I need for a SQL query to transform an int with a value between 1 to 300000 to a number which has this pattern : always 8 number.
For example:
1 becomes 00000001,
123 becomes 00000123,
123456 becomes 00123456.
I have no idea how to do that... How can I do it?
In Standard SQL, you can use this trick:
select substring(cast( (num + 100000000) as varchar(255)) from 2)
Few databases actually support this syntax. Any given database can do what you want, but the method depends on the database you are using.
For MS SQL Server
You could use FORMAT function, like this:
SELECT FORMAT(123,'00000000')
https://database.guide/how-to-format-numbers-in-sql-server/#:~:text=Starting%20from%20SQL%20Server%202012,the%20output%20should%20be%20formatted.
Read at the link Leading Zeroes
For MySql/Oracle
You could use LPAD, like this:
SELECT LPAD('123',8,'0')
https://database.guide/how-to-add-leading-zeros-to-a-number-in-mysql/
Related
I'm trying to figure out a way, using SQL, to query for values that go out to, say, 5 or more decimal places. In other words, I want to see only results that have 5+ decimal places (e.g. 45.324754) - the numbers before the decimal are irrelevant, however, I still need to see the full number. Is this possible? Any help if appreciated.
Assuming your DBMS supports FLOOR and your datatype conversion model supports this multiplication, you can do this:
SELECT *
FROM Table
WHERE FLOOR(Num*100000)!=Num*100000
This has the advantage of not requiring a conversion to a string datatype.
On SQL Server, you can specify:
SELECT *
FROM Table
WHERE Value <> ROUND(Value,4,1);
For an ANSI method, you can use:
SELECT *
FROM Table
WHERE Value <> CAST(Value*100000.0 AS INT) / 100000.0;
Although this method might cause an overflow if you're working with large numbers.
I imagine most DBMSs have a round function
SELECT *
FROM YourTable
WHERE YourCol <> ROUND(YourCol,4)
This worked for me in SQL Server:
SELECT *
FROM YourTable
WHERE YourValue LIKE '%._____%';
select val
from tablename
where length(substr(val,instr(val, '.')+1)) > 5
This is a way to do it in oracle using substr and instr
You can use below decode statement to identify maximum decimal present in database table
SELECT max(decode(INSTR(val,'.'), 0, 0, LENGTH(SUBSTR(val,INSTR(val,'.')+1)))) max_decimal
FROM tablename A;
I am currently using SQL Server 2008, and i have a list of numbers. There are some that are associated with letters, and I would like to pull only the numbers that are associated with these values. I thought a right statement might work, but I'm kind of a noob, and havent been able to find a way that works.
For example:
columnName
==============
12
13
122b
11a
10z
9
609
I want to view:
Result
==============
122b
11a
10z
try this:
WHERE YourColumn like '%[a-z]'
SQL Server 2008 doesn't support regexes directly (it requires a CLR extension), but it's pattern matching supports ranges:
WHERE column LIKE '%[a-zA-Z]%'
SELECT *
FROM tableName
WHERE ISNUMERIC(columnName) = 0
SQLFIddle Demo
Other Source(s)
ISNUMERIC
I have a zipcode stored in a text field (string) and would like to select only the last 3 digits of the value in my select statement. is this possible? Is there a standard way of doing this so that the SQL is interchangeable accross databases? I will be using it in production on Oracle, but i test on Interbase (yes, yes, i know, two totally diff DBs, but thats what i am doing)
thanks for any help you can offer
Assuming the zipcodes all have the same length, you can use substr.
If they don't have the same length, you have to do similar things with the strlen function.
Interbase does not have a built-in substring function, but it does have a UDF (user defined function) called SUBSTR in lib_udf.dll that works like this:
select substr(clients.lastname, 1, 10)
from clients
You declare the UDF like this:
DECLARE EXTERNAL FUNCTION SUBSTR
CSTRING(80),
SMALLINT,
SMALLINT
RETURNS CSTRING(80) FREE_IT
ENTRY_POINT 'IB_UDF_substr' MODULE_NAME 'ib_udf';
Oracle does have a built-in substr function that you use like this:
select substr(clients.lastname, 1, 10)
from clients
--jeroen
This depends on how your storing the zip code. If you are using 5 digits only
then this should work for Oracle and may work for Interbase.
select * from table where substr(zip,3,3) = '014'
IF you store Zip + 4 and you want the last 3 digits and some are 5 digits and some are 9 digits you would have to do the following.
select * from table where substr(zip,length(zip) -2,3) = '014'
and one option that may work better in both databases is
select * from table where zip like '%014'
I need to filter out junk data in SQL (SQL Server 2008) table. I need to identify these records, and pull them out.
Char[0] = A..Z, a..z
Char[1] = 0..9
Char[2] = 0..9
Char[3] = 0..9
Char[4] = 0..9
{No blanks allowed}
Basically, a clean record will look like this:
T1234, U2468, K123, P50054 (4 record examples)
Junk data looks like this:
T12.., .T12, MARK, TP1, SP2, BFGL, BFPL (7 record examples)
Can someone please assist with a SQL query to do a LEFT and RIGHT method and extract those characters, and do a LIKE IN or something?
A function would be great though!
The following should work in a few different systems:
SELECT *
FROM TheTable
WHERE Data LIKE '[A-Za-z][0-9][0-9][0-9][0-9]%'
AND Data NOT LIKE '% %'
This approach will indeed match P2343, P23423JUNK, and other similar text but requires that the format is A0000*.
Now, if the OP implies a format of 1st position is a character and all succeeding positions are numeric, as in A0+, then use the following (in SQL Server and a good deal of other database systems):
SELECT *
FROM TheTable
WHERE SUBSTRING(Data, 1, 1) LIKE '[A-Za-z]'
AND SUBSTRING(Data, 2, LEN(Data) - 1) NOT LIKE '%[^0-9]%'
AND LEN(Data) >= 5
To incorporate this into a SQL Server 2008 function, since this appears to be what you'd like most, you can write:
CREATE FUNCTION ufn_IsProperFormat(#data VARCHAR(50))
RETURNS BIT
AS
BEGIN
RETURN
CASE
WHEN SUBSTRING(#Data, 1, 1) LIKE '[A-Za-z]'
AND SUBSTRING(#Data, 2, LEN(#Data) - 1) NOT LIKE '%[^0-9]%'
AND LEN(#Data) >= 5 THEN 1
ELSE 0
END
END
...and call into it like so:
SELECT *
FROM TheTable
WHERE dbo.ufn_IsProperFormat(Data) = 1
...this query needs to change for Oracle queries because Oracle doesn't appear to support bracket notation in LIKE clauses:
SELECT *
FROM TheTable
WHERE REGEXP_LIKE(Data, '^[A-za-z]\d{4,}$')
This is the expansion gbn is doing in his answer, but these versions allow for varying string lengths without the OR conditions.
EDIT: Updated to support examples in SQL Server and Oracle for ensuring the format A0+, so that A1324, A2342388, and P2342 match but A2342JUNK and A234 do not.
The Oracle REGEXP_LIKE code was borrowed from Mark's post but updated to support 4 or more numeric digits.
Added a custom SQL Server 2008 approach which implements these techniques.
Depends on your database. Many have regex functions (note examples not tested so check)
e.g. Oracle
SELECT x
FROM table
WHERE REGEXP_LIKE(x, '^[A-za-z][:digit:]{4}$')
Sybase uses LIKE
Given that you're allowing between 3 and 6 digits for the number in your examples then it's probably better to use the ISNUMERIC() function on the 2nd character onwards:
SELECT *
FROM TheTable
-- start with a letter
WHERE Data LIKE '[A-Za-z]%'
-- everything from 2nd character onwards is a number
AND ISNUMERIC( SUBSTRING( Data, 2, 50 ) ) = 1
-- number doesn't have a decimal place
AND Data NOT LIKE '%.%'
For more information look at the ISNUMERIC function on MSDN.
Also note that:
I've limited the 2nd part with the number to 50 characters maximum, change this to suit your needs.
Strictly speaking you should check for currency symbols etc, as ISNUMERIC allows them, as well as +/- and some others
A better option might be to create a function that checks that each character after the first is between 0 and 9 (or 1 and 0 if you're using ASCII codes).
You can't use Regular Expressions in SQL Server, so you have to use OR. Correcting David Andres' answer...
WHERE
(
Data LIKE '[A-Za-z][0-9][0-9][0-9]'
OR
Data LIKE '[A-Za-z][0-9][0-9][0-9][0-9]'
OR
Data LIKE '[A-Za-z][0-9][0-9][0-9][0-9][0-9]'
)
David's answer allows "D1234junk" through
You also only need "[A-Z]" if you don't have case sensitivity
I have stored values in my database that look like 5XXXXXX, where X can be any digit. In other words, I need to match incoming SQL query strings like 5349878.
Does anyone have an idea how to do it?
I have different cases like XXXX7XX for example, so it has to be generic. I don't care about representing the pattern in a different way inside the SQL Server.
I'm working with c# in .NET.
You can write queries like this in SQL Server:
--each [0-9] matches a single digit, this would match 5xx
SELECT * FROM YourTable WHERE SomeField LIKE '5[0-9][0-9]'
stored value in DB is: 5XXXXXX [where x can be any digit]
You don't mention data types - if numeric, you'll likely have to use CAST/CONVERT to change the data type to [n]varchar.
Use:
WHERE CHARINDEX(column, '5') = 1
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
CHARINDEX
ISNUMERIC
i have also different cases like XXXX7XX for example, so it has to be generic.
Use:
WHERE PATINDEX('%7%', column) = 5
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
PATINDEX
Regex Support
SQL Server 2000+ supports regex, but the catch is you have to create the UDF function in CLR before you have the ability. There are numerous articles providing example code if you google them. Once you have that in place, you can use:
5\d{6} for your first example
\d{4}7\d{2} for your second example
For more info on regular expressions, I highly recommend this website.
Try this
select * from mytable
where p1 not like '%[^0-9]%' and substring(p1,1,1)='5'
Of course, you'll need to adjust the substring value, but the rest should work...
In order to match a digit, you can use [0-9].
So you could use 5[0-9][0-9][0-9][0-9][0-9][0-9] and [0-9][0-9][0-9][0-9]7[0-9][0-9][0-9]. I do this a lot for zip codes.
SQL Wildcards are enough for this purpose. Follow this link: http://www.w3schools.com/SQL/sql_wildcards.asp
you need to use a query like this:
select * from mytable where msisdn like '%7%'
or
select * from mytable where msisdn like '56655%'