你猜怎么着?@BenCreasy完全正确!!

从jQuery 1.12.0版本开始，我们可以这样做:

$.post({
    url: yourURL,
    data: yourData,
    contentType: 'application/json; charset=utf-8'
})
    .done(function (response) {
        //Do something on success response...
    });

我刚刚测试了一下，它工作了!!

只使用

jQuery.ajax ({
    url: myurl,
    type: "POST",
    data: mydata,
    dataType: "json",
    contentType: "application/json; charset=utf-8",
    success: function(){
        //
    }
});

更新@JK:如果你在你的问题中只写一个带有$的代码示例。你在答案中找到一个相应的例子。我不想重复同样的信息，你已经学习了，直到知道:$。Post和$。Get是$.ajax的缩写形式。所以只需使用$。Ajax，您可以使用它的全部参数集，而无需更改任何全局设置。

By the way I wouldn't recommend overwriting the standard $.post. It's my personal opinion, but for me it's important, not only that the program works, but also that all who read your program understand it with the same way. Overwriting standard methods without having a very important reason can follow to misunderstanding in reading of the program code. So I repeat my recommendation one more time: just use the original $.ajax form jQuery instead of jQuery.get and jQuery.post and you receive programs which not only perfectly work, but can be read by people without any misunderstandings.

我最终在我的脚本中添加了以下方法:

jQuery["postJSON"] = function( url, data, callback ) {
    // shift arguments if data argument was omitted
    if ( jQuery.isFunction( data ) ) {
        callback = data;
        data = undefined;
    }

    return jQuery.ajax({
        url: url,
        type: "POST",
        contentType:"application/json; charset=utf-8",
        dataType: "json",
        data: data,
        success: callback
    });
};

然后使用它

$.postJSON('http://url', {data: 'goes', here: 'yey'}, function (data, status, xhr) {
    alert('Nailed it!')
});

这是通过简单地从原始JQuery源代码中复制“get”和“post”的代码并硬编码一些参数来强制JSON post来实现的。

谢谢!

目前的文档显示，在3.0，$。Post将接受Settings对象，这意味着您可以使用$。ajax选项。3.0还没有发布，在提交时，他们说要在文档中隐藏对它的引用，但以后再找吧!

$.ajax({
  url:url,
  type:"POST",
  data:data,
  contentType:"application/json; charset=utf-8",
  dataType:"json",
  success: function(){
    ...
  }
})

参见:jQuery.ajax()

问题的核心是，在编写本文时，JQuery还没有postJSON方法，而getJSON存在并做正确的事情。

postJSON方法会做以下事情:

postJSON = function(url,data){
    return $.ajax({url:url,data:JSON.stringify(data),type:'POST', contentType:'application/json'});
};

并且可以这样使用:

postJSON( 'path/to/server', my_JS_Object_or_Array )
    .done(function (data) {
        //do something useful with server returned data
        console.log(data);
    })
    .fail(function (response, status) {
        //handle error response
    })
    .always(function(){  
      //do something useful in either case
      //like remove the spinner
    });