如何通过 symfony2 控制器获取 Ajax 发布请求
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How to get Ajax post request by symfony2 Controller
提问by whitebear
I send text message like this
我发送这样的短信
html markup
html标记
<textarea id="request" cols="20" rows="4"></textarea>
javascript code
javascript代码
var data = {request : $('#request').val()};
$.ajax({
type: "POST",
url: "{{ path('acme_member_msgPost') }}",
data: data,
success: function (data, dataType) {
alert(data);
},
error: function (XMLHttpRequest, textStatus, errorThrown) {
alert('Error : ' + errorThrown);
}
});
symfony2 controller code
symfony2 控制器代码
$request = $this->container->get('request');
$text = $request->request->get('data');
but $textis null ...
但$text为空...
I have tried normal post request (not Ajax) by firefox http request tester.
我已经尝试过 firefox http 请求测试器的正常发布请求（不是 Ajax）。
/app_dev.php/member/msgPost
Controller works and $texthas a value.
控制器工作并$text具有价值。
So I think the php code is OK, there is the problem on Ajax side, however
所以我认为php代码是可以的，但是Ajax方面存在问题
'success:function' is called as if succeeded.
'success:function' 就像成功一样被调用。
How can you get the contents of javascript data structure?
如何获取javascript数据结构的内容？
回答by sf_tristanb
First, you don't need to access the container in your controller as it already implements ContainerAware
首先，您不需要访问控制器中的容器，因为它已经实现了 ContainerAware
So basically your code should look like this in your Controller.php
所以基本上你的代码应该是这样的 Controller.php
public function ajaxAction(Request $request)
{
$data = $request->request->get('request');
}
Also, make sure by the data you are sending is not null by using console.log(data)in the JS of your application.
此外，通过console.log(data)在应用程序的 JS 中使用，确保您发送的数据不为空。
And finally the answer of your question: you are not using the right variable, you need to access the value of $('#request').val()but you stored it in a requestvariable and you used a datavariable name in your controller.
最后是你的问题的答案：你没有使用正确的变量，你需要访问的值，$('#request').val()但你将它存储在一个request变量中，并且data在你的控制器中使用了一个变量名。
Consider changing the name of the variable, because it's confusing.
考虑更改变量的名称，因为它令人困惑。
回答by Elnur Abdurrakhimov
If you're sending the data as JSON — not as form urlencoded — you need to access the request body directly:
如果您以 JSON 格式发送数据——而不是 urlencoded 格式——则需要直接访问请求正文：
$data = json_decode($request->getContent());
回答by mangel.snc
You are doing it wrong when obtaining the value, you must use:
获取值时你做错了，你必须使用：
$data = $request->request->get('request');
'cause requestis the name of your parameter.
因为request是您的参数名称。
