Use Ajax response script into a Django HttpResponse

Question

I am trying to pass the ajax response obtained from view to the template using HttpResponse but I don't have any idea, how to do that?

view.py

 analyzer=SentimentIntensityAnalyzer()
    def index(request):
        return render(request, "gui/index.html")

    @csrf_exempt
    def output(request):
        sentences = request.POST.get('name',None)
        senti = analyzer.polarity_scores(sentences)
        context_dict = {'sentiment': senti}
        return render(request,"gui/index.html", context = context_dict

I want the senti to be printed after the score on the page but I am unable to obtain it.

template file

<!doctype html>
<html>
     <head>
       <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js">
       </script>
     </head>
     <body>
       <form action = Post>
                    Enter Sentence:<input id = "name" type = "text" name = "EnterSentence" encoding = "utf-8"><br>
         <input onclick = "testfunction()" type = "button" value = "Submit" >
        </form>
        <div><strong>Score is {{ sentiment }}</strong></div>
    </body>
    <script>
    var testfunction = () => {
    var test = document.getElementById("name").value
    console.log(test)

    $.ajax({
             type: "POST",
             dataType: "json",
             url: 'output/',
             data:{
                    csrfmiddlewaretoken: '{{ csrf_token }}',
                   'name': test
                    },
                    success: function(response) {
                    console.log("Succesful return firm ajax call");
                    },
                    error: function(result){
                    console.log("Failure");
                    }
             });

    }
    </script>
</html>

So, you want to show retrieved data from in a specific place right? — Rarblack
– Rarblack, Commented Oct 12, 2018 at 10:22
what is the intended use? I'm assuming the view shown corresponds to the URL of your Ajax call - and it responds with html (the rendered template). You can easily just dump this html into a particular container, in your success callback, — Robin Zigmond
– Robin Zigmond, Commented Oct 12, 2018 at 10:23
ah, thanks for the edit. In that case don't bother with the {{ sentiment }} template variable. Just put an empty div here and populate it with the response to the Ajax call. — Robin Zigmond
– Robin Zigmond, Commented Oct 12, 2018 at 10:24
Why have you asked this question again? As I told you last time, Ajax is completely pointless here and you should remove it. — Daniel Roseman
– Daniel Roseman, Commented Oct 12, 2018 at 10:30

Rarblack · Accepted Answer · 2018-10-12 13:02:54Z

1

In your view.py return render(request,"gui/index.html", context = context_dict code is missing ending paranthesis.

This is the correct order of jQuery ajax:

$.ajax({
  dataType: "json",
  url: url,
  data: data,
  success: success
});

Your success and error fields are inside of data.

<script>
    $(document).ready(function(){
        $("button").click(function(){
            $.ajax({url: "demo_test.txt", success: function(result){
                $("#div1").html(result);
            }});
        });
    });
    </script>
    </head>
    <body>

    <div id="div1"><h2>Let jQuery AJAX Change This Text</h2></div>

    <button>Get External Content</button>

    </body>

This is an example of how to use .html method of ajax jquery. You can adjust for your own.

Additionally, use below code to loop through response:

  $.each( data, function( key, val ) {
    HTMLString += <li id='" + key + "'>" + val + "</li>
  });

and this should be inside of the function of success and then pass HTMLString into .html method

To make it clearer how to $.each works:

var numbers = [1, 2, 3, 4, 5, 6];
$.each(numbers , function (index, value){
  console.log(index + ':' + value); 
});

edited Oct 12, 2018 at 13:02

answered Oct 12, 2018 at 10:26

Rarblack

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9 Comments

Rarblack Over a year ago

Can you post the output

Beginner Over a year ago

Now ,its same as the screenshot ,i pasted above but without the values after score in preview method

Rarblack Over a year ago

Don't forget that you are getting a dictionary as a response. You need to loop through it

Rarblack Over a year ago

How did you use the suggested .html method? It shouldn't be the same even you don't loop.

Rarblack Over a year ago

Because your ajax doesn't return with success. Look at your image it is consoling error message 'Failure'

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