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
Related
I have a table that I save some data include list of numbers.
like this:
numbers
(null)
،42593
،42593،42594،36725،42592،36725،42592
،42593،42594،36725،42592
،31046،36725،42592
I would like to count the number elements in every row in SQL Server
count
0
1
6
4
3
You could use a replacement trick here:
SELECT numbers,
COALESCE(LEN(numbers) - LEN(REPLACE(numbers, ',', '')), 0) AS num_elements
FROM yourTable;
The above trick works by counting the number of commas (assuming your data really has commas as separators). For example, your last sample data point was:
,31046,36725,42592 => length is 18
310463672542592 => length is 15
Hence the difference in lengths correctly yields the right number of elements.
Another idea is to useSTRING_SPLIT:
SELECT y.numbers,
(SELECT COUNT(Value) - 1
FROM string_split(COALESCE(y.numbers,''),',')) AS num_elements
FROM yourtable AS y;
I know this looks a bit unhandy on first glance due to this strange -1 in the second line and the COALESCE in the third line. So why do I talk about this option?
Well, the strange thing in your case which causes these difficulties in my query is that your rows always start with a comma.
This is quite weird and it would be much easier without this first comma in every row.
Let's assume you remove this comma in future. Then this will become really easy and good readable:
SELECT y.numbers,
(SELECT COUNT(Value)
FROM string_split(y.numbers,',')) AS num_elements
FROM yourtable AS y;
Try out: db<>fiddle
your data
CREATE TABLE yourtable(
numbers VARCHAR(max)
);
INSERT INTO yourtable
(numbers) VALUES
(null),
('،42593'),
('،42593،42594،36725،42592،36725،42592'),
('،42593،42594،36725،42592'),
('،31046،36725،42592');
you need ISNULL and len
select
ISNULL(len(numbers) - len(replace(numbers,'،','')) ,0) count
from yourtable
the other way is by using IIF and string_split as follows
SELECT IIF(count < 0, 0, count) count
FROM (SELECT (SELECT Count(*) - 1
FROM STRING_SPLIT (Replace(Replace(numbers, 'R', ''), '،',
'R'), 'R'
)) AS
'count'
FROM yourtable) A
dbfiddle
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;
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 pretty hard request to do in SQL Server 2008, but I'm not able to do the whole...
I have two kind of records :
16HENFC******** (8 numbers after more 'FC')
16HEN******* (7 numbers after more 'EN')
I have to select the * (which are in fact numbers), and add a 0 at the beginning of the second form of record to just have 8 long selected values.
Then I have to insert the result in a empty table.
I think I did the first part which is :
SUBSTRING(SELECT mycolumn1 FROM mytable1 WHERE mycolumn1 LIKE '16HENFC%', 5, 8) ;
In summary,
I have those records in my column :
'16HENFC071052'
'16HEN5130026'
I want to select them and transform them to insert those ones in an other column :
'05130026'
'FC071052'
[EDIT]=>
CREATE TABLE nom_de_la_table
(
colonne1 VARCHAR(250),
colonne2 VARCHAR(250)
)
INSERT INTO nom_de_la_table (colonne1)
VALUES
('16HEN5138745'),
('16HENFC071052v2'),
('16HENFC78942878'),
('16HEN4830026'),
('16HEN7815934'),
('16HENFC74859422'),
('16HEN9687326'),
('16HENFC74889639'),
('16HEN9798556');
[etc...]
So two different types of records, and I want to insert the result of what you did first with just two records in an other column but for the 956 records of my table. And this is the result with the two examples :
'05130026'
'FC071052'
Left-Filling a string is a relatively easy request. Here's an example:
select right(replicate('0',8) + right(test,len(test)-len('16HEN')),8)
from (
select '16HENFC071052' as test
union all
select '16HEN5130026' as test
) z
Use replicate to left-fill your string with the amount of digits you wish to end up with. Append your desired string, in this case, slice your prefix off by taking the right X characters where X = len(target) - len(prefix). Finally, take the right characters of the whole string equal to your desired 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;