I would like to get all the characters in a field before a space
For example, if field1 is "chara ters"
I want it to return "chara"
What would this select statement look like?
SELECT LEFT(field1,LOCATE(' ',field1) - 1)
Note that if the string in question contains no spaces, this will return an empty string.
Below is another method that works and may seem a bit simpler to some. It uses the SUBSTRING_INDEX MySQL function. A 1 returns everything before the first space, and a -1 returns everything after the last space.
This returns 'chara':
SELECT SUBSTRING_INDEX( field1, ' ', 1 )
This returns 'ters':
SELECT SUBSTRING_INDEX( field1, ' ', -1 )
Details
A positive value will look for your specified character from the start of the string, and a negative value will start from the end of the string. The value of the number indicates the quantity of your specified character to look for before returning the remaining piece of the string. If the character you are searching for does not exist, the entire field value will be returned.
In this case, a -2 would return everything to the right of the second to last space, which doesn't exist in this example, so the entire field value will be returned.
You would need some string operations for that. Assuming every field has at least one space character:
SELECT SUBSTR(field1, 0, LOCATE(' ', field1)) FROM your_table;
Safe approach:
SELECT IF(
LOCATE(' ', field1),
SUBSTR(field1, 0, LOCATE(' ', field1)),
field1
) FROM your_table;
For a generalized approach that returns the nth value (one-based index) from a column containing delimited values:
select SUBSTRING_INDEX(SUBSTRING_INDEX(my_column,#delimiter,#index),#delimiter,-1) from my_table;
This will return the entire column contents if the delimiter does not exist, and the last field if the index value exceeds the number of fields.
For instance, if my_column contains 'foo bar baz' and you want the 2nd field:
select SUBSTRING_INDEX(SUBSTRING_INDEX(my_column,' ',2),' ',-1) from my_table;
will return bar, and if you specify the 4th field (which does not exist):
select SUBSTRING_INDEX(SUBSTRING_INDEX(my_column,' ',4),' ',-1) from my_table;
will return baz (the last field).
Related
So I need to filter column which contains either one, two or three whitespace character.
CREATE TABLE a
(
[col] [char](3) NULL,
)
and some inserts like
INSERT INTO a VALUES (' ',' ', ' ')
How do I get only the row with one white space?
Simply writing
SELECT *
FROM a
WHERE column = ' '
returns all rows irrespective of one or more whitespace character.
Is there a way to escape the space? Or search for specific number of whitespaces in column? Regex?
Use like clause - eg where column like '%[ ]%'
the brackets are important, like clauses provide a very limited version of regex. If its not enough, you can add a regex function written in C# to the DB and use that to check each row, but it won't be indexed and thus will be very slow.
The other alternative, if you need speed, is to look into full text search indexes.
Here is one approach you can take:
DECLARE #data table ( txt varchar(50), val varchar(50) );
INSERT INTO #data VALUES ( 'One Space', ' ' ), ( 'Two Spaces', ' ' ), ( 'Three Spaces', ' ' );
;WITH cte AS (
SELECT
txt,
DATALENGTH ( val ) - ( DATALENGTH ( REPLACE ( val, ' ', '' ) ) ) AS CharCount
FROM #data
)
SELECT * FROM cte WHERE CharCount = 1;
RETURNS
+-----------+-----------+
| txt | CharCount |
+-----------+-----------+
| One Space | 1 |
+-----------+-----------+
You need to use DATALENGTH as LEN ignores trailing blank spaces, but this is a method I have used before.
NOTE:
This example assumes the use of a varchar column.
Trailing spaces are often ignored in string comparisons in SQL Server. They are treated as significant on the LHS of the LIKE though.
To search for values that are exactly one space you can use
select *
from a
where ' ' LIKE col AND col = ' '
/*The second predicate is required in case col contains % or _ and for index seek*/
Note with your example table all the values will be padded out to three characters with trailing spaces anyway though. You would need a variable length datatype (varchar/nvarchar) to avoid this.
The advantage this has over checking value + DATALENGTH is that it is agnostic to how many bytes per character the string is using (dependant on datatype and collation)
DB Fiddle
How to get only rows with one space?
SELECT *
FROM a
WHERE col LIKE SPACE(1) AND col NOT LIKE SPACE(2)
;
Though this will only work for variable length datatypes.
Thanks guys for answering.
So I converted the char(3) column to varchar(3).
This seemed to work for me. It seems sql server has ansi padding that puts three while space in char(3) column for any empty or single space input. So any search or len or replace will take the padded value.
I have column with software versions. I was trying to remove dot from 2nd occurrence in column, like
select REGEXP_REPLACE('12.5.7.8', '.','');
expected out is 12.578
sample data is here
Is it possible to remove dot from 2nd occurrence
One option is to break this into two pieces:
Get the first number.
Get the rest of the numbers as an array.
Then convert the array to a string with no separator and combine with the first:
select (split_part('12.5.7.8', '.', 1) || '.' ||
array_to_string((REGEXP_SPLIT_TO_ARRAY('12.5.7.8', '[.]'))[2:], '')
)
Another option is to replace the first '.' with something else, then get rid of the '.'s and replace the something else with a '|':
select translate(regexp_replace(version, '^([^.]+)[.](.*)$', '\1|\2'), '|.', '.')
from software_version;
Here is a db<>fiddle with the three versions, including the version a_horse_with_no_name mentions in the comment.
I'd just take the left and right:
concat(
left(str, position('.' in str)),
replace(right(str, -position('.' in str)), '.', '')
)
For a str of 12.34.56.78, the left gives 12. and the right using a negative position gives 34.56.78 upon which the replace then removes the dots
Basically translate will change character to character and Replace string to string , and here i have tried to remove spaces using translate to count the number words .
select translate(' #',' ','') from dual;
select replace(' #',' ','') from dual;
select ename , nvl(length(replace(TRANSLATE(upper(trim(ename)),'ABCDEFGHIJKLMNOPQRSTUVWXYZ'' ',' # '),' ',''))+1,1) NOOFWORDs
from emp;
Unfortunately Oracle has made many bizarre choices around null vs. empty string.
One of those has to do with TRANSLATE. TRANSLATE will return NULL if any of its arguments (including the last one) is NULL, no matter what the logical behavior should be.
So, to remove spaces (say) with TRANSLATE, you must add a character you do NOT want to be removed to both the second and the third argument. I added the lower-case letter z, but you could add anything (a dot, the digit 0, whatever - just make sure you add the same character at the beginning of both arguments)
... translate (input_string, 'z ', 'z') ....
For example:
select translate(' #','z ','z') from dual;
TRANSLATE('#','Z','Z')
------------------------
#
select translate(' #',' ','') from dual;
Returns NULL because in Oracle empty strings unfortunately yield NULLs. Therefore it's equivalent to
SELECT translate(' #', ' ', NULL)
FROM dual;
and translate() returns NULL when an argument is null. Actually this is well documented in "TRANSLATE":
(...)
You cannot use an empty string for to_string to remove all characters in from_string from the return value. Oracle Database interprets the empty string as null, and if this function has a null argument, then it returns null.
If you want to replace one character, use replace() as you already did. For a few but more than one characters you can nest the replace()s.
This however gets unhandy, when you want to replace quite a lot of characters. In such a situation, if the replacement character is only one character or the empty string regexp_replace() using a character class or alternates may come in handy.
For example
SELECT regexp_replace('a12b478c01', '[0-9]', '')
FROM dual;
replaces all the digits so just 'abc' remains and
SELECT regexp_replace('ABcc1233', 'c|3', '')
FROM dual;
removes any '3' or 'c' and results in 'AB12'. In your very example
SELECT regexp_replace(' #', ' ', '')
FROM dual;
would also work and give you '#'. Though in the simple case of your example a simple replace() is enough.
I'm sure this is super easy, but how would I go about converting LastName,FirstName to LastName,FirstInitial?
For example changing Smith,John to Smith,J or Johnson,John to Johnson,J etc.
Thank You!
In case of LastName and FirstName columns:
select LastName,substr(FirstName,1,1)
from mytable
;
In case of a fullname saved in a single column:
select substr(fullname,1,instr(fullname || ',',',')-1) || substr(fullname,instr(fullname || ',',','),2)
from mytable
;
or
select regexp_replace (fullname,'([^,]*,?)(.).*','\1\2')
from mytable
;
Here is one way, using just "standard" instr and substr. Assuming your input is a single string in the format 'Smith,John':
select substr(fullname, 1, instr(fullname, ',')+1) from yourtable;
yourtable is the name of the table, and fullname is the name of the column.
instr(fullname, ',') finds the position of the comma within the input string (it would be 6 in 'Smith,John'); thensubstrtakes the substring that begins at the first position (the1in the function call) and ends at the position calculated byinstr`, PLUS 1 (to get the first initial as well).
Given data in a column which look like this:
00001 00
00026 00
I need to use SQL to remove anything after the space and all leading zeros from the values so that the final output will be:
1
26
How can I best do this?
Btw I'm using DB2
This was tested on DB2 for Linux/Unix/Windows and z/OS.
You can use the LOCATE() function in DB2 to find the character position of the first space in a string, and then send that to SUBSTR() as the end location (minus one) to get only the first number of the string. Casting to INT will get rid of the leading zeros, but if you need it in string form, you can CAST again to CHAR.
SELECT CAST(SUBSTR(col, 1, LOCATE(' ', col) - 1) AS INT)
FROM tab
In DB2 (Express-C 9.7.5) you can use the SQL standard TRIM() function:
db2 => CREATE TABLE tbl (vc VARCHAR(64))
DB20000I The SQL command completed successfully.
db2 => INSERT INTO tbl (vc) VALUES ('00001 00'), ('00026 00')
DB20000I The SQL command completed successfully.
db2 => SELECT TRIM(TRIM('0' FROM vc)) AS trimmed FROM tbl
TRIMMED
----------------------------------------------------------------
1
26
2 record(s) selected.
The inner TRIM() removes leading and trailing zero characters, while the outer trim removes spaces.
This worked for me on the AS400 DB2.
The "L" stands for Leading.
You can also use "T" for Trailing.
I am assuming the field type is currently VARCHAR, do you need to store things other than INTs?
If the field type was INT, they would be removed automatically.
Alternatively, to select the values:
SELECT (CAST(CAST Col1 AS int) AS varchar) AS Col1
I found this thread for some reason and find it odd that no one actually answered the question. It seems that the goal is to return a left adjusted field:
SELECT
TRIM(L '0' FROM SUBSTR(trim(col) || ' ',1,LOCATE(' ',trim(col) || ' ') - 1))
FROM tab
One option is implicit casting: SELECT SUBSTR(column, 1, 5) + 0 AS column_as_number ...
That assumes that the structure is nnnnn nn, ie exactly 5 characters, a space and two more characters.
Explicit casting, ie SUBSTR(column,1,5)::INT is also a possibility, but exact syntax depends on the RDBMS in question.
Use the following to achieve this when the space location is variable, or even when it's fixed and you want to make a more robust query (in case it moves later):
SELECT CAST(SUBSTR(LTRIM('00123 45'), 1, CASE WHEN LOCATE(' ', LTRIM('00123 45')) <= 1 THEN LEN('00123 45') ELSE LOCATE(' ', LTRIM('00123 45')) - 1 END) AS BIGINT)
If you know the column will always contain a blank space after the start:
SELECT CAST(LOCATE(LTRIM('00123 45'), 1, LOCATE(' ', LTRIM('00123 45')) - 1) AS BIGINT)
both of these result in:
123
so your query would
SELECT CAST(SUBSTR(LTRIM(myCol1), 1, CASE WHEN LOCATE(' ', LTRIM(myCol1)) <= 1 THEN LEN(myCol1) ELSE LOCATE(' ', LTRIM(myCol1)) - 1 END) AS BIGINT)
FROM myTable1
This removes any content after the first space character (ignoring leading spaces), and then converts the remainder to a 64bit integer which will then remove all leading zeroes.
If you want to keep all the numbers and just remove the leading zeroes and any spaces you can use:
SELECT CAST(REPLACE('00123 45', ' ', '') AS BIGINT)
While my answer might seem quite verbose compared to simply SELECT CAST(SUBSTR(myCol1, 1, 5) AS BIGINT) FROM myTable1 but it allows for the space character to not always be there, situations where the myCol1 value is not of the form nnnnn nn if the string is nn nn then the convert to int will fail.
Remember to be careful if you use the TRIM function to remove the leading zeroes, and actually in all situations you will need to test your code with data like 00120 00 and see if it returns 12 instead of the correct value of 120.