Can I combine like and equal to get data? - sql

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

Related

Query to find if a column contains both number and decimal only

I have a column to check if contains number from 0-9 and a decimal. Since in the version of SQL am using the below does not seem working
select *
from tablename
whwere columnname like '%[^.0-9]%'
Also tried using column name like '%[0-9]%' and columnname not like '%.%' but if there is a negative sign it is not getting captured. Please advise.
The column data type is float. So can someone provide me a query to check if the column contains values from 0-9 and also it can contain decimal values these two are permitted. If say for example if I have value 9,9.99 ,-1.24 the query should output -1.24 I need this value other than decimal and number –
The issue with your LIKE clause is bad predicate logic ...like '%[^.0-9]%'should be NOT LIKE '%[^0-9.]%'
Take this sample data.
DECLARE #table TABLE (SomeNbr VARCHAR(32));
INSERT #table VALUES ('x'),('0'),('0.12'),('999'),('-29.33'),('88.33.22'),('9-9-'),('11-');
What you were trying to do would be accomplished like this:
SELECT t.someNbr
FROM #table AS t
WHERE someNbr NOT LIKE '%[^0-9.]%';
The problem here is we'll also return "88.33.22" and miss "-29.33", both valid float values. You can handle hyphens by adding a hyphen to your LIKE pattern:
SELECT t.someNbr, LEN(t.SomeNbr)-LEN(REPLACE(t.SomeNbr,'.',''))
FROM #table AS t
WHERE someNbr NOT LIKE '%[^0-9.-]%';
But now we also pick up "9-9-" and stuff with 2+ dots. To ensure that each starts with a number OR a hyphen, to ensure hyphens only exist in the front of the string (if at all) and that we a maximum of one dot:
--==== This will do a good job but can still be broken
SELECT t.someNbr
FROM #table AS t
WHERE someNbr NOT LIKE '%[^0-9.-]%' -- Can only contain numbers, dots and hyphens
AND LEN(t.SomeNbr)-LEN(REPLACE(t.SomeNbr,'.','')) < 2 -- can have up to 1 dot
AND LEN(t.SomeNbr)-LEN(REPLACE(t.SomeNbr,'-','')) < 2 -- can have up to 1 hyphen
AND PATINDEX('%-%',t.SomeNbr) < 2 -- hyphen can only be in the front
This does the trick and returns:
someNbr
--------------------------------
0
0.12
999
-29.33
All that said - **DONT DO THIS ANY OF THIS ^^^ **. There is no need to parse numbers in this way except to show others why not to. I can still break this. They way I return valid floats in a scenario like this is with TRY_CAST or TRY_CONVERT. This returns what you need and will perform better.
--==== Best Solution
SELECT t.someNbr
FROM #table AS t
WHERE TRY_CAST(t.SomeNbr AS float) IS NOT NULL;

SQL ORACLE : Is it possible to convert NUMBER with CHAR (varchar2 datatype) to NUMBER datatype

I have a big problem right now and I really need your help, because I can't find the right answer.
I am currently writing a script that triggers a migration process from a table with raw data (data we received from an excel file) to a new normalized schema.
My problem is that there is a column PRICE (varchar2 datatype) with a bunch of traps. For example: 540S, 25oo , I200 , S000 .
And I need to convert it to the correct NUMBER(9,2) format so I can get: 5405, 2500, 1200, 5000 as NUMBER for the previous examples and INSERT INTO my_new_table.
Is there any way I can parse every CHAR of these strings that verify certain conditions?
Or others better way?
Thank you :)!
One of the wonderful things about Oracle that some other DBs lack, is the TRANSLATE function:
SELECT TRANSLATE(number, 'SsIilOoxyz', '5511100') FROM t
This will convert:
S, s to 5
I, i and l to 1
O, o to 0
Remove any x, y or z from the number
The second and third arguments to translate define what characters are to be mapped. If the first string is longer than the second then anything over the length of the second is deleted from the resulting string. Mapping is direct based on position:
'SsIilOoxyz',
'5511100'
Look at the columns of the characters; the character above is mapped to the character below:
S->5,
s->5,
I->1,
i->1,
l->1,
O->0,
o->0,
x->removed,
y->removed,
z->removed`
You can use translate() and along with to_number(). Your rules are not exactly clear, but something like this:
select to_number(translate(price, '0123456789IoS', '012345678910'))
from t;
This replaces I with 1, o with 0, and removes S.

Access SQL: like function on a number field

I have ,for example, this table in a Microsoft Access database:
id numeric
context text
numberfield numeric
I want to select every record that ends with 9 in the column"numberfield". This gives a problem because it is a numeric field and as a result I can not use the following SQL:
select * from table where numberfield like "%9"
A solution is that I change the numberfield to a text. But this gives a problem because there are several users and the change might give a problem in the future. Is there an option to select on the ending when it is a number field?
That sound a little fishy.. are you sure you can use that query? Don't know about Access but almost any other DBMS allows it.
If it really doesn't work, you can do this:
select * from table where STR(numberfield) like "*9"
EDIT: Maybe it didn't work because you used % which is used with * in Access :
select * from table where numberfield like "*9"
Numbers are numbers, so use Mod for this:
select * from table where numberfield mod 10 = 9
Instead of casting to string and comparing, just extract the rightmost digit with a MOD operation.
Edit your query as follows:
SELECT *
FROM table
WHERE ((([numberfield] Mod 10)=9));

Make column values in a table have equal number of characters

I have a table DomainDetail having a column fieldID.
It has values like A1,B22,A567,D7779,B86759 .. i.e. from two characters to max six characters.
I want these values have equal number of characters
A000001,B000022,A000567,D07779 and B86759 .
This is how I think I should proceed
Estimate size of field value = LEN(fieldID)
Insert number of zeros equal to (6 - number of characters) .
How can I insert 0's sandwiched inside original value . How can do in SQL ?
like this
select
left(fieldID, 1) +
right('000000' + right(fieldID, len(fieldID) - 1), 5)
from DomainDetail
take a look at SQL FIDDLE example
It sounds like a problem better solved by business logic, i.e. the layer of code above your database. Whenever you insert, do the padding in that code - then always use that code/layer to insert into the table.
It seems to be a business logic requirement anyway - "ID must have a maximum 6 characters". Because a database wouldn't impose such a limit.
(unless you are using stored procedures as your business logic layer? in which case, PadLeft function in T-SQL)
select
stuff(fieldId,2,0,
LEFT('0000000000000000000000000',
(select max(LEN(FieldID)) from DomainDetail)
-LEN(fieldId)))
from DomainDetail
If you need a fixed output length just replace inner select (select max(LEN(FieldID)) from DomainDetail) with for example 6
Here is a SQLFiddle demo
If you want to UPDATE, then use this
UPDATE DomainDetail
SET fieldId=
SUBSTRING(fieldId,1,1)+
REPLICATE('0',6-LEN(id))+
SUBSTRING(fieldId,2,LEN(id)-1)
If you want to just SELECT without altering the values in the table, then this should work
SELECT SUBSTRING(id,1,1)+
REPLICATE('0',6-LEN(id))+
SUBSTRING(id,2,LEN(id)-1)
FROM DomainDetail
Hope this helps,
Raj
select stuff(fieldid, 2, 0, replicate('0', 6-len(fieldid)))
from DomainDetail

determine DB2 text string length

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;