I have got a table with a column that is type of VARCHAR2(255 BYTE). I would like to select only these rows that have numbers as a value, so I discard any other values as for example "lala","1z". I just want to have pure numbers from 1 to ..... 999999999 (just digital numbers in other words) :P
Could you tell me how to make it?
if you're using Oracle 12c r2 or later then use the built-in validate_conversion() function:
select *
from your_table
where validate_conversion(cast(your_column as number)) = 0
validate_conversion() returns 0 when the proposed conversion would succeed and 1 when it wouldn't. It also supports date and timestamp conversions. Find out more.
Something like this is the usual option. You could use regexp, but it's usually a bit slower.
select column1
from tableA
where translate(column1, '1234567890', '') is null;
Here's the regexp version kfinity referred to. The regex matches a line consisting of 1 or more digits.
select column1
from tableA
where regexp_like(column1, '^\d+$');
You don't want zero to start a number. So it seems like regular expressions are the way to go:
where regexp_like(column1, '^[1-9][0-9]*$');
Related
My question is very similar to this one: removing all the rows from a table with columns A and B, where some records include non-numeric characters (looking like '1234#5' or '1bbbb'). However, the solutions I read around don't seem to work for me. For example,
SELECT count(*) FROM tbl
--962060;
SELECT count(*)
FROM tbl
WHERE (REGEXP_like(A,'[^0-9]') OR REGEXP_like(B,'[^0-9]') ) ;
--17
SELECT count(*)
FROM tbl
WHERE (REGEXP_like(A,'[0-9]') and REGEXP_like(B,'[0-9]') )
;
--962060
From the 3rd query, I'd expect to see (962060-17)=962043. Why is it still 962060? An alternative query like this also gives the same answer:
SELECT count(*)
FROM tbl
WHERE (REGEXP_like(A,'[[:digit:]]')and REGEXP_like(B,'[[:digit:]]') )
;
--962060
Of course, I could bypass the problem by doing query1 minus query2, but I'd like to learn how to do that using regular expressions.
If you use regexp you should take in account that any part of string may be matched as regexp. According your example you should specify that whole string should cntain only numbers ^ - is the beginig of string $ - is the end. And you may use \d- is digits
SELECT count(*)
FROM tbl
WHERE (REGEXP_like(A,'^[0-9]+$') and REGEXP_like(B,'^[0-9]+$') )
or
SELECT count(*)
FROM tbl
WHERE (REGEXP_like(A,'^\d+$') and REGEXP_like(B,'^\d+$') )
I know you specifically asked for a regex solution, but translate can solve these kind of questions as well (and usually faster because regexes use more processing power):
select count(1)
from tbl
where translate(a, 'x0123456789', 'x') is null
and translate(b, 'x0123456789', 'x') is null;
What this does: translate the characters 0123456789 to null, and if the result is null, then the input must have been all digits. The 'x' is just there because the third argument to translate can not be null.
Thought I should add this here, might be helpful to other readers.
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 have a data set that looks something like this:
A6177PE
A85506
A51SAIO
A7918F
A810004
A11483ON
A5579B
A89903
A104F
A9982
A8574
A8700F
And I need to find all the ENDings where they are non-numeric. In this example, that means PE, AIO, F, ON, B and F.
In pseudocode, I'm imagining I need something like
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,[SOME_CLEVER_LOGIC]) AS X FROM TABLE);
Any ideas? Can I solve this without learning regexp?
EDIT: To clarify, my data set is a lot larger than this example. Also, I'm only interested in the part of the string AFTER the numeric part. If the string is "A6177PE" I want "PE".
Disclaimer: I don't know Oracle SQL. But, I think something like this should work:
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,REGEXP_INSTR(COL, "[[:ALPHA:]]+$")) AS X FROM TABLE);
REGEXP_INSTR(COL, "[[:ALPHA:]]+$") should return the position of the first of the characters at the end of the field.
For readability, I'd recommend using the REGEXP_SUBSTR function (If there are no performance issues of course, as this is definitely slower than the accepted solution).
...also similar to REGEXP_INSTR, but instead of returning the position of the substring, it returns the substring itself
SELECT DISTINCT SUBSTR(MY_COLUMN,REGEXP_SUBSTR("[a-zA-Z]+$")) FROM MY_TABLE;
(:alpha: is supported also, as #Audun wrote )
Also useful: Oracle Regexp Support (beginning page)
For example
SELECT SUBSTR(col,INSTR(TRANSLATE(col,'A0123456789','A..........'),'.',-1)+1)
FROM table;
I need to order a select query using a varchar column, using numerical and text order. The query will be done in a java program, using jdbc over postgresql.
If I use ORDER BY in the select clause I obtain:
1
11
2
abc
However, I need to obtain:
1
2
11
abc
The problem is that the column can also contain text.
This question is similar (but targeted for SQL Server):
How do I sort a VARCHAR column in SQL server that contains words and numbers?
However, the solution proposed did not work with PostgreSQL.
Thanks in advance, regards,
I had the same problem and the following code solves it:
SELECT ...
FROM table
order by
CASE WHEN column < 'A'
THEN lpad(column, size, '0')
ELSE column
END;
The size var is the length of the varchar column, e.g 255 for varying(255).
You can use regular expression to do this kind of thing:
select THECOL from ...
order by
case
when substring(THECOL from '^\d+$') 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 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%'