在下拉表单中更改选定值时提交Rails

I have a Rails app with listing of leads in a table. In one of the collumns I display status of a lead in a drop down menu. I want to enable changing this status of the lead on changing the value selected in the drop down.

我有一个Rails应用程序，其中包含表格中的潜在客户列表。在其中一个列中，我在下拉菜单中显示潜在客户的状态。我想在更改下拉列表中选择的值时启用更改此状态。

This is what I tried:

这是我试过的：

The code to display the form in a table cell:

在表格单元格中显示表单的代码：

      <% @leads.each do |lead| %>
  <tr>
    <td><%= lead.id %></td>
<td><%= form_for(lead,:url => 'update_lead_status') do |f| %>
              <div class="field">
                <%= f.select :status, ["to_call","called","confirmed","lite"], :selected => lead.status, onchange: "this.form.submit();" %>
              </div>
            <% end %>
        </td>

my update_lead_status method in leads controller:

我在引导控制器中的update_lead_status方法：

#PUT
  def update_lead_status
    @lead = Lead.find(params[:id])
    respond_to do |format|
      # format.js
      if @lead.update_attributes(params[:lead])
        format.html { redirect_to leads_url, notice: 'Lead was successfully updated.' }
        format.json { head :no_content }
      else
        format.html { render action: "edit" }
        format.json { render json: @lead.errors, status: :unprocessable_entity }
      end
    end
  end

Also I want the submission to be Ajax style without refreshing.

此外，我希望提交是Ajax样式而不刷新。

1 个解决方案

#1

Set form id and then submit form

设置表单ID，然后提交表单

<%= form_for(lead,:url => 'update_lead_status',:html=>{:id=>'lead_form'}) do |f| %>

 <%= f.select :status, ["to_call","called","confirmed","lite"], :selected => lead.status, onchange: "$('#lead_form').submit();" %>
<% end %>