scala system call to mysql fails - mysql

Question

I have a shell script
mysql --local-infile=1 -h myhost -P myport -u me < ./loader_file.sql
When I run this directly in command line, it works great. data gets loaded into the database.
but inside a scala code
val myScript = "mysql --local-infile=1 -h myhost -P myport -u me" #< "loader_file.sql"
myScript !
I get a error message saying the mysql commands in my shell, such as "use" and "load" are not found. Obviously, shell is interpreting the loader_file.sql as bash command rather then mysql commands.
But if I incorporate the whole string as one command
val myScript = "mysql --local-infile=1 -h myhost -P myport -u me < loader_file.sql"
Then I get a mysql dump with all its valid options. Apparently, mysql thinks I am feeding it illegal input parameters.
Anyone knows how to fix this?

Answer 1

Here's what you need to do it from scala code:
import scala.sys.process._
val cmd = Seq("bash","-c","mysql --local-infile=1 -h myhost -P myport -u me < loader_file.sql")
for( line <- cmd.lineStream_! ){
printf("%s\n",line)
}
That should work under linux, OSX, cygwin, or similar (sounds like you have one of those environments, since bash is present).
There are lots of variations on this particular theme, allowing you to set the default working directory, specify environment variables, separately process STDOUT and STDERR streams, and capture the return value, etc.
Here's a version that does everything except specify environment variables:
import scala.sys.process._
val cmd = Seq("bash","-c","command with args, blah, blah, blah")
val proc = Process( cmd, new java.io.File(".") )
val exitValue = proc ! ProcessLogger (
(out) => System.out.printf("stdout:%s\n",out)
,
(err) => System.err.printf("stderr:%s\n",err)
)
printf("exit value: %d\n",exitValue)
The documentation for scala.sys.process is pretty good, although it assumes a linux-type environment. A shell command line will sometimes work without being wrapped with Seq("bash","-c", ...), but it isn't as portable (Windows java doesn't pass your command line to bash, for example).