I want to perform comparison for the specified columns in database, the comparison logic should compare the numbers regardless of their signs and will retrieve the result original with its sign.
For example, below code works well but as can be seen in the select block it returns the absolute value of columns. Is there any trick, cheat in Firebird 2.1 to overcome that?
SELECT a.ELM_NUM,a.COMBO, maxvalue(abs(a.N_1),abs(a.N_2)) as maxN from ntm a order by a.ELM_NUM
You can use a CASE condition:
SELECT a.ELM_NUM,a.COMBO,
CASE WHEN abs(a.N_1) > abs(a.N_2) THEN a.N_1 ELSE a.N_2 END as maxN
from ntm a
order by a.ELM_NUM
Related
I am implementing trigram similarity for word matching in column comum1. similarity() returns real. I have converted 0.01 to real and rounded to 2 decimal digits. Though there are rank values greater than 0.01, I get no results on screen. If I remove the WHERE condition, lots of results are available. Kindly guide me how to overcome this issue.
SELECT *,ROUND(similarity(comum1,"Search_word"),2) AS rank
FROM schema.table
WHERE rank >= round(0.01::real,2)
I have also converted both numbers to numeric and compared, but that also didn't work:
SELECT *,ROUND(similarity(comum1,"Search_word")::NUMERIC,2) AS rank
FROM schema.table
WHERE rank >= round(0.01::NUMERIC,2)
LIMIT 50;
The WHERE clause can only reference input column names, coming from the underlying table(s). rank in your example is the column alias for a result - an output column name.
So your statement is illegal and should return with an error message - unless you have another column named rank in schema.table, in which case you shot yourself in the foot. I would think twice before introducing such a naming collision, while I am not completely firm with SQL syntax.
And round() with a second parameter is not defined for real, you would need to cast to numeric like you tried. Another reason your first query is illegal.
Also, the double-quotes around "Search_word" are highly suspicious. If that's supposed to be a string literal, you need single quotes: 'Search_word'.
This should work:
SELECT *, round(similarity(comum1,'Search_word')::numeric,2) AS rank
FROM schema.table
WHERE similarity(comum1, 'Search_word') > 0.01;
But it's still pretty useless as it fails to make use of trigram indexes. Do this instead:
SET pg_trgm.similarity_threshold = 0.01; -- set once
SELECT *
FROM schema.table
WHERE comum1 % 'Search_word';
See:
Finding similar strings with PostgreSQL quickly
That said, a similarity of 0.01 is almost no similarity. Typically, you need a much higher threshold.
In a hive table how can I add the '-' sign in a field, but for random records? If I use the syntax below it changes all the records in the field to negative, but I want to change random records to negative.
This is the syntax I used which changed all the records to negative:
CAST(CAST(-1 AS DECIMAL(1,0)) AS DECIMAL(19,2))
*CAST(regexp_replace(regexp_replace(TRIM(column name),'\\-',''),'-','') as decimal(19,2)),
If you want to change random values to negative, why not use a case expression?
select (case when rand() < 0.5 then - column_name else column_name end)
Despite your query, this assumes that the column is a number of some sort, because negating strings doesn't make much sense.
I have a data source like following
If I ran the following sql query it removes all the records with "Seg Type" MOD and ignores the Fnn range given.
select * from NpsoQueue
where SegmentType not in ('MOD')
and Fnn not between 0888452158 and 0888452158
I want the query to consider both conditions. So, if I ran the query it should remove only the first record
The logic in your where clause is incorrect
Use
select * from NpsoQueue
where NOT (
SegmentType = 'MOD'
and Fnn between '0888452158' and '0888452158'
)
Also, a number with a leading zero is a string literal so you need to put single quotes around it to preserve the leading zero and stop implicit casts happening
As mentioned by #TriV you could also use OR. These are fundamental boolean logic concepts, i.e. not related to SQL Server or databases
I have data following data structure..
_ID _BEGIN _END
7003 99210 99217
7003 10225 10324
7003 111111
I want to look through every _BEGIN and _END and return all rows where the input value is between the range of values including the values themselves (i.e. if 10324 is the input, row 2 would be returned)
I have tried this filter but it does not work..
where #theInput between a._BEGIN and a._END
--THIS WORKS
where convert(char(7),'10400') >= convert(char(7),a._BEGIN)
--BUT ADDING THIS BREAKS AND RETURNS NOTHING
AND convert(char(7),'10400') < convert(char(7),a._END)
Less than < and greater than > operators work on xCHAR data types without any syntactical error, but it may go semantically wrong. Look at examples:
1 - SELECT 'ab' BETWEEN 'aa' AND 'ac' # returns TRUE
2 - SELECT '2' BETWEEN '1' AND '10' # returns FALSE
Character 2 as being stored in a xCHAR type has greater value than 1xxxxx
So you should CAST types here. [Exampled on MySQL - For standard compatibility change UNSIGNED to INTEGER]
WHERE CAST(#theInput as UNSIGNED)
BETWEEN CAST(a._BEGIN as UNSIGNED) AND CAST(a._END as UNSIGNED)
You'd better change the types of columns to avoid ambiguity for later use.
This would be the obvious answer...
SELECT *
FROM <YOUR_TABLE_NAME> a
WHERE #theInput between a._BEGIN and a._END
If the data is string (assuming here as we don't know what DB) You could add this.
Declare #searchArg VARCHAR(30) = CAST(#theInput as VARCHAR(30));
SELECT *
FROM <YOUR_TABLE_NAME> a
WHERE #searchArg between a._BEGIN and a._END
If you care about performance and you've got a lot of data and indexes you won't want to include function calls on the column values.. you could in-line this conversion but this assures that your predicates are Sargable.
SELECT * FROM myTable
WHERE
(CAST(#theInput AS char) >= a._BEGIN AND #theInput < a.END);
I also saw several of the same type of questions:
SQL "between" not inclusive
MySQL "between" clause not inclusive?
When I do queries like this, I usually try one side with the greater/less than on either side and work from there. Maybe that can help. I'm very slow, but I do lots of trial and error.
Or, use Tony's convert.
I supposed you can convert them to anything appropriate for your program, numeric or text.
Also, see here, http://technet.microsoft.com/en-us/library/aa226054%28v=sql.80%29.aspx.
I am not convinced you cannot do your CAST in the SELECT.
Nick, here is a MySQL version from SO, MySQL "between" clause not inclusive?
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