我有jQuery 1.7.2和jsonp传递的ajax params对象的问题

i need a help with this code:

我需要这个代码的帮助:

i need send service, data and content params... but does not work ((

我需要发送服务,数据和内容参数...但不起作用((

var addons = {};
$.ajax({
    'url': 'http://site.com/index.php',
    dataType:'jsonp'
    success:function()
    {
        addons.name = 'Some name';
        addons.caption = 'Some caption';
        addons.description = 'Some description';
    }
});
var data = {
    'services'  :   1,
    'data'      :   addons,
    'content'   :   $(self).find('textarea').val()
};
$.ajax({
    'url'       :   'index.php',
    'dataType'  :   'json',
    'type'      :   'post',
    'data'      :   data,
    success:function(response){
        //...
    }
});

Send Post:

services=1&content=hi

1 个解决方案

#1

You need to define the object´keys and values witin the brackets or you define an class like function and than save the values. But for your target the following should be working fine.

您需要在括号中定义对象的键和值,或者您定义类似函数的类,而不是保存值。但是对于你的目标,以下应该可以正常工作。

var addons = {
                 name: 'Some name', 
                 caption: 'Some caption',
                 description: 'Some description'
             };

Kind Regards