Dynamically adding a form to a Django formset

Django

Django Problem Overview


I want to dynamically add new forms to a Django formset, so that when the user clicks an "add" button it runs JavaScript that adds a new form (which is part of the formset) to the page.

Django Solutions


Solution 1 - Django

This is how I do it, using jQuery:

My template:

<h3>My Services</h3>
{{ serviceFormset.management_form }}
{% for form in serviceFormset.forms %}
    <div class='table'>
    <table class='no_error'>
        {{ form.as_table }}
    </table>
    </div>
{% endfor %}
<input type="button" value="Add More" id="add_more">
<script>
    $('#add_more').click(function() {
        cloneMore('div.table:last', 'service');
    });
</script>

In a javascript file:

function cloneMore(selector, type) {
    var newElement = $(selector).clone(true);
    var total = $('#id_' + type + '-TOTAL_FORMS').val();
    newElement.find(':input').each(function() {
        var name = $(this).attr('name').replace('-' + (total-1) + '-','-' + total + '-');
        var id = 'id_' + name;
        $(this).attr({'name': name, 'id': id}).val('').removeAttr('checked');
    });
    newElement.find('label').each(function() {
        var newFor = $(this).attr('for').replace('-' + (total-1) + '-','-' + total + '-');
        $(this).attr('for', newFor);
    });
    total++;
    $('#id_' + type + '-TOTAL_FORMS').val(total);
    $(selector).after(newElement);
}

What it does:

cloneMore accepts selector as the first argument, and the type of formset as the 2nd one. What the selector should do is pass it what it should duplicate. In this case, I pass it div.table:last so that jQuery looks for the last table with a class of table. The :last part of it is important because the selector is also used to determine what the new form will be inserted after. More than likely you'd want it at the end of the rest of the forms. The type argument is so that we can update the management_form field, notably TOTAL_FORMS, as well as the actual form fields. If you have a formset full of, say, Client models, the management fields will have IDs of id_clients-TOTAL_FORMS and id_clients-INITIAL_FORMS, while the form fields will be in a format of id_clients-N-fieldname with N being the form number, starting with 0. So with the type argument the cloneMore function looks at how many forms there currently are, and goes through every input and label inside the new form replacing all the field names/ids from something like id_clients-(N)-name to id_clients-(N+1)-name and so on. After it is finished, it updates the TOTAL_FORMS field to reflect the new form and adds it to the end of the set.

This function is particularly helpful to me because the way it is setup it allows me to use it throughout the app when I want to provide more forms in a formset, and doesn't make me need to have a hidden "template" form to duplicate as long as I pass it the formset name and the format in which the forms are laid out. Hope it helps.

Solution 2 - Django

Simplified version of Paolo's answer using empty_form as a template.

<h3>My Services</h3>
{{ serviceFormset.management_form }}
<div id="form_set">
    {% for form in serviceFormset.forms %}
        <table class='no_error'>
            {{ form.as_table }}
        </table>
    {% endfor %}
</div>
<input type="button" value="Add More" id="add_more">
<div id="empty_form" style="display:none">
    <table class='no_error'>
        {{ serviceFormset.empty_form.as_table }}
    </table>
</div>
<script>
    $('#add_more').click(function() {
        var form_idx = $('#id_form-TOTAL_FORMS').val();
        $('#form_set').append($('#empty_form').html().replace(/__prefix__/g, form_idx));
        $('#id_form-TOTAL_FORMS').val(parseInt(form_idx) + 1);
    });
</script>

Solution 3 - Django

Paolo's suggestion works beautifully with one caveat - the browser's back/forward buttons.

The dynamic elements created with Paolo's script will not be rendered if the user returns to the formset using the back/forward button. An issue that may be a deal breaker for some.

Example:

  1. User adds two new forms to the formset using the "add-more" button

  2. User populates the forms and submits the formset

  3. User clicks the back button in the browser

  4. Formset is now reduced to the original form, all dynamically added forms are not there

This is not a defect with Paolo's script at all; but a fact of life with dom manipulation and browser's cache.

I suppose one could store the values of the form in the session and have some ajax magic when the formset loads to create the elements again and reload the values from the session; but depending on how anal you want to be about the same user and multiple instances of the form this may become very complicated.

Anyone has a good suggestion for dealing with this?

Thanks!

Solution 4 - Django

Simulate and imitate:

  • Create a formset which corresponds to the situation before clicking the "add" button.
  • Load the page, view the source and take a note of all <input> fields.
  • Modify the formset to correspond to the situation after clicking the "add" button (change the number of extra fields).
  • Load the page, view the source and take a note of how the <input> fields changed.
  • Create some JavaScript which modifies the DOM in a suitable way to move it from the before state to the after state.
  • Attach that JavaScript to the "add" button.

While I do know formsets use special hidden <input> fields and know approximately what the script must do, I don't recall the details off the top of my head. What I described above is what I would do in your situation.

Solution 5 - Django

For the coders out there who are hunting resources to understand the above solutions a little better:

Django Dynamic Formsets

After reading the above link, the Django documentation and previous solutions should make a lot more sense.

Django Formset Documentation

As a quick summary of what I was getting confused by: The Management Form contains an overview of the forms within. You must keep that information accurate in order for Django to be aware of the forms you add. (Community, please give me suggestions if some of my wording is off here. Im new to Django.)

Solution 6 - Django

One option would be to create a formset with every possible form, but initially set the unrequired forms to hidden - ie, display: none;. When it's necessary to display a form, set it's css display to block or whatever is appropriate.

Without know more details of what your "Ajax" is doing, it's hard to give a more detailed response.

Solution 7 - Django

Another cloneMore version, which allows for selective sanitization of fields. Use it when you need to prevent several fields from being erased.

$('table tr.add-row a').click(function() {
    toSanitize = new Array('id', 'product', 'price', 'type', 'valid_from', 'valid_until');
    cloneMore('div.formtable table tr.form-row:last', 'form', toSanitize);
});

function cloneMore(selector, type, sanitize) {
    var newElement = $(selector).clone(true);
    var total = $('#id_' + type + '-TOTAL_FORMS').val();
    newElement.find(':input').each(function() {
        var namePure = $(this).attr('name').replace(type + '-' + (total-1) + '-', '');
        var name = $(this).attr('name').replace('-' + (total-1) + '-','-' + total + '-');
        var id = 'id_' + name;
        $(this).attr({'name': name, 'id': id}).removeAttr('checked');
        
        if ($.inArray(namePure, sanitize) != -1) {
            $(this).val('');
        }
            
    });
    newElement.find('label').each(function() {
        var newFor = $(this).attr('for').replace('-' + (total-1) + '-','-' + total + '-');
        $(this).attr('for', newFor);
    });
    total++;
    $('#id_' + type + '-TOTAL_FORMS').val(total);
    $(selector).after(newElement);
}

Solution 8 - Django

There is a small issue with the cloneMore function. Since it's also cleaning the value of the django auto-generated hidden fields, it causes django to complain if you try to save a formset with more than one empty form.

Here is a fix:

function cloneMore(selector, type) {
    var newElement = $(selector).clone(true);
    var total = $('#id_' + type + '-TOTAL_FORMS').val();
    newElement.find(':input').each(function() {
        var name = $(this).attr('name').replace('-' + (total-1) + '-','-' + total + '-');
        var id = 'id_' + name;

        if ($(this).attr('type') != 'hidden') {
            $(this).val('');
        }
        $(this).attr({'name': name, 'id': id}).removeAttr('checked');
    });
    newElement.find('label').each(function() {
        var newFor = $(this).attr('for').replace('-' + (total-1) + '-','-' + total + '-');
        $(this).attr('for', newFor);
    });
    total++;
    $('#id_' + type + '-TOTAL_FORMS').val(total);
    $(selector).after(newElement);
}

Solution 9 - Django

Because all answers above use jQuery and make some things a bit complex I wrote following script:

function $(selector, element) {
    if (!element) {
        element = document
    }
    return element.querySelector(selector)
}

function $$(selector, element) {
    if (!element) {
        element = document
    }
    return element.querySelectorAll(selector)
}

function hasReachedMaxNum(type, form) {
    var total = parseInt(form.elements[type + "-TOTAL_FORMS"].value);
    var max = parseInt(form.elements[type + "-MAX_NUM_FORMS"].value);
    return total >= max
}

function cloneMore(element, type, form) {
    var totalElement = form.elements[type + "-TOTAL_FORMS"];
    total = parseInt(totalElement.value);
    newElement = element.cloneNode(true);
    for (var input of $$("input", newElement)) {
        input.name = input.name.replace("-" + (total - 1) + "-", "-" + total + "-");
        input.value = null
    }
    total++;
    element.parentNode.insertBefore(newElement, element.nextSibling);
    totalElement.value = total;
    return newElement
}
var addChoiceButton = $("#add-choice");
addChoiceButton.onclick = function() {
    var choices = $("#choices");
    var createForm = $("#create");
    cloneMore(choices.lastElementChild, "choice_set", createForm);
    if (hasReachedMaxNum("choice_set", createForm)) {
        this.disabled = true
    }
};

First you should set auto_id to false and so disable the duplication of id and name. Because the input names have to be unique in there form, all identification is done with them and not with id's. You also have to replace the form, type and the container of the formset. (In the example above choices)

Solution 10 - Django

Yea I'd also recommend just rendering them out in the html if you have a finite number of entries. (If you don't you'll have to user another method).

You can hide them like this:

{% for form in spokenLanguageFormset %}
    <fieldset class="languages-{{forloop.counter0 }} {% if spokenLanguageFormset.initial_forms|length < forloop.counter and forloop.counter != 1 %}hidden-form{% endif %}">

Then the js is really simple:

addItem: function(e){
    e.preventDefault();
    var maxForms = parseInt($(this).closest("fieldset").find("[name*='MAX_NUM_FORMS']").val(), 10);
    var initialForms = parseInt($(this).closest("fieldset").find("[name*='INITIAL_FORMS']").val(), 10);
    // check if we can add
    if (initialForms < maxForms) {
        $(this).closest("fieldset").find("fieldset:hidden").first().show();
        if ($(this).closest("fieldset").find("fieldset:visible").length == maxForms ){
            // here I'm just hiding my 'add' link
            $(this).closest(".control-group").hide();
        };
    };
}

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionBrian TolView Question on Stackoverflow
Solution 1 - DjangoPaolo BergantinoView Answer on Stackoverflow
Solution 2 - DjangoDaveView Answer on Stackoverflow
Solution 3 - DjangocethegeekView Answer on Stackoverflow
Solution 4 - DjangoakaiholaView Answer on Stackoverflow
Solution 5 - DjangoRyan BuchmeierView Answer on Stackoverflow
Solution 6 - DjangoDaniel NaabView Answer on Stackoverflow
Solution 7 - DjangoxaralisView Answer on Stackoverflow
Solution 8 - DjangoCesar CanassaView Answer on Stackoverflow
Solution 9 - Djangouser6216224View Answer on Stackoverflow
Solution 10 - DjangoBob SprynView Answer on Stackoverflow