I have this query
SELECT *
FROM [RawData$]
WHERE 'Temperature[°C]' <= 100
But at the execution I get this error:
Data type mismatch in criteria expression.
The data in this column is 100% integer so I guess there is no problem.
Further this works fine:
SELECT *
FROM [RawData$]
WHERE 'Temperature[°C]'
I also tried this too but then I get no values at all:
SELECT *
FROM [RawData$]
WHERE 'Temperature[°C]' <= '100'
Actually the final question would be:
What query do i need to search a column which name is: Temperature[°C]
[Temperature[°C]][Temperature[[]°C]]
do not work.
This WHERE 'Temperature[°C]' <= 100 compares the literal string Temperature[°C] to the Integer 100.
Use WHERE [Temperature(°C)] < 100 instead.
Note:
square brackets [] are reserved to be wrapped around fieldnames (much like you tried to use single quotes ')
in this particular case, the square brackets in your header get interpreted as normal brackets ().
I think the issue is the column name has square brackets, so you have to maybe do it like this:
SELECT *
FROM [RawData$]
WHERE "Temperature[°C]" <= '100'
With the double quotes, instead. Let me know if that works.
Related
I my DB i want to select data which having A15-A19 using LIKE operator but couldnt get required result.
the code i made as SELECT * FROM MASTER_RULE WHERE VALUE BETWEEN LIKE 'A15%' AND 'A19%' and also tried regular expression as SELECT * FROM MASTER_RULE WHERE REGEXP_LIKE(value, 'A[1-9]') . But regexp gives all records not specified range 15-19.
How to achieve the solution for this?
Your first query is not ok, it has one extra keyword that you do not need.
Here is the regexp_like solution:
SELECT * FROM MASTER_RULE WHERE REGEXP_LIKE(value, '^A[1][5-9]')
Here is a demo
UPDATE:
Here is the "BETWEEN SOLUTION":
SELECT *
FROM MASTER_RULE
WHERE substr(value, 2,length(value)-1) between 15 AND 19
You could just use regular string comparisons:
where value >= 'A15' and
value < 'A20'
Not only is this simple, but the code can also take advantage of an index on value.
As you mentioned in the comments your data is like A15, A16, A17. etc you can achive your requirement with simple in clause also.
SELECT * FROM MASTER_RULE WHERE VALUE in ('A15','A16','A17','A18,'A19');
Using SQL in PostgreSQL I need to select all the rows from my table called "crop" when the first digit of the integer numbers in column "field_id" is 7.
select *
from crop
where (left (field_id,1) = 7)
First, you know that the column is a number, so I would be inclined to explicitly convert it, no matter what you do:
where left(crop::text, 1) = '7'
where crop::text like '7%'
The conversion to text is simply to be explicit about what is happening and it makes it easier for Postgres to parse the query.
More importantly, if the value has a fixed number of digits, then I would suggest using a numeric range; something like this:
where crop >= 700000 and crop < 800000
This makes it easier for Postgres to use an index on the column.
Try with cast, like this:
select *
from crop
where cast(substring(cast(field_id as varchar(5)),1,1) as int) = 7
where 5 in varchar(5) you should put number how long is your integer.
I have ,for example, this table in a Microsoft Access database:
id numeric
context text
numberfield numeric
I want to select every record that ends with 9 in the column"numberfield". This gives a problem because it is a numeric field and as a result I can not use the following SQL:
select * from table where numberfield like "%9"
A solution is that I change the numberfield to a text. But this gives a problem because there are several users and the change might give a problem in the future. Is there an option to select on the ending when it is a number field?
That sound a little fishy.. are you sure you can use that query? Don't know about Access but almost any other DBMS allows it.
If it really doesn't work, you can do this:
select * from table where STR(numberfield) like "*9"
EDIT: Maybe it didn't work because you used % which is used with * in Access :
select * from table where numberfield like "*9"
Numbers are numbers, so use Mod for this:
select * from table where numberfield mod 10 = 9
Instead of casting to string and comparing, just extract the rightmost digit with a MOD operation.
Edit your query as follows:
SELECT *
FROM table
WHERE ((([numberfield] Mod 10)=9));
A column has a string values like "1/200", "3.5" or "6". How can I convert this String to numeric value in single SQL query?
My actual SQL is more complicated, here is a simple example:
SELECT number_value_in_string FROM table
number_value_in_string's format will be one of:
##
#.##
#/###
I need to sort by the numeric value of this column. But of course postgres doesn't agree with me that 1/200 is a proper number.
Seeing your name I cannot but post a simplification of your answer:
SELECT id, number_value_in_string FROM table
ORDER BY CASE WHEN substr(number_value_in_string,1,2) = '1/'
THEN 1/substr(number_value_in_string,3)::numeric
ELSE number_value_in_string::numeric END, id;
Ignoring possible divide by zero.
I would define a stored function to convert the string to a numeric value, more or less like this:
CREATE OR REPLACE FUNCTION fraction_to_number(s CHARACTER VARYING)
RETURN DOUBLE PRECISION AS
BEGIN
RETURN
CASE WHEN s LIKE '%/%' THEN
CAST(split_part(s, '/', 1) AS double_precision)
/ CAST(split_part(s, '/', 2) AS double_precision)
ELSE
CAST(s AS DOUBLE PRECISION)
END CASE
END
Then you can ORDER BY fraction_to_number(weird_column)
If possible, I would revisit the data design. Is it all this complexity really necessary?
This postgres SQL does the trick:
select (parts[1] :: decimal) / (parts[2] :: decimal) as quotient
FROM (select regexp_split_to_array(number_value_in_string, '/') as parts from table) x
Here's a test of this code:
select (parts[1] :: decimal) / (parts[2] :: decimal) as quotient
FROM (select regexp_split_to_array('1/200', '/') as parts) x
Output:
0.005
Note that you would need to wrap this in a case statement to protect against divide-by-zero errors and/or array out of bounds issues etc if the column did not contain a forward slash
Note also that you could do it without the inner select, but you would have to use regexp_split_to_array twice (once for each part) and you would probably incur a performance hit. Nevertheless, it may be easier to code in-line and just accept the small performance loss.
I managed to solve my problem. Thanks all.
It goes something like this, in a single SQL. (I'm using POSTGRESQL)
It will sort a string coming in as either "#", "#.#" or "1/#"
SELECT id, number_value_in_string FROM table ORDER BY CASE WHEN position('1/' in number_value_in_string) = 1
THEN 1/substring(number_value_in_string from (position('1/' in number_value_in_string) + 2) )::numeric
ELSE number_value_in_string::numeric
END ASC, id
Hope this will help someone outhere in the future.
I have the following data:
[sequences]
/a1
/a2
/a3
...
/a10
The query SELECT * FROM sequences WHERE nbr <= '/a10' should return the list above, instead it returns:
[results]
/a1
/a10
How do I make it return all the rows in the above list?
It works as it should. To compare the numeric value, you'll have to convert these to numbers somehow. A good start would be to use substr(yourfieldname, 3) to cut of the/a. Then you can useconvert` to typecast it to int, so your final query will look something like:
select * from sequences where convert(int, substr(nbr, 3)) <= 10
Mind that the exact functions and rules for converting strings to ints may very per dbms. This illustrates the general idea, though.
SELECT *
FROM sequences
WHERE toInt(substring (nbr, 2)) <= 10;
Name and syntax of 'substring'-function and 'toInt' will vary from db-implementation to db-implementation.