Calculating New Values
Overview
Teaching: 5 min
Exercises: 5 minQuestions
How can I calculate new values on the fly?
Objectives
Write queries that calculate new values for each selected record.
After carefully re-reading the expedition logs, we realize that the radiation measurements they report may need to be corrected upward by 5%. Rather than modifying the stored data, we can do this calculation on the fly as part of our query:
SELECT 1.05 * reading FROM Survey WHERE quant = 'rad';
1.05 * reading |
---|
10.311 |
8.19 |
8.8305 |
7.581 |
4.5675 |
2.2995 |
1.533 |
11.8125 |
When we run the query,
the expression 1.05 * reading
is evaluated for each row.
Expressions can use any of the fields,
all of usual arithmetic operators,
and a variety of common functions.
(Exactly which ones depends on which database manager is being used.)
For example,
we can convert temperature readings from Fahrenheit to Celsius
and round to two decimal places:
SELECT taken, round(5 * (reading - 32) / 9, 2) FROM Survey WHERE quant = 'temp';
taken | round(5*(reading-32)/9, 2) |
---|---|
734 | -29.72 |
735 | -32.22 |
751 | -28.06 |
752 | -26.67 |
As you can see from this example, though, the string describing our new field (generated from the equation) can become quite unwieldy. SQL allows us to rename our fields, any field for that matter, whether it was calculated or one of the existing fields in our database, for succinctness and clarity. For example, we could write the previous query as:
SELECT taken, round(5 * (reading - 32) / 9, 2) as Celsius FROM Survey WHERE quant = 'temp';
taken | Celsius |
---|---|
734 | -29.72 |
735 | -32.22 |
751 | -28.06 |
752 | -26.67 |
We can also combine values from different fields,
for example by using the string concatenation operator ||
:
SELECT personal || ' ' || family FROM Person;
personal | ’ ‘ | family | ||
---|---|---|---|---|
William Dyer | ||||
Frank Pabodie | ||||
Anderson Lake | ||||
Valentina Roerich | ||||
Frank Danforth |
Fixing Salinity Readings
After further reading, we realize that Valentina Roerich was reporting salinity as percentages. Write a query that returns all of her salinity measurements from the
Survey
table with the values divided by 100.Solution
SELECT taken, reading / 100 FROM Survey WHERE person = 'roe' AND quant = 'sal';
taken reading / 100 752 0.416 837 0.225
Unions
The
UNION
operator combines the results of two queries:SELECT * FROM Person WHERE id = 'dyer' UNION SELECT * FROM Person WHERE id = 'roe';
id personal family dyer William Dyer roe Valentina Roerich The
UNION ALL
command is equivalent to theUNION
operator, except thatUNION ALL
will select all values. The difference is thatUNION ALL
will not eliminate duplicate rows. Instead,UNION ALL
pulls all rows from the query specifics and combines them into a table. TheUNION
command does aSELECT DISTINCT
on the results set. If all the records to be returned are unique from your union, useUNION ALL
instead, it gives faster results since it skips theDISTINCT
step. For this section, we shall use UNION.Use
UNION
to create a consolidated list of salinity measurements in which Valentina Roerich’s, and only Valentina’s, have been corrected as described in the previous challenge. The output should be something like:
taken reading 619 0.13 622 0.09 734 0.05 751 0.1 752 0.09 752 0.416 837 0.21 837 0.225 Solution
SELECT taken, reading FROM Survey WHERE person != 'roe' AND quant = 'sal' UNION SELECT taken, reading / 100 FROM Survey WHERE person = 'roe' AND quant = 'sal' ORDER BY taken ASC;
Selecting Major Site Identifiers
The site identifiers in the
Visited
table have two parts separated by a ‘-‘:SELECT DISTINCT site FROM Visited;
site DR-1 DR-3 MSK-4 Some major site identifiers (i.e. the letter codes) are two letters long and some are three. The “in string” function
instr(X, Y)
returns the 1-based index of the first occurrence of string Y in string X, or 0 if Y does not exist in X. The substring functionsubstr(X, I, [L])
returns the substring of X starting at index I, with an optional length L. Use these two functions to produce a list of unique major site identifiers. (For this data, the list should contain only “DR” and “MSK”).Solution
SELECT DISTINCT substr(site, 1, instr(site, '-') - 1) AS MajorSite FROM Visited;
Key Points
Queries can do the usual arithmetic operations on values.
Use UNION to combine the results of two or more queries.