Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
450 views
in Technique[技术] by (71.8m points)

working shell command fails to execute with scala.sys.process

Following command executes successfully when ran in a shell

csvsql --db "postgresql://user:password@localhost:5432/samples" --table sample_table --insert /absolute.path.csv

but fails to execute when ran from scala like this:

s"""csvsql --db "postgresql://user:password@localhost:5432/samples" --table sample_table --insert /absolute.path.csv""" !!

with following error:

ArgumentError: Could not parse rfc1738 URL from string '"postgresql://user:password@localhost:5432/samples"'

question from:https://stackoverflow.com/questions/65918983/working-shell-command-fails-to-execute-with-scala-sys-process

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The shell will have removed the quotes before passing the argument to the process, so you need to remove the quotes from the db string. It is probably better to use a Seq of arguments rather than a string:

Seq("csvsql", "--db", "postgresql://user:password@localhost:5432/samples", "--table", "sample_table", "--insert", "/absolute.path.csv") !!

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...