How to transform $.post to $.ajax?

Question

I have this $.post peace of code:

$.post("../admin-login",
{
   dataName:JSON.stringify({
   username:uname,
   password:pass,
   })

}, function(data,status){
    console.log("Data:"+data);
    answer = data;
    }
);

and I wont to transform it to $.ajax. On the servlet side I am demanding request.getParamter("dataName") but I do not know how to write data: section in $.ajax so that I can get parameters like that(request.getParamter("dataName"))? Also, it seems to be problem with this type of code, I am asuming cause of async, that I cannot do this:

var answer="";

function(data,status){
        console.log("Data:"+data);
        answer = data;
}

And that answer is keeping empty(""), even though in console is written in deed "true" or "false" as my server answers. What is this about? Thanks in advance.

I found out that problem is in the click() event. Ajax finishes when click() finishes, so I am not able to get data before event is done. What is bad in that is that I cannot fetch data because it is finished. Does anyone know how to solve this?

The second parameter to `$.post` in your sample is how you are already defining your data. Why not try that? — Brandon Buck, Jun 23 '15 at 11:42

jsam · Accepted Answer · 2015-06-23T12:00:28.693

1

$.post("../admin-login",
{
   dataName:JSON.stringify({
   username:uname,
   password:pass,
   })

}, function(data,status){
    console.log("Data:"+data);
    answer = data;
    }
);

becomes

function getResult(data) {
  // do something with data
  // you can result = data here
  return data;
}

$.ajax({
  url: "../admin-login",
  type: 'post',
  contentType: "application/x-www-form-urlencoded",
  data: {
   dataName:JSON.stringify({
   username:uname,
   password:pass,
   })    
},
  success: function (data, status) {
    getResult(data);
    console.log(data);
    console.log(status);
  },
  error: function (xhr, desc, err) {
    console.log(xhr);
  }
});

edited Jun 23 '15 at 12:00

answered Jun 23 '15 at 11:42

jsam

301
1
7

Do you know why I cannot write: var result=""; success: function (data, status) { console.log(data); console.log(status); result = data; } ? Thanks. – executos Jun 23 '15 at 11:49
If you want to do that you need to do `var result="";` before the `$.ajax` call and then inside the success function you can do `result = data`. But be carefull because that is async. – jsam Jun 23 '15 at 11:51
Yeah, that async is making me headaces. I tried using async:false, but it still won't wait for server response and result is still empty. – executos Jun 23 '15 at 11:55
You can also take a look at Promises since they will sove your problem. This is a great response http://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-asynchronous-call – jsam Jun 23 '15 at 11:58
I found out that problem is in the click() event. Ajax finishes when click() finishes, so I am not able to get data before event is done. What is bad in that is that I cannot fetch data because it is finished. Do you know how to prevent this kind of behavior? Thanks. – executos Jun 23 '15 at 12:26

score 0 · Answer 2 · edited May 23 '17 at 11:43

0

You need to see how the information os arriving to your servlet as query parameter or payload.

See this HttpServletRequest get JSON POST data

edited May 23 '17 at 11:43

Community

1
1

answered Jun 23 '15 at 11:39

arterzatij

82
7

NightOwlPrgmr · Answer 3 · 2015-06-23T11:53:31.650

0

You could try structuring your AJAX request like the below:

var dataName = username:uname, password:pass;

$.ajax({
    url: "../admin-login",
    data: JSON.stringify(dataName),
    type: "POST",
    cache: false,
    dataType: "json"
}).done(function(data, status) {
    console.log("Data:"+data);
    answer = data;
});

edited Jun 23 '15 at 11:53

answered Jun 23 '15 at 11:40

NightOwlPrgmr

1,322
3
21
31

How to transform $.post to $.ajax?

3 Answers3