AJAX调用并在Node中获取返回数据

Question

我想从客户端到后端进行ajax调用。 我从成功函数中获得了成功的调用，但是，我不明白如何从服务器获取数据以从客户端返回。

目前我尝试使用res.send的错误是：

Error: Can't set headers after they are sent.

AJAX

function getProfessorResults() {
      var textData = $('#inputsm').val();

      var data = {user:"gopal@gmail.com"};
      $.ajax({
          url: 'http://localhost:3000',
          data: { theme: "somevalue", snippet: { name: "somename", content: "somevalue" } },
          method: 'POST',
          async: false,
          cache: false,
          timeout: 5000,
          contentType: "application/json",
          success: function(data) {
              console.log("success");
          },
          complete: function(data) {
            console.log("completed");
          },
          error: function(jqXHR, textStatus, errorThrown) {

              alert('Error connecting to the Node.js server... ' + textStatus + " " + errorThrown);
          }
      });
    }

JS后端

exports.home = function(req, res) {


  function passList(profArray, callback) {
    setTimeout(function () {
      callback(profArray);
    }, 1000);
  }


  function getProfs(teacher_name, successCallback) {

    google.resultsPerPage = 10
    var nextCounter = 0

    google(teacher_name, function (err, res){
      if (err) console.error(err)
      var teacher_results = []; //Hold all the teachers returned from the function
      for (var i = 0; i < res.links.length; ++i) {
        var link = res.links[i];
        if (!link.title.includes('Add') || !link.title.includes('RATINGS') || !link.title.includes("Hint")) {

            teacher_results.push(link.title);

        }//End if for comparisons ||
      } //End For
      successCallback(teacher_results);
    }); //End google function
      teacher_results = ['tester1', 'tester2'];
      successCallback(teacher_results);
  } //End searchForProfessor


  getProfs(teacher_name, function(data) {
    prof_list = data;
    console.log(prof_list);
    return true;
  });


  if (req.method == 'POST'){
      console.log("true");
      // dataReceived = JSON.parse(req);
      // console.log(dataReceived);
      var obj = {
          tid: 'ryan'
      };

      res.send(JSON.stringify(obj));
   }



  res.render('home', {
    profs: prof_list,
    dataStuff : dataReceived
  });
};

Answer 1

在后端，您应该有一条到达AJAX呼叫的路由。 在这里，您可以调用send响应。

在node.js / express中，这看起来像

app.get('/ajaxURL', function (req, res) {
  res.send('I want this string to return to the client');
});

要从前端访问数据，请在AJAX回调中访问它：

$.ajax({url: '/ajaxURL'}).done(function (data) {
    console.log(data);
});

Answer 2

我没有正确地了解上下文，但是您可以通过此示例弄清楚。

从服务器response.send("Your data");发送数据response.send("Your data"); 通过AJAX的成功方法在客户端中访问此数据： success:function(data){console.log(data)};