I have two strings. I would like to know the upto how many characters are similar in both the strings.
E.x: lets say 'xyzabc', 'xyzadh'. I would like to know if there is a function that can give the index at which the similarity is breaking. In this case it would be 4 because upto 'xyza' the strings are same. If the strings are 'xyzabc', 'xymabc' then the result should be 2.
I would like to use it as select func('xyzabc', 'xyzwer'); to get the required answer. Kinldy let me know if there is a function existing in SQL.
Thanks a lot in advance!!!
One way to do this is with regular expressions. Here is the idea. You would use regexp_substr(). The regular expression would, in your example, be 'xyzwer|xyzwe|xyzw|xyz|xy|x'. You would then measure the length of the matching substring. In other words, you need to convert one of the strings to a regular expression.
An alternative is to use a giant case statement:
(case when left(str1, 10) = left(str2, 10) then 10
when left(str1, 9) = left(str2, 9) then 9
...
when left(str1, 1) = left(str2, 1) then 1
else 0
end)
This assumes that 10 is the longest string.
Related
I have a column in my table with these values:
PING_TO_ME_20100828_Any87
TO_THESE_D_COLUMN_ENTRY_20200825
TO_THESE_D_20100829_COLUMN_ENTRY
201901_ARE_YOU_TRYING_TO_REACH47
ASK_TO_UOU_201008
I need to separate date values in a separate column.
My output should be:
20100828
20200825
20100829
201901
201008
Any help is very much appreciated.
You will (and already have) likely get comments about this telling you to fix your design. And while that is likely true...I won't try to pick apart why you are doing this, and I'll just give you the answer you came here for.
Your goal is to pick out either an 8 digit string of integers, or a 6 digit string of integers.
Here is one way you could do it:
SELECT x.y
, COALESCE(SUBSTRING(x.y, NULLIF(PATINDEX('%[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]%', x.y), 0), 8)
, SUBSTRING(x.y, NULLIF(PATINDEX('%[0-9][0-9][0-9][0-9][0-9][0-9]%', x.y), 0), 6))
FROM (
VALUES ('PING_TO_ME_20100828_Any87'),
('TO_THESE_D_COLUMN_ENTRY_20200825'),
('TO_THESE_D_20100829_COLUMN_ENTRY'),
('201901_ARE_YOU_TRYING_TO_REACH47'),
('ASK_TO_UOU_201008')
) x(y)
Explanation:
Since you are looking for both 8 and 6 digit values, you need to check for the longer of the two first. So first I search for the occurrence of a string of 8 integers using:
NULLIF(PATINDEX('%[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]%', x.y), 0)
This returns the first position of a string of 8 integers. The reason I wrap it in a NULLIF() is because if the value is not found, then PATINDEX will return 0.
I use NULLIF() to return NULL in that case, essentially indicating nothing was found. If you pass a NULL value to SUBSTRING() then it also returns NULL.
This is all just a nice way of "failing over" to the 6 character string check.
So there I do the same thing again:
NULLIF(PATINDEX('%[0-9][0-9][0-9][0-9][0-9][0-9]%', x.y), 0)
Except this time, I only repeat [0-9] six times. And again, I use the NULLIF() trick, so that it returns NULL if no string is found.
Throw that all into SUBSTRING() and COALESCE() and you've got a function that returns the results you're looking for.
Potential downsides
There are a couple down sides to this method.
It is not checking for a valid date, it's simply looking for a string of either 8 integers, or 6 integers. It could be 12345678 and it would still detect and return that.
If there are strings of integers longer than 8 digits, it will grab only the first 8 characters.
If there are multiple occurrences of 6 or 8 character integer strings...it will only return the first one.
There are much more robust ways you could write this, but it all depends on your data and what you need to do.
Other methods
Another way it could be done depending on which version of SQL Server you are using, is using STRING_SPLIT().
SELECT x.y, s.[value]
FROM (
VALUES ('PING_TO_ME_20100828_Any87'),('TO_THESE_D_COLUMN_ENTRY_20200825'),('TO_THESE_D_20100829_COLUMN_ENTRY'),('201901_ARE_YOU_TRYING_TO_REACH47'),('ASK_TO_UOU_201008')
) x(y)
CROSS APPLY (
SELECT [value]
FROM STRING_SPLIT(x.y, '_')
WHERE [value] LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]'
OR [value] LIKE '[0-9][0-9][0-9][0-9][0-9][0-9]'
) s
This method handles a couple of the downsides mentioned earlier. For example, it will ONLY return integer strings of length 6 or 8. It will also return ALL integer strings of length 6 or 8 and not just the first one.
And there's other ways to identify the strings as well, like using ISNUMERIC(x.[value]) or TRY_CONVERT(int, s.[value]).
It all depends on how you are using this code...if it's runs fast enough, and it's a one off script, then it really doesn't matter. If it's running for millions of records at a time, then yeah you should play around with other methods.
If I have a number (such as 88) and I want to perform a LIKE query in Rails on a primary ID column to return all records that contain that number at the end of the ID (IE: 88, 288, etc.), how would I do that? Here's the code to generate the result, which works fine in SQLLite:
#item = Item.where("id like ?", "88").all
In PostgreSQL, I'm running into this error:
PG::Error: ERROR: operator does not exist: integer ~~ unknown
How do I do this? I've tried converting the number to a string, but that doesn't seem to work either.
Based on Erwin's Answer:
This is a very old question, but in case someone needs it, there is one very simple answer, using ::text cast:
Item.where("(id::text LIKE ?)", "%#{numeric_variable}").all
This way, you find the number anywhere in the string.
Use % wildcard to the left only if you want the number to be at the end of the string.
Use % wildcard to the right also, if you want the number to be anywhere in the string.
Simple case
LIKE is for string/text types. Since your primary key is an integer, you should use a mathematical operation instead.
Use modulo to get the remainder of the id value, when divided by 100.
Item.where("id % 100 = 88")
This will return Item records whose id column ends with 88
1288
1488
1238872388
862388
etc...
Match against arbitrary set of final two digits
If you are going to do this dynamically (e.g. match against an arbitrary set of two digits, but you know it will always be two digits), you could do something like:
Item.where(["id % 100 = ?", last_two_digits)
Match against any set or number of final digits
If you wanted to match an arbitrary number of digits, so long as they were always the final digits (as opposed to digits appearing elsewhere in the id field), you could add a custom method on your model. Something like:
class Item < ActiveRecord
...
def find_by_final_digits(num_digits, digit_pattern)
# Where 'num_digits' is the number of final digits to match
# and `digit_pattern` is the set of final digits you're looking fo
Item.where(["id % ? = ?", 10**num_digits, digit_pattern])
end
...
end
Using this method, you could find id values ending in 88, with:
Item.find_by_final_digits(2, 88)
Match against a range of final digits, of any length
Let's say you wanted to find all id values that end with digits between 09 and 12, for whatever reason. Maybe they represent some special range of codes you're looking up. To do this you could do another custom method to use Postgres' BETWEEN to find on a range.
def find_by_final_digit_range(num_digits, start_of_range, end_of_range)
Item.where(["id % ? BETWEEN ? AND ?", 10**num_digits, start_of_range, end_of_range)
end
...and could be called using:
Item.find_by_final_digit_range(2, 9, 12)
...of course, this is all just a little crazy, and probably overkill.
The LIKE operator is for string types only.
Use the modulo operator % for what you are trying to do:
#item = Item.where("(id % 100) = ?", "88").all
I doubt it "works" in SQLite, even though it coerces the numeric types to strings. Without leading % the pattern just won't work.
-> sqlfiddle demo
Cast to text and use LIKE as you intended for arbitrary length:
#item = Item.where("(id::text LIKE ('%'::text || ?)", "'12345'").all
Or, mathematically:
#item = Item.where("(id % 10^(length(?)) = ?", "'12345'", "12345").all
LIKE operator does not work with number types and id is the number type so you can use it with concat
SELECT * FROM TABLE_NAME WHERE concat("id") LIKE '%ID%'
I need to order a select query using a varchar column, using numerical and text order. The query will be done in a java program, using jdbc over postgresql.
If I use ORDER BY in the select clause I obtain:
1
11
2
abc
However, I need to obtain:
1
2
11
abc
The problem is that the column can also contain text.
This question is similar (but targeted for SQL Server):
How do I sort a VARCHAR column in SQL server that contains words and numbers?
However, the solution proposed did not work with PostgreSQL.
Thanks in advance, regards,
I had the same problem and the following code solves it:
SELECT ...
FROM table
order by
CASE WHEN column < 'A'
THEN lpad(column, size, '0')
ELSE column
END;
The size var is the length of the varchar column, e.g 255 for varying(255).
You can use regular expression to do this kind of thing:
select THECOL from ...
order by
case
when substring(THECOL from '^\d+$') is null then 9999
else cast(THECOL as integer)
end,
THECOL
First you use regular expression to detect whether the content of the column is a number or not. In this case I use '^\d+$' but you can modify it to suit the situation.
If the regexp doesn't match, return a big number so this row will fall to the bottom of the order.
If the regexp matches, convert the string to number and then sort on that.
After this, sort regularly with the column.
I'm not aware of any database having a "natural sort", like some know to exist in PHP. All I've found is various functions:
Natural order sort in Postgres
Comment in the PostgreSQL ORDER BY documentation
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%'