ruby on rails - undefined method `val' for #<Arel::Nodes::BindParam:0x007fa13ef32028> -


i new rails, , cannot figure out why error specified above when try create new application after login.

also, while looking @ solutions online, found there wrong foreign key. therefore, removed foreign key , added again using migration. however, schema.rb file shown below not show index, show student_id column.

questions :

  1. should creation of has_one , belongs_to association require 1 mention in routes.rb . ( read online require, why ? )
  2. how decide place belongs_to , has_one.
  3. how find out above error generated ?

model student ( created using devise )

class student < activerecord::base       devise :database_authenticatable, :registerable,          :recoverable, :rememberable, :trackable, :validatable        has_one :student_application  end

studentapplication ( generate using scaffold)

class studentapplication < activerecord::base   belongs_to :student  end

routes.rb

rails.application.routes.draw    resources :student_applications    devise_for :students    root to: "student_applications#index" end

in controller student_application_controller.rb

def new   #@student_application = studentapplication.new   @student_application = current_student.build_student_application() end  def create   #@student_application = studentapplication.new(student_application_params)   #@new_student = student.find(params[:id])   @student_application = current_student.create_student_application(student_application_params)    respond_to |format|     if @student_application.save       format.html { redirect_to @student_application, notice: 'student application created.' }       format.json { render :show, status: :created, location: @student_application }     else       format.html { render :new }       format.json { render json: @student_application.errors, status: :unprocessable_entity }     end   end end  private def student_application_params   params.require(:student_application).permit(:student_id, :name, :phone, :email_id, :gpa) end

schema.rb

  create_table "student_applications", force: :cascade |t|     t.string   "name"     t.string   "phone"     t.string   "email_id"     t.decimal  "gpa"     t.datetime "created_at", null: false     t.datetime "updated_at", null: false     t.integer  "student_id"   end    create_table "students", force: :cascade |t|     t.string   "email",                  default: "", null: false     t.string   "encrypted_password",     default: "", null: false     t.string   "reset_password_token"     t.datetime "reset_password_sent_at"     t.datetime "remember_created_at"     t.integer  "sign_in_count",          default: 0,  null: false     t.datetime "current_sign_in_at"     t.datetime "last_sign_in_at"     t.string   "current_sign_in_ip"     t.string   "last_sign_in_ip"     t.datetime "created_at"     t.datetime "updated_at"   end    add_index "students", ["email"], name: "index_students_on_email", unique: true   add_index "students", ["reset_password_token"], name: "index_students_on_reset_password_token", unique: true  end


Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more