Skip to content

vertical-blank/scala-sql-formatter

Folders and files

NameName
Last commit message
Last commit date

Latest commit

 

History

81 Commits
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Repository files navigation

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.

About

SQL Formatter for Scala

Topics

Resources

License

Stars

Watchers

Forks

Packages

No packages published

Contributors 3

  •  
  •  
  •  

Languages