[英]How to update two parameters using Ajax in Ruby on Rails?
我正在尝试使用Ajax更新此辅助函数:
def radio_buttons_collection(types, invoice_type)
types_html = types.map do |type|
content_tag :span, :class => "radio_option" do
radio_button_tag("project[invoice_type]", type, type == invoice_type) + type
end
end
safe_join(types_html)
end
但是,我很难在其中获取两个参数。
# get_invoice_types.js.erb:
$('#project_invoice_type').html("<%= escape_javascript(radio_buttons_collection(@types, @invoice_type)) %>");
# projects_controller.rb:
def get_invoice_types
person = Person.find(params[:person_id])
@types = person.address_types
project = Project.find(params[:id])
@invoice_type = project.invoice_type
end
# application.js:
$("#project_person_id").change(function() {
$.ajax({
url: '/projects/get_invoice_types',
data: 'person_id=' + this.value,
dataType: 'script'
})
});
有人可以帮忙吗?
现在,只有@types
参数可以正确传递。
谢谢你的帮助!
帮助Anezio Campos:
$("#project_person_id").change(function() {
var person_id = $(this).val();
var project_id = $("#project_id").val();
$.ajax({
url: '/projects/get_invoice_types',
data: {person_id: person_id, id: project_id},
dataType: 'script'
})
});
尝试一下(如果可行)。
在projects#get_invoice_types
您尝试通过以下方式获取项目:
project = Project.find(params[:id])
但是您没有在ajax调用中传递id
,而是仅传递person_id
您还应该在ajax调用中传递id
参数
添加一个id为project_id
且值为project.id的hidden_field。
然后
$("#project_person_id").change(function() {
$.ajax({
url: '/projects/get_invoice_types',
data: 'person_id=' + this.value + '&id=' + $("#project_id").value(),
dataType: 'script'
})
});
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