praxmon
praxmon

Reputation: 5121

Django data updation in the table

I am using Django to enter some data into my database. After entering the data I want to edit it. Now, what I am trying is, the user should not go to any other page to change the data. So I have implemented a javascript method which edits the text on the front end.

How do I reflect the changes made by the user in the database?

The related code is given below:

<html>
{% csrf_token  %}
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js">
</script>

        <table id="table">
                <tr>
                        <th>Name</th>
                        <th>Phone Number</th>
                </tr>
                {% for record in queryset  %}
                <tr>
                        <td onClick="clickable(this)"> {{record.first}} </td>
                        <td onClick="clickable(this)"> {{record.second}}</td>
                </tr>
                {%endfor%}
        </table>

<script>
function clickable(ele)
{
        var value = prompt("Enter the details");
        if(value)
        {
                ele.id='edited'
                ele.innerHTML = value;
                //I want to send the request to my django view to edit the database here
                //The data has been updated.

        }

Upvotes: 0

Views: 1526

Answers (1)

Prashant Gaur
Prashant Gaur

Reputation: 9828

You should send a Ajax request to your server using jQuery you are using. with Ajax request request you should send your updated data .
Simple Ajax request can be .

$('#click_place').click(function() { // when click is placed
            $.ajax({ // create an AJAX call...
                data: $(this).serialize(), // get the content here Ex. {'name': 'test', 'phone': '123'}
                type: $(this).attr('method'), // GET or POST
                url: $(this).attr('action'), // request url
                success: function(response) { // on success..
                    // display your message
                }
            });
            return false;
        });

You can follow How to POST a django form with AJAX & jQuery .
http://coreymaynard.com/blog/performing-ajax-post-requests-in-django/ .

Edit : You can simply call below function at any event .

function  myajaxhit() {
                $.ajax({ // create an AJAX call...
                    data: $(this).serialize(), // get the content here Ex. {'name': 'test', 'phone': '123'}
                    type: $(this).attr('method'), // GET or POST
                    url: $(this).attr('action'), // request url
                    success: function(response) { // on success..
                        // display your message
                    }
                });
}

just call myajaxhit() at any place . Please change it as per your requirement .

Upvotes: 1

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