I want to replace special characters from two different words as shown in the image below. From the first word, I want to replace a special character with "I" and from the second word, I want to replace a special character with "U".
My query is like below: It works for the first word. Can you pls assist?
SELECT
distinct ABC REGEXP_REPLACE(REGEXP_REPLACE(ABC, r'([^\p{ASCII}]+)', 'I') ,r'\&', 'U')
FROM Table
where ABC like '%B??RD%' or ABC like '%M??D%';
Instead of using REGEX, you can use CASE for this scenario:
WITH CTE as (
SELECT 'B��RD'as ABC,
UNION ALL SELECT 'M��D'as ABC)
SELECT
ABC,
CASE ABC
WHEN 'B��RD' THEN 'BIRD'
WHEN 'M��D' THEN 'MUD'
END AS output
FROM CTE
Output:
Related
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`
Can anyone help me, I have a problem regarding on how can I get the below result of data. refer to below sample data. So the logic for this is first I want delete the letters before the number and if i get that same thing goes on , I will delete the numbers before the letter so I can get my desired result.
Table:
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Desired Output:
PIX32BLU
A9CARBONGRY
PIXL128BLK
You need to use a combination of the SUBSTRING and PATINDEX Functions
SELECT
SUBSTRING(SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99),PATINDEX('%[^0-9]%',SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99)),99) AS youroutput
FROM yourtable
Input
yourtable
fielda
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Output
youroutput
PIX32BLU
A9CARBONGRY
PIXL128BLK
SQL Fiddle:http://sqlfiddle.com/#!6/5722b6/29/0
To do this you can use
PATINDEX('%[0-9]%',FieldName)
which will give you the position of the first number, then trim off any letters before this using SUBSTRING or other string functions. (You need to trim away the first letters before continuing with the next step because unlike CHARINDEX there is no starting point parameter in the PATINDEX function).
Then on the remaining string use
PATINDEX('%[a-z]%',FieldName)
to find the position of the first letter in the remaining string. Now trim off the numbers in front using SUBSTRING etc.
You may find this other solution helpful
SQL to find first non-numeric character in a string
Try this it may helps you
;With cte (Data)
AS
(
SELECT 'SALV3000640PIX32BLU' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470B9CARBONGRY' UNION ALL
SELECT 'SALV3334470D9CARBONGRY' UNION ALL
SELECT 'TP3000620PIXL128BLK'
)
SELECT * , CASE WHEN CHARINDEX('PIX',Data)>0 THEN SUBSTRING(Data,CHARINDEX('PIX',Data),LEN(Data))
WHEN CHARINDEX('A9C',Data)>0 THEN SUBSTRING(Data,CHARINDEX('A9C',Data),LEN(Data))
ELSE NULL END AS DesiredResult FROM cte
Result
Data DesiredResult
-------------------------------------
SALV3000640PIX32BLU PIX32BLU
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470B9CARBONGRY NULL
SALV3334470D9CARBONGRY NULL
TP3000620PIXL128BLK PIXL128BLK
I need to display the whole number in a field if it starts with "AB" otherwise do not show/display the number.
Your question is missing code of how you display this (since you wrote you need to display it to the field) so i can't answer you with actual code but here is solution.
If you want to select only rows which column1 starts with AB then use LIKE function. So condition at selecting command is Select * from yourtable where column1 LIKE 'AB%'
If you already selected and displayed data, let's say in datagridview, and you want to fill textbox with string that contains AB then you would go through all rows at specific column and look for it with string.Contains("AB");
So basically you put this command in foreach loop and you have it.
I was wrong. You can use a LIKE, just not in the WHERE clause.
;WITH testdata AS (
SELECT 'aw12354' AS val UNION ALL
SELECT 'a12b344' UNION ALL
SELECT 'AB11111' UNION ALL
SELECT '11AB111' UNION ALL
SELECT '11111AB' UNION ALL
SELECT 'ab22222'
)
SELECT
CASE WHEN val LIKE 'AB%' THEN val ELSE NULL END AS valFull
, CASE WHEN val LIKE 'AB%' THEN SUBSTRING(val,3,len(val)) ELSE NULL END AS valNums
FROM testdata
;
You can also use CLR to build a regex solution, but that is a LOT more involved.
Getting Examples from similar Stack Overflow threads,
Remove all characters after a specific character in PL/SQL
and
How to Select a substring in Oracle SQL up to a specific character?
I would want to retrieve only the first characters before the occurrence of a string.
Example:
STRING_EXAMPLE
TREE_OF_APPLES
The Resulting Data set should only show only STRING_EXAM and TREE_OF_AP because PLE is my delimiter
Whenever i use the below REGEXP_SUBSTR, It gets only STRING_ because REGEXP_SUBSTR treats PLE as separate expressions (P, L and E), not as a single expression (PLE).
SELECT REGEXP_SUBSTR('STRING_EXAMPLE','[^PLE]+',1,1) from dual;
How can i do this without using numerous INSTRs and SUBSTRs?
Thank you.
The problem with your query is that if you use [^PLE] it would match any characters other than P or L or E. You are looking for an occurence of PLE consecutively. So, use
select REGEXP_SUBSTR(colname,'(.+)PLE',1,1,null,1)
from tablename
This returns the substring up to the last occurrence of PLE in the string.
If the string contains multiple instances of PLE and only the substring up to the first occurrence needs to be extracted, use
select REGEXP_SUBSTR(colname,'(.+?)PLE',1,1,null,1)
from tablename
Why use regular expressions for this?
select substr(colname, 1, instr(colname, 'PLE')-1) from...
would be more efficient.
with
inputs( colname ) as (
select 'FIRST_EXAMPLE' from dual union all
select 'IMPLEMENTATION' from dual union all
select 'PARIS' from dual union all
select 'PLEONASM' from dual
)
select colname, substr(colname, 1, instr(colname, 'PLE')-1) as result
from inputs
;
COLNAME RESULT
-------------- ----------
FIRST_EXAMPLE FIRST_EXAM
IMPLEMENTATION IM
PARIS
PLEONASM
I want to select names from a table where the 'name' column contains '%' anywhere in the value. For example, I want to retrieve the name 'Approval for 20 % discount for parts'.
SELECT NAME FROM TABLE WHERE NAME ... ?
You can use like with escape. The default is a backslash in some databases (but not in Oracle), so:
select name
from table
where name like '%\%%' ESCAPE '\'
This is standard, and works in most databases. The Oracle documentation is here.
Of course, you could also use instr():
where instr(name, '%') > 0
One way to do it is using replace with an empty string and checking to see if the difference in length of the original string and modified string is > 0.
select name
from table
where length(name) - length(replace(name,'%','')) > 0
Make life easy on yourselves and just use REGEXP_LIKE( )!
SQL> with tbl(name) as (
select 'ABC' from dual
union
select 'E%FS' from dual
)
select name
from tbl
where regexp_like(name, '%');
NAME
----
E%FS
SQL>
I read the documentation mentioned by Gordon. The relevent sentence is:
An underscore (_) in the pattern matches exactly one character (as opposed to one byte in a multibyte character set) in the value
Here was my test:
select c
from (
select 'a%be' c
from dual) d
where c like '_%'
The value a%be was returned.
While the suggestions of using instr() or length in the other two answers will lead to the correct answer, they will do so slowly. Filtering on function results simply take longer than filtering on fields.