# FAKE Sequel connestion string that includes hash characters in psswd
schema = 'SCHEMA_FAKE'
psswd = 'aaa###bbbccc'
host = 'my.host.el'
port = '1521'
sid = 'SIDDB'
conn_string = "oracle://#{ schema }" + ':' + psswd + '@' + host + ':' +
port + '/' + sid
The returning error:
C:/Ruby193/lib/ruby/1.9.1/uri/common.rb:176:in `split': bad URI(is not
URI?): oracle://SCHEMA_FAKE:aaa###[email protected]:1521/SIDDB
(URI::InvalidURIError)
from C:/Ruby193/lib/ruby/1.9.1/uri/common.rb:211:in `parse'
from C:/Ruby193/lib/ruby/1.9.1/uri/common.rb:747:in `parse'
from
C:/Ruby193/lib/ruby/gems/1.9.1/gems/sequel-3.41.0/lib/sequel/database/connecting.rb:52:in
`connect'
from
C:/Ruby193/lib/ruby/gems/1.9.1/gems/sequel-3.41.0/lib/sequel/core.rb:147:in
`connect'
from C:/Users/etypaldos/Desktop/prod-wrk.rb:35:in `<main>'
[Finished in 0.2s with exit code 1]
This is due to lib/sequel/database/connecting.rb: in line 52:
uri = URI.parse(conn_string)
One way to resolve this could be to url-encode(escape) the conn_string
uri = URI.parse(URI.escape(conn_string))
and properly URI.unescape where is appropriate.
require 'uri'
conn_string="oracle://SCHEMA_FAKE:###[email protected]:1521/SIDDB"
uri = URI.parse(URI.escape(conn_string))
print "uri: ", uri, "\n"
print "uri.scheme: ", URI.unescape(uri.scheme), "\n"
print "uri.password: ", URI.unescape(uri.password), "\n"
now it seems to work:
uri: oracle://SCHEMA_FAKE:%23%23%[email protected]:1521/SIDDB
uri.scheme: oracle
uri.password: ###aaabbbccc
[Finished in 0.2s]
I had no time to check further as it seems to affect more parts. Please
suggest how it could be handled in an elegant way.
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