我想附加一些数据options
在select
特定类的菜单使用jquery data()
来自jquery ajax()
调用。但是当我尝试调用它时,我收到了错误,它没有被连接。或者,更准确地说,它是没有得到所有在下面的代码的antipenultimate线应用,为console.log($(this).data('address'.vendorName));
给我的错误Uncaught TypeError: Cannot read property 'vendorName' of undefined
jquery数据()不附加值
内$.each()
循环,是因为我想也许只是用 最后刺尝试$('.vendor_address_id_' + id).data....
未应用于所有具有该类的元素。但我并不认为这是必要的。我究竟做错了什么?
$.each(returnedData, function (key, val) {
var id = val.id;
//attach address information to each select option for display in .vendor_full_address_table
$('.vendor_address_id_' + id).each(function (k, v) {
$(this).data('address', {
'vendorName': val.vendor_name,
'address1': val.address1,
'address2': val.address2,
'city': val.city,
'state': val.state,
'zip': val.zip
});
//gives error: Uncaught TypeError: Cannot read property 'vendorName' of undefined
console.log($(this).data('address').vendorName);
});
});
//console.log(returnedData);
Object
address_0: Object
address1: "street address1"
address2: ""
city: "Kalamazoo"
id: "15"
state: "MI"
vendor_name: "companyA"
zip: "123456"
address_1: Object
etc...
etc...
etc...
address_2: Object
etc...
etc...
etc...
让我们轻松:在jsFiddle上发布此代码并删除您认为不需要的代码。 – jdigital