I am trying to find out how to write an SQL statement that will grab fields where the string is not 12 characters long. I only want to grab the string if they are 10 characters.
What function can do this in DB2?
I figured it would be something like this, but I can't find anything on it.
select * from table where not length(fieldName, 12)
From similar question DB2 - find and compare the lentgh of the value in a table field - add RTRIM since LENGTH will return length of column definition. This should be correct:
select * from table where length(RTRIM(fieldName))=10
UPDATE 27.5.2019: maybe on older db2 versions the LENGTH function returned the length of column definition. On db2 10.5 I have tried the function and it returns data length, not column definition length:
select fieldname
, length(fieldName) len_only
, length(RTRIM(fieldName)) len_rtrim
from (values (cast('1234567890 ' as varchar(30)) ))
as tab(fieldName)
FIELDNAME LEN_ONLY LEN_RTRIM
------------------------------ ----------- -----------
1234567890 12 10
One can test this by using this term:
where length(fieldName)!=length(rtrim(fieldName))
This will grab records with strings (in the fieldName column) that are 10 characters long:
select * from table where length(fieldName)=10
Mostly we write below statement
select * from table where length(ltrim(rtrim(field)))=10;
Related
Here is my problem statement:
I have single column table having the data like as :
ROW-1>> 7302-2210177000-XXXX-XXXXXX-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-U-XXXXXXXXX-XXXXXX
ROW-2>> 0311-1130101-XXXX-000000-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-X-XXXXXXXXX-WIPXXX
Here i want to separate these values from '-' and load into a new table. There are 11 segments in this string separated by '-', therefore, 11 columns. The problem is:
A. The length of these values are changing, however, i have to keep it as the length of these values in the standard format or the length which it has
e.g 7302- (should have four values, if the value less then that then keep that value eg. 73 then it should populate 73.
Therefore, i have to separate as well as mentation the integrity. The code which i am writing is :
select
SUBSTR(PROFILE_ID,1,(case when length(instr(PROFILE_ID,'-')<>4) THEN (instr(PROFILE_ID,'-') else SUBSTR(PROFILE_ID,1,4) end)
)AS [RQUIRED_COLUMN_NAME]
from [TABLE_NAME];
getting right parenthesis error
Please help.
I used the regex_substr SQL function to solve the above issue. Here below is an example:
select regex_substr('7302-2210177000-XXXX-XXXXXX-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-U-XXXXXXXXX-XXXXXX ROW-2>> 0311-1130101-XXXX-000000-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-X-XXXXXXXXX-WIPXXX',[^-]+,1,1);
Output is: 7302 --which is the 1st segment of the string
Similarly, the send string segment which is separated by "-" in the string can be obtained by just replacing the 1 with 2 in the above query at the end.
Example : select regex_substr('7302-2210177000-XXXX-XXXXXX-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-U-XXXXXXXXX-XXXXXX ROW-2>> 0311-1130101-XXXX-000000-XXX-XXXXXXXXXX-XXXXXX-XXXXXX-X-XXXXXXXXX-WIPXXX',[^-]+,1,2);
output: 2210177000 which is the 2nd segment of the string
Say I have a table named 'Parts'. I am looking to create a SQL query that compares the first X characters of two of the fields, let's call them 'PartNum1' and 'PartNum2'. For example, I would like to return all records from 'Parts' where the first 6 characters of 'PartNum1' equals the first 6 characters of 'PartNum2'.
Parts
PartNum1
PartNum2
12345678
12345600
12388888
12345000
12000000
14500000
the query would only return row 1 since the first 6 characters match. MS SQL Server 2017 in case that makes a difference.
If they are strings, use left():
left(partnum1, 6) = left(partnum2, 6)
This would be appropriate in a where, on, or case expression. Note that using left() would generally prevent the use of indexes. If this is for a join and you care about performance, you might want to include a computed column with the first six characters.
you can try something like this. I am assuming datatype as integer. You can set size of varchar based on length of fields.
select *
from Parts
WHERE SUBSTRING(CAST(PartNum1 AS VARCHAR(max)), 1,6) = SUBSTRING(CAST(PartNum2 AS VARCHAR(max)), 1,6)
You can go for simple division to see if the numerator matches for those partnumbers.
DECLARE #table table(partnum int, partnum2 int)
insert into #table values
(12345678, 12345600)
,(12388888, 12345000)
,(12000000, 14500000);
select * from #table where partnum/100 = partnum2/100
partnum
partnum2
12345678
12345600
I have data like this
1234500010
1234500020
1234500021
12345600010
12345600011
123456700010
123456700020
123456710010
The pattern is
1-data(varian 3-7 digit number) + 2-data(any 3 digit number) + 3-data (any 2 digit number)
I want to create SQL to get 1-data only.
For example I want to get data 12345
I want the result only
1234500010
1234500020
1234500021
If I using "like",
select *
FROM data
where ID like '12345%' `
I will get all the data with 12345, 123456 and 1234567
If I using equal, I will only get one specific data.
Can I combine like and equal together to get result like what I want?
select * FROM data where data = '12345 + any 2-data(3 digit) + any 3-data(2 digit)'
Anyone can help?
Addition : Sorry if I didn't mention the data type and make some miss communication. The data type is in char. #Gordon answers and the others not wrong. It works for number and varchar. but not works for char type. Here I post some pic for char data type. Oracle specification for char data type is a fixed lenght. So if I input less than lenght the remain of it will be change into a space.
Thank you very much. Hope someone can help for this
Since your datatype is CHAR, Gordon's answer is not working for you. CHAR adds trailing spaces for the strings less than maximum limit. You could use TRIM to fix this as shown. But, you should preferably store numbers in the NUMBER type and not CHAR or VARCHAR2, which will create other problems sooner or later.
select *
from data
where trim(ID) like '12345_____';
I think you want:
select *
from data
where ID like '12345_____' -- exactly 5 _
Here is a rextester demonstrating the answer.
You really can't combine equality and LIKE. But you can use a regular expression to do this kind of searching, with the REGEXP_LIKE function:
SELECT *
FROM DATA
WHERE REGEXP_LIKE(ID, '^12345[0-9]{3}[0-9]{2}');
But if I understand correctly, for your 1-data you really want a 3 to 7 digit number:
SELECT *
FROM DATA
WHERE REGEXP_LIKE(ID, '^[0-9]{3,7}[0-9]{3}[0-9]{2}');
Oracle regular expression docs here
SQLFiddle here
Best of luck.
I think this gives you the solution you want,
create table data(ID number(15));
insert into data values(1234500010);
insert into data values(1234500020);
insert into data values(1234500021);
insert into data values(12345600010);
insert into data values(12345600011);
insert into data values(123456700010);
insert into data values(123456700020);
insert into data values(123456710010);
select * from data where ID like '12345_____'
// After 5_ underscore are exactly 5 , any 3 digits from 2-data(3 underscores) and 2 digits from 3-data(2 underscores)
You'll be getting(OUTPUT) :
ID
1234500010
1234500020
1234500021
3 rows returned in 0.00 seconds
I am trying to write a SQL query that only returns rows where a specific column (let's say 'amount' column) contains numbers comprising of only one digit, e.g. only '1's (1111111...) or only '2's (2222222...), etc.
In addition, 'amount' column contains numbers with decimal points as well and these kind of values should also be returned, e.g. 1111.11, 2222.22, etc
If you want to make the query generic that you don't have to specify each possible digit you could change the where to the following:
WHERE LEN(REPLACE(REPLACE(amount,LEFT(amount,1),''),'.','') = 0
This will always use the first digit as comparison for the rest of the string
If you are using SQL Server, then you can try this script:
SELECT *
FROM (
SELECT CAST(amount AS VARCHAR(30)) AS amount
FROM TableName
)t
WHERE LEN(REPLACE(REPLACE(amount,'1',''),'.','') = 0 OR
LEN(REPLACE(REPLACE(amount,'2',''),'.','') = 0
I tried like this in place of 1111111 replace with column name:
Select replace(Str(1111111, 12, 2),0,left(11111,1))
I have a column Property in a table Order. We used to store a dictionary(varchar) in this column.
We now have a new column (int) called ShoeSize which has a default value of 0.
So what I want to achieve is to retrieve only the numeric value out of the Property column and update the ShoeSize column with that value.
The Property column value looks like this:
ShoeSize<|$à&£#>15<|#ç§~#>
or
ShoeSize<|$à&£#>3<|#ç§~#>
My question is:
How can I manipulate my SQL in select statement to select only the numeric value of the Property column? So in other words, How would I be able to only end up with 15 or 3
Thanks in advance.
If you are using MSSQL and your format is fixed then you can try like below
DECLARE #str VARCHAR(50)
SET #str = 'ShoeSize<|$à&£#>15<|#ç§~#>'
SELECT SUBSTRING(#str,CHARINDEX('£#>',#str) + 3, CHARINDEX('<|#',#str) - (CHARINDEX('£#>',#str) + 3))
You can use REGEXP_SUBSTR to extract the numbers from your string.
For example,
SELECT REGEXP_SUBSTR('ShoeSize<|$à&£#>15<|#ç§~#>','[[:digit:]]+') FROM dual;
will return 15 to you. Of course, my assumption is that the shoe size in your string will be a number, and your intention is not to collate all digits which occur in your string to come at a number which you will save as shoe size