Scala Cookbook

1.4. Substituting Variables into Strings

Problem

You want to perform variable substitution into a string, like you can do with other languages, such as Perl, PHP, and Ruby.

Solution

Beginning with Scala 2.10 you can use string interpolation in a manner similar to other languages like Perl, PHP, and Ruby.

To use basic string interpolation in Scala, precede your string with the letter s and include your variables inside the string, with each variable name preceded by a $ character. This is shown in the println statement in the following example:

scala> val name = "Fred"
name: String = Fred

scala> val age = 33
age: Int = 33

scala> val weight = 200.00
weight: Double = 200.0

scala> println(s"$name is $age years old, and weighs $weight pounds.")
Fred is 33 years old, and weighs 200.0 pounds.

According to the official Scala string interpolation documentation, when you precede your string with the letter s, you’re creating a processed string literal. This example uses the “s string interpolator,” which lets you embed variables inside a string, where they’re replaced by their values. As stated in the documentation, “Prepending s to any string literal allows the usage of variables directly in the string.”

Using expressions in string literals

In addition to putting variables inside strings, you can include expressions inside a string by placing the expression inside curly braces. According to the official string interpolation documentation, “Any arbitrary expression can be embedded in ${}.”

In the ...