This seems ludicrously simple but I cannot figure out how to convert a hash-string to a hash.
When I do a Answer.find_by_sql I get a string like this
deepthought = "\"answertolife\"=>\"42\""
But I cannot figure out how to turn that into a hash.
I have tried:
pry(main)> Hash[deepthought]
ArgumentError: odd number of arguments for Hash
pry(main)> JSON.parse deepthought
JSON::ParserError: 757: unexpected token at '"answertolife"=>"42"'
pry(main)> deepthought.to_json
=> "\"\\\"answertolife\\\"=>\\\"42\\\"\""
I saw How do I convert a String object into a Hash object? , but I still cannot figure it out.
Try this
eval("{ #{deepthought} }")
It wraps the deepthought string with curly brace { }, and then use eval
A bit late but if you need to convert a multiple entries this works great.
def hstore_to_hash(hstore)
values = {}
hstore.gsub(/"/, '').split(",").each do |hstore_entry|
each_element = hstore_entry.split("=>")
values[each_element[0]] = each_element[1]
end
values
end
这似乎可行,但感觉很脏。
JSON.parse "{ #{deepthought} }".gsub('=>', ':')
Rails4 supports hstore out of the box so I'd probably handle the string casting the same way Rails4 does it. If you look inside the Rails4 PostgreSQL-specific casting code, you'll find string_to_hstore
:
def string_to_hstore(string)
if string.nil?
nil
elsif String === string
Hash[string.scan(HstorePair).map { |k, v|
v = v.upcase == 'NULL' ? nil : v.gsub(/\A"(.*)"\Z/m,'\1').gsub(/\\(.)/, '\1')
k = k.gsub(/\A"(.*)"\Z/m,'\1').gsub(/\\(.)/, '\1')
[k, v]
}]
else
string
end
end
and a little lower down in the same file, you'll find HstorePair
:
HstorePair = begin
quoted_string = /"[^"\\]*(?:\\.[^"\\]*)*"/
unquoted_string = /(?:\\.|[^\s,])[^\s=,\\]*(?:\\.[^\s=,\\]*|=[^,>])*/
/(#{quoted_string}|#{unquoted_string})\s*=>\s*(#{quoted_string}|#{unquoted_string})/
end
Stash that somewhere convenient (probably somewhere in lib/
) and send your hstore strings through that string_to_hstore
to unpack them into Hashes.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.