I need help wit removing "'pp'" from search results which appear at the biginning of text. Values in search resuls contain spaces and also '. I need to remove only 'pp from bigginig
This sounds like:
select regexp_replace(col, '^pp', '')
Or a case expression:
select (case when col like 'pp%' then substr(col, 3) else col end)
You don't need regular expressions and can use simple string functions.
If you want to use SELECT then:
SELECT value,
CASE
WHEN value LIKE 'pp%'
THEN SUBSTR( value, 3 )
ELSE value
END AS replaced_value
FROM table_name
Outputs:
VALUE | REPLACED_VALUE
:---- | :-------------
pp123 | 123
pp1pp | 1pp
123pp | 123pp
12345 | 12345
and, if you want to UPDATE the table:
UPDATE table_name
SET value = SUBSTR( value, 3 )
WHERE value LIKE 'pp%';
Then:
SELECT * FROM table_name;
Outputs:
| VALUE |
| :---- |
| 123 |
| 1pp |
| 123pp |
| 12345 |
db<>fiddle here
Related
How can I check whether a certain substring (for instance 18UT) is part of a string in a column?
Redshifts' SUBSTRING function allows me to "cut" a certain substring based on a starting index + length of the subtring, but not check whether a specific substring exists is in the column's value.
Example:
+------------------+
| col |
+------------------+
| 14TH, 14KL, 18AB |
| 14LK, 18UT, 15AK |
| 14AB, 08ZT, 18ZH |
| 14GD, 52HG, 18UT |
+------------------+
Desired result:
+------------------+------+
| col | 18UT |
+------------------+------+
| 14TH, 14KL, 18AB | No |
| 14LK, 18UT, 15AK | Yes |
| 14AB, 08ZT, 18ZH | No |
| 14GD, 52HG, 18UT | Yes |
+------------------+------+
Here is one option:
select col,
case when ', ' || col || ', ' like '%, 18UT, %' then 'yes' else 'no' end has_18ut
from mytable
While this will solve your immediate, problem, it should be note that storing delimited lists in a database table is bad practice, and should be avoided. Each value should go to a separate row instead.
I am a beginner in SQL language and I am using postgre sql and doing little exercices to learn. I have a column of strings named acronym from a destination table:
DO1
ES1
ES2
FR1
FR10
FR2
FR3
FR4
FR5
FR6
FR7
FR8
FR9
GP1
GP2
IN1
IN2
MU1
RU1
TR1
UA1
I would like to add a padding zero for acronym numbers that have only one digit, output:
DO01
ES01
ES02
FR01
FR02
FR03
FR04
FR05
FR06
FR07
FR08
FR09
FR10
GP01
GP02
IN01
IN02
MU01
RU01
TR01
UA01
How can I get to the left of the first number in the string? There is some regex I think but I did not figure it out
You can use the rpad() function to add characters to the end of the value:
select rpad(col, '0', 4)
In your case, though, you want a value in-between. On simple method is -- assuming that the first two characters are strings -- is:
(case when length(col) = 3
then left(col, 2) || '0' || right(col, 1)
else col
end)
Another possibility is using regexp_replace():
regexp_replace(col, '^([^0-9]{2})([0-9])$', '\10\2')
Both of these assume that the strings to be padded are three characters, which is consistent with your data. It is unclear what you want for other lengths.
try with below:
to_char() function
select to_char(column1, 'fm000') as column2
from Test_table;
fm "fill mode"prefix avoids leading spaces in the resulting var char.
000 it defines the number of digits you want to have.
You can use string functions like lpad(), substr(), left():
select
concat(left(columnname, 2), lpad(substr(columnname, 3), 2, '0')) result
from tablename
See the demo.
Results:
| result |
| ------ |
| DO01 |
| ES01 |
| ES02 |
| FR01 |
| FR10 |
| FR02 |
| FR03 |
| FR04 |
| FR05 |
| FR06 |
| FR07 |
| FR08 |
| FR09 |
| GP01 |
| GP02 |
| IN01 |
| IN02 |
| MU01 |
| RU01 |
| TR01 |
| UA01 |
I have table like this
-----------------------------------
| id | col_name | colname_suf |
-----------------------------------
| 1 | textSuff_ix| null |
| 2 | strSuff_ix2| null |
| ... |
-----------------------------------
The main idea of my solution is to move suffix to another column. But suffix and text before can be in known range(text1, text2, str, Suff_ix1, Suff_ix2, Suff_ix3...). How i can fix it using sql query?
Use string functions like left(), right() and charindex() to split the column and update:
update tablename
set [col_name] = left([col_name], charindex('_', [col_name]) - 1),
[colname_suf] = right([col_name], len([col_name]) - charindex('_', [col_name]))
where [col_name] like '%[_]%';
See the demo.
Results:
> id | col_name | colname_suf
> -: | :------- | :----------
> 1 | textSuff | ix
> 2 | strSuff | ix2
You can use string operations:
select left(col_name, charindex('_', col_name) - 1) as prefix,
stuff(col_name, 1, charindex('_', col_name), '') as suffix
I am trying to left pad with a single zero after the '-'.
I did check the other answers here but didnt help me.
Here is the table :
+---------+
| Job |
+---------+
| 3254-1 |
| 3254-25 |
| 3254-6 |
+---------+
I need to left pad with single zero after '-' if the value is between 1 and 9 in the end
I want the results to be :
+---------+
| Job |
+---------+
| 3254-01 |
| 3254-25 |
| 3254-06 |
+---------+
You can use CHARINDEX(), SUBSTRING() and REPLACE() as:
CREATE TABLE Jobs(
Job VARCHAR(45)
);
INSERT INTO Jobs VALUES
('3254-1'),
('3254-25'),
('3254-6');
SELECT CASE
WHEN CHARINDEX('-', Job, 1)+1 < LEN(Job) THEN Job
ELSE
REPLACE(Job, '-', '-0')
END AS Job
FROM Jobs;
Results:
+----+---------+
| | Job |
+----+---------+
| 1 | 3254-01 |
| 2 | 3254-25 |
| 3 | 3254-06 |
+----+---------+
If you want an update, I think this is the simplest method:
update t
set job = replace(job, '-', '-0')
where job like '%-_';
This problem is simplified greatly because you are only adding a single padding character.
If you have version 2012+, then format function may be used as :
select concat(nr1, '-', format( cast ( q2.nr2 as int ), '00')) as result
from
(
select substring(q1.str,1,charindex('-',q1.str,1)-1) as nr1,
substring(q1.str,charindex('-',q1.str,1)+1,len(q1.str)) as nr2
from
(
select '3254-1' as str union all
select '3254-25' as str union all
select '3254-6' as str
) q1
) q2;
result
------
3254-01
3254-25
3254-06
Rextester Demo
I have a column with data as abc.123, def.345 and so on. It is basically a name followed by a . and then a code. I am able to divide them using
select
LEFT(Campaign, ISNULL(NULLIF(CHARINDEX('.', Campaign + ' ') -1, -1), LEN(Campaign))),
STUFF(Campaign, 1, Len(Campaign) +1- CHARINDEX('.',Reverse(Campaign)), '')
from myTable;
Input set:
| myColumn |
| abc.123 |
| def.345 |
| 444 |
However, for data like '444' it shows:
|Name| Code |
|abc | 123 |
|def | 345 |
|444 | | (SHOULD BE => | | 444 |)
Assumptions:
Data can be without a '.' In such case we can insert the data in either Name or Code depending on datatype ie Name=>Alphanumeric, Code=>Numeric.
eg: Data like
999 => | | 999 |
a24.345 => | a24 | 345 |
a72 => | a72 | |
Use CASE expressions together with the LIKE operand in order to differentiate the cases:
SELECT
CASE
WHEN Campaign LIKE '[a-z]%' THEN LEFT(Campaign, CHARINDEX('.', Campaign + '.') - 1)
ELSE null
END AS Name,
CASE
WHEN Campaign LIKE '[0-9]%' THEN Campaign
WHEN Campaign LIKE '%.[0-9]%' THEN
RIGHT(Campaign, LEN(Campaign) - CHARINDEX('.', Campaign))
ELSE null
END AS Code
FROM myTable
You can see an example here: http://sqlfiddle.com/#!3/ae7ef7/1
Use a case statement:
select (case when myColumn like '%.%'
then <your code here>
else myColumn
end) as code