Date_select with Searchlogic

I am aware that date selects params are supposed to supported by
Searchlogic.

But I’m battling to get them working.

I am using one of the named_scopes that it makes, and it works perfectly
in the console:

User.created_at_gte(1.year.ago)

this behaves as expected.

In my view I have this:

  • form_for @search do |f|
    Surname
    =f.text_field :surname_like
    Age between
    =f.text_field :age_gte
    =f.text_field :age_lte
    Joined after
    =f.date_select :created_at_gte

    =f.submit “Search”

the error I get is when actually using the field, I get this error:
The is not a valid condition. You may only use conditions that map to a
named scope

and the params are:

{“commit”=>“Search”,
“search”=>{“created_at_gte(1i)”=>“2011”,
“created_at_gte(2i)”=>“2”,
“created_at_gte(3i)”=>“25”,
“age_gte”=>“5”,
“order”=>"",
“surname_like”=>"",
“age_lte”=>“25”}}

Now unless I’m wrong, searchlogic should be able to parse the params
with their date groups. Or it’s just being stupid and thinking
“created_at_gte(1i)” is a named scope. In which case I’m going to have
to alter the params to make it friendly for searchlogic.

anythink I’m missing?

+1, hit this nail right on today :frowning:

William Y. wrote:

+1, hit this nail right on today :frowning:

Sorry for the late reply, u probably moved on from this.

I got a method that takes all the date params from the params and
changes them to a friendlier version.

def parse_date_params(param_hash)
hash = param_hash.clone
regexp = /(.+)((\d)[a-zA-Z])/
dates_arrays = hash.select{|key, value| key.to_s.match(regexp) }

hash.delete_if{|key, value| key.to_s.match(regexp) }

add_to_hash = {}

dates_arrays.each do |a|
  matchdata = a[0].match(regexp)
  # check for all three.
  add_to_hash[matchdata[1]] ? 

add_to_hash[matchdata[1]][matchdata[2]] = a[1] :
add_to_hash[matchdata[1]] = {matchdata[2] => a[1]}
end

add_to_hash.each_pair { |key, value| hash[key] = 

Date.civil(value[“1”].to_i, value[“2”].to_i,value[“3”].to_i)}

return hash

end

let me know if that helps.

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