# Calculating New Values

## Overview

Teaching: 5 min
Exercises: 5 min
Questions
• 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';
``````
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';
``````
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 the `UNION` operator, except that `UNION ALL` will select all values. The difference is that `UNION ALL` will not eliminate duplicate rows. Instead, `UNION ALL` pulls all rows from the query specifics and combines them into a table. The `UNION` command does a `SELECT DISTINCT` on the results set. If all the records to be returned are unique from your union, use `UNION ALL` instead, it gives faster results since it skips the `DISTINCT` 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:

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 function `substr(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.