The Wayback Machine - https://web.archive.org/web/20201005184611/https://github.com/vertical-blank/scala-sql-formatter
Skip to content
master
Go to file
Code

Latest commit

 

Git stats

Files

Permalink
Failed to load latest commit information.
Type
Name
Latest commit message
Commit time
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

readme.md

scala-sql-formatter

Build Status Maven Central

SQL Formatter for Scala.

This is a bridge to these libraries.

Usage

Scala (on JVM)

libraryDependencies += "com.github.vertical-blank" %% "scala-sql-formatter" % "1.0.1"

Scala.js

libraryDependencies += "com.github.vertical-blank" %%% "scala-sql-formatter" % "1.0.1"

scalaJSLinkerConfig ~= { _.withModuleKind(ModuleKind.CommonJSModule) }

Examples

You can easily use com.github.vertical_blank.sqlformatter.scala.SqlFormatter :

SqlFormatter.format("SELECT * FROM table1")

This will output:

SELECT
  *
FROM
  table1

Dialect

You can pass dialect with FormatConfig :

SqlFormatter.format(
  "SELECT *",
  FormatConfig(dialect = SqlDialect.CouchbaseN1QL))

Currently just four SQL dialects are supported:

Format

Defaults to two spaces. You can pass indent string with FormatConfig to format :

SqlFormatter.format(
  "SELECT * FROM table1",
  FormatConfig(indent = "    "))

This will output:

SELECT
    *
FROM
    table1

Placeholders replacement

You can pass Seq to formatWithIndexedParams, or Map to formatWithNamedParams :

// Named placeholders
SqlFormatter.formatWithNamedParams("SELECT * FROM tbl WHERE foo = @foo", params = Map("foo" -> "'bar'"))

// Indexed placeholders
SqlFormatter.formatWithIndexedParams("SELECT * FROM tbl WHERE foo = ?", params = Seq("'bar'"))

Both result in:

SELECT
  *
FROM
  tbl
WHERE
  foo = 'bar'

Same as the format method, both formatWithNamedParams and formatWithIndexedParams accept FormatConfig.

You can’t perform that action at this time.