
)
select Sn[array_length(Sn,1)-1] from as_array
The above extends S(x,y) to S(a,b,...,x,y) the task remains to extracting x from Sn. If it is the case that all original sub-strings S are formatted S(x,y) then the last select reduces to select Sn[1]
How can we convert 12345 into 1,2,3,4,5 .
I can do the reverse by using replace command and i can replace comma by null. But I am not able to do the above. Could you please help on the same.
Thanks in Advance
You can use regexp_replace():
rtrim(regexp_replace('12345', '([0-9])', '\1,'), ',')
The rtrim() is necessary because the last digit is also replaced.
Online example: https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=724adecda03305b281ad3bf0f380ca58
If you have a fixed template with consecutive, one-digit numbers like 1234 or 12345 or 123456789
you may try the following by using listagg function of Oracle :
with t as
(
select '12345' as col from dual
)
select listagg(level,',') within group (order by level) as str
from t
connect by level <= length(col);
STR
---------
1,2,3,4,5
SQL Fiddle Demo 1
OR
For more generalized solution, use the following :
with t as
(
select 'abcde' as col from dual
)
select listagg(substr(col,level,1),',') within group (order by level) as str
from t
connect by level <= length(col);
STR
---------
a,b,c,d,e
SQL Fiddle Demo 2
So I'm working with BigQuery SQL right now trying to figure out how to remove letters but keep numeric numbers. For example:
XXXX123456
AAAA123456789
XYZR12345678
ABCD1234567
1111
2222
All have the same amount of letters in front of the numbers along with regular numbers no letters. I want the end result to look like:
123456
123456789
12345678
1234567
1111
2222
I tried using PATINDEX but BigQuery doesn't support the function. I've also tried using LEFT but that function will get rid of any value and I don't want to get rid of any numeric value only letter values. Any help would be much appreciated!
-Maykid
You can use regexp_replace():
select regexp_replace(str, '[^0-9]', '')
Below example is for BigQuery Standard SQL
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'XXXX123456' str UNION ALL
SELECT 'AAAA123456789' UNION ALL
SELECT 'XYZR12345678' UNION ALL
SELECT 'ABCD1234567' UNION ALL
SELECT '1111' UNION ALL
SELECT '2222'
)
SELECT str, REGEXP_REPLACE(str, r'[a-zA-Z]', '') str_adjusted
FROM `project.dataset.table`
I have a string as follows: first, last (123456) the expected result should be 123456. Could someone help me in which direction should I proceed using Oracle?
It will depend on the actual pattern you care about (I assume "first" and "last" aren't literal hard-coded strings), but you will probably want to use regexp_substr.
For example, this matches anything between two brackets (which will work for your example), but you might need more sophisticated criteria if your actual examples have multiple brackets or something.
SELECT regexp_substr(COLUMN_NAME, '\(([^\)]*)\)', 1, 1, 'i', 1)
FROM TABLE_NAME
Your question is ambiguous and needs clarification. Based on your comment it appears you want to select the six digits after the left bracket. You can use the Oracle instr function to find the position of a character in a string, and then feed that into the substr to select your text.
select substr(mycol, instr(mycol, '(') + 1, 6) from mytable
Or if there are a varying number of digits between the brackets:
select substr(mycol, instr(mycol, '(') + 1, instr(mycol, ')') - instr(mycol, '(') - 1) from mytable
Find the last ( and get the sub-string after without the trailing ) and convert that to a number:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE test ( str ) AS
SELECT 'first, last (123456)' FROM DUAL UNION ALL
SELECT 'john, doe (jr) (987654321)' FROM DUAL;
Query 1:
SELECT TO_NUMBER(
TRIM(
TRAILING ')' FROM
SUBSTR(
str,
INSTR( str, '(', -1 ) + 1
)
)
) AS value
FROM test
Results:
| VALUE |
|-----------|
| 123456 |
| 987654321 |
Here is a sample of my data:
ABC*12345ABC
BCD*234()
CDE*3456789(&(&
DEF*4567A*B*C
Using SQL Server 2008 or SSIS, I need to parse this data and return the following result:
12345
234
3456789
4567
As you can see, the asterisk (*) is my first delimiter. The second "delimiter" (I use this term loosely) is when the sequence of numbers STOP.
So, basically, just grab the sequence of numbers after the asterisk...
How can I accomplish this?
EDIT:
I made a mistake in my original post. An example of another possible value would be:
XWZ*A12345%$%
In this case, I would like to return the following:
A12345
The value can START with an alpha character, but it will always END with a number. So, grab everything after the asterisk, but stop at the last number in the sequence.
Any help with this will be greatly appreciated!
You could do this with a little patindex and charindex trickery, like:
; with YourTable(col1) as
(
select 'ABC*12345ABC'
union all select 'BCD*234()'
union all select 'CDE*3456789(&(&'
union all select 'DEF*4567A*B*C'
union all select 'XWZ*A12345%$%'
)
select left(AfterStar, len(Leader) + PATINDEX('%[^0-9]%', AfterLeader) - 1)
from (
select RIGHT(AfterStar, len(AfterStar) - PATINDEX('%[0-9]%', AfterStar) + 1)
as AfterLeader
, LEFT(AfterStar, PATINDEX('%[0-9]%', AfterStar) - 1) as Leader
, AfterStar
from (
select RIGHT(col1, len(col1) - CHARINDEX('*', col1)) as AfterStar
from YourTable
) as Sub1
) as Sub2
This prints:
12345
234
3456789
4567
A12345
If you ignore that this is in SQL then the first thing that comes to mind is Regex:
^.*\*(.*[0-9])[^0-9]*$
The capture group there should get what you want. I don't know if SQL has a regex function.