blocks|key|1367105|text|$.getJSON()方法执行HTTP+GET而不是POST。您需要使用$.post()|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1367106|$.post(url,+dataToBeSent,+function(data,+textStatus)+{
++//data+contains+the+JSON+object
++//textStatus+contains+the+status:+success,+error,+etc
},+"json");|code-block|syntax|javascript|1367107|在调用中，dataToBeSent可以是您想要任何东西，尽管如果要发送html表单的内容，您可以使用serialize方法从表单创建帖子的数据。|style|CODE|1367108|var+dataToBeSent+=+$("form").serialize();|1367109|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.jquery.com/Ajax/jQuery.post|1|http://docs.jquery.com/Ajax/serialize^0|10|8|0|0|0|5|C|1E|9|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]]|9|@$A|16|B|17|1|18]]|C|$]]|$1|M|3|N|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|O|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

The $.getJSON() method does an HTTP GET and not POST. You need to use <a href="http://docs.jquery.com/Ajax/jQuery.post" rel="noreferrer">$.post()</a>

<pre><code>$.post(url, dataToBeSent, function(data, textStatus) {
 //data contains the JSON object
 //textStatus contains the status: success, error, etc
}, "json");
</code></pre>

In that call, <code>dataToBeSent</code> could be anything you want, although if are sending the contents of a an html form, you can use the <a href="http://docs.jquery.com/Ajax/serialize" rel="noreferrer">serialize</a> method to create the data for the POST from your form.

<pre><code>var dataToBeSent = $("form").serialize();
</code></pre>

blocks|key|1367156|text|如果你只有两个参数，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1367157|$.getJSON('/url-you-are-posting-to',data,function(result){

++++//do+something+useful+with+returned+result//
++++result.variable-in-result;
});|code-block|syntax|javascript|1367158|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

if you have just two parameters you can do this:

<pre><code>$.getJSON('/url-you-are-posting-to',data,function(result){

 //do something useful with returned result//
 result.variable-in-result;
});
</code></pre>

blocks|key|1367244|text|这是我的“一行”解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1367245|$.postJSON+=+function(url,+data,+func)+{+$.post(url%2B(url.indexOf("?")+==+-1+?+"?"+:+"&")%2B"callback=?",+data,+func,+"json");+}|code-block|syntax|javascript|1367246|为了使用jsonp和POST方法，此函数将"callback“GET参数添加到URL。这是它的使用方法：|1367247|$.postJSON("http://example.com/json.php",{+id+:+287+},+function+(data)+{
+++console.log(data.name);
});|1367248|服务器必须准备好处理回调GET参数，并将json字符串返回为：|1367249|jsonp000000+({"name":"John",+"age":+25});|1367250|其中"jsonp000000“为回调GET的值。|1367251|在PHP中，该实现如下所示：|1367252|print_r($_GET['callback']."(".json_encode($myarr).");");|1367253|我做了一些跨域测试，它似乎起作用了。不过，仍然需要更多的测试。|1367254|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

This is my "one-line" solution:

<pre><code>$.postJSON = function(url, data, func) { $.post(url+(url.indexOf("?") == -1 ? "?" : "&amp;")+"callback=?", data, func, "json"); }
</code></pre>

In order to use jsonp, and POST method, this function adds the "callback" GET parameter to the URL. This is the way to use it:

<pre><code>$.postJSON("http://example.com/json.php",{ id : 287 }, function (data) {
 console.log(data.name);
});
</code></pre>

The server must be prepared to handle the callback GET parameter and return the json string as:

<pre><code>jsonp000000 ({"name":"John", "age": 25});
</code></pre>

in which "jsonp000000" is the callback GET value.

In PHP the implementation would be like:

<pre><code>print_r($_GET['callback']."(".json_encode($myarr).");");
</code></pre>

I made some cross-domain tests and it seems to work. Still need more testing though.

blocks|key|1582340|text|我有做getJSON的代码。我只是简单地将其替换为post。令我惊讶的是，它成功了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1582341|+++$.post("@Url.Action("Command")",+{+id:+id,+xml:+xml+})
++++++.done(function+(response)+{
+++++++++++//+stuff
++++++++})
++++++++.fail(function+(jqxhr,+textStatus,+error)+{
+++++++++++//+stuff
++++++++});



++++[HttpPost]
++++public+JsonResult+Command(int+id,+string+xml)
++++{
++++++++++//+stuff
++++}+|code-block|syntax|javascript|1582342|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I had code that was doing getJSON. I simply replaced it with post. To my surprise, it worked

<pre><code> $.post("@Url.Action("Command")", { id: id, xml: xml })
 .done(function (response) {
 // stuff
 })
 .fail(function (jqxhr, textStatus, error) {
 // stuff
 });



 [HttpPost]
 public JsonResult Command(int id, string xml)
 {
 // stuff
 } 
</code></pre>

blocks|key|1367354|text|对于发送AJAX请求和返回$.getJSON()数据作为响应，JSON非常方便。遗憾的是，jQuery文档缺少一个应该命名为$.postJSON()的姊妹函数。为什么不直接使用$.getJSON()，然后就完了呢？嗯，也许您想发送大量数据，或者在我的例子中，IE7只是不想正确处理GET请求。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1367355|确实，目前还没有$.postJSON()方法，但是您可以通过在$.post()函数中指定第四个参数(类型)来完成相同的任务：|1367356|我的代码看起来像这样：|1367357|$.post('script.php',+data,+function(response)+{
++//+Do+something+with+the+request
},+'json');|code-block|syntax|javascript|1367358|entityMap^0|D|B|1Q|C|2G|B|0|8|C|V|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|13|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

<code>$.getJSON()</code> is pretty handy for sending an AJAX request and getting back JSON data as a response. Alas, the jQuery documentation lacks a sister function that should be named <code>$.postJSON()</code>. Why not just use <code>$.getJSON()</code> and be done with it? Well, perhaps you want to send a large amount of data or, in my case, IE7 just doesn’t want to work properly with a GET request.

It is true, there is currently no <code>$.postJSON()</code> method, but you can accomplish the same thing by specifying a fourth parameter (type) in the <code>$.post()</code> function:

My code looked like this:

<pre><code>$.post('script.php', data, function(response) {
 // Do something with the request
}, 'json');
</code></pre>

blocks|key|1367420|text|只需将以下代码行添加到您的<script>+(在加载jQuery之后但在发布任何内容之前的某个位置)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1367421|$.postJSON+=+function(url,+data,+func)
{
++++$.post(url,+data,+func,+'json');
}|code-block|syntax|javascript|1367422|用$.postJSON替换(部分/全部)+$.getJSON，尽情享受吧！|1367423|您可以使用与$.getJSON相同的Javascript回调函数。不需要对服务器端进行更改。(嗯，我总是推荐在PHP中使用$_REQUEST。http://php.net/manual/en/reserved.variables.request.php，Among+$_REQUEST,+$_GET+and+$_POST+which+one+is+the+fastest?)|1367424|这比@lepe的解决方案更简单。|1367425|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/en/reserved.variables.request.php|1|https://stackoverflow.com/questions/1924939/php-request-vs-get-and-post^0|D|8|0|0|1|A|L|9|0|6|9|1P|9|1Z|1J|0|3J|1N|1|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]|$9|1H|A|1I|1|1J]]|E|$]]|$1|O|3|P|5|6|7|1K|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

Just add these lines to your <code>&lt;script&gt;</code> (somewhere after jQuery is loaded but before posting anything): 

<pre><code>$.postJSON = function(url, data, func)
{
 $.post(url, data, func, 'json');
}
</code></pre>

Replace (some/all) <code>$.getJSON</code> with <code>$.postJSON</code> and enjoy!

You can use the same Javascript callback functions as with <code>$.getJSON</code>.
No server-side change is needed. (Well, I always recommend using <code>$_REQUEST</code> in PHP. <a href="http://php.net/manual/en/reserved.variables.request.php" rel="noreferrer">http://php.net/manual/en/reserved.variables.request.php</a>, <a href="https://stackoverflow.com/questions/1924939/php-request-vs-get-and-post">Among $_REQUEST, $_GET and $_POST which one is the fastest?</a>)

This is simpler than @lepe's solution.

blocks|key|1582469|text|我只用了post和一个if：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1582470|data+=+getDataObjectByForm(form);
var+jqxhr+=+$.post(url,+data,+function(){},+'json')
++++.done(function+(response)+{
++++++++if+(response+instanceof+Object)
++++++++++++var+json+=+response;
++++++++else
++++++++++++var+json+=+$.parseJSON(response);
++++++++//+console.log(response);
++++++++//+console.log(json);
++++++++jsonToDom(json);
++++++++if+(json.reload+!=+undefined+&&+json.reload)
++++++++++++location.reload();
++++++++$("body").delay(1000).css("cursor",+"default");
++++})
++++.fail(function+(jqxhr,+textStatus,+error)+{
++++++++var+err+=+textStatus+%2B+",+"+%2B+error;
++++++++console.log("Request+Failed:+"+%2B+err);
++++++++alert("Fehler!");
++++});|code-block|syntax|javascript|1582471|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I just used post and an if:

<pre><code>data = getDataObjectByForm(form);
var jqxhr = $.post(url, data, function(){}, 'json')
 .done(function (response) {
 if (response instanceof Object)
 var json = response;
 else
 var json = $.parseJSON(response);
 // console.log(response);
 // console.log(json);
 jsonToDom(json);
 if (json.reload != undefined &amp;&amp; json.reload)
 location.reload();
 $("body").delay(1000).css("cursor", "default");
 })
 .fail(function (jqxhr, textStatus, error) {
 var err = textStatus + ", " + error;
 console.log("Request Failed: " + err);
 alert("Fehler!");
 });
</code></pre>

I am using above method &amp; it works well with one parameter in URL.

e.g. <code>Students/getstud/1</code> where controller/action/parameter format is applied.

Now I have an action in Students controller that accepts two parameters and return a JSON object.

So how do I post data with <code>$.getJSON()</code> using post method?

Similar methods are also acceptable.

The point is to call an action of the controller with AJAX.

How to use getJSON, sending data with post method?

服务器

jQuery

JSON

我正在使用上面的方法&它与URL中的一个参数一起工作得很好。例如，应用控制器/动作/参数格式的Students/getstud/1。现在我在学生控制器中有了一个动作，它接受两个参数并返回一个JSON对象。那么，如何使用post方法通过$.getJSON()发布数据呢？类似的方法也是可以接受的。重点是使用AJAX调用控制...

问如何使用getJSON，通过post方法发送数据？
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问如何使用getJSON，通过post方法发送数据？
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