django MultiValueDictKeyError error, how do I deal with it

django MultiValueDictKeyError error, how do I deal with it

I'm trying to save a object to my database, but it's throwing a MultiValueDictKeyError error.
The problems lies within the form, the is_private is represented by a checkbox. If the check box is NOT selected, obviously nothing is passed. This is where the error gets chucked.
How do I properly deal with this exception, and catch it? 
The line is
is_private = request.POST['is_private']

Solutions/Answers:

Answer 1:

Use the MultiValueDict’s get method. This is also present on standard dicts and is a way to fetch a value while providing a default if it does not exist.

is_private = request.POST.get('is_private', False)

Generally,

my_var = dict.get(<key>, <default>)

Answer 2:

Choose what is best for you:

1

is_private = request.POST.get('is_private', False);

If is_private key is present in request.POST the is_private variable will be equal to it, if not, then it will be equal to False.

2

if 'is_private' in request.POST:
    is_private = request.POST['is_private']
else:
    is_private = False

3

from django.utils.datastructures import MultiValueDictKeyError
try:
    is_private = request.POST['is_private']
except MultiValueDictKeyError:
    is_private = False

Answer 3:

You get that because you’re trying to get a key from a dictionary when it’s not there. You need to test if it is in there first.

try:

is_private = 'is_private' in request.POST

or

is_private = 'is_private' in request.POST and request.POST['is_private']

depending on the values you’re using.

Answer 4:

Why didn’t you try to define is_private in your models as default=False?

class Foo(models.Models):
    is_private = models.BooleanField(default=False)

Answer 5:

Another thing to remember is that request.POST['keyword'] refers to the element identified by the specified html name attribute keyword.

So, if your form is:

<form action="/login/" method="POST">
  <input type="text" name="keyword" placeholder="Search query">
  <input type="number" name="results" placeholder="Number of results">
</form>

then, request.POST['keyword'] and request.POST['results'] will contain the value of the input elements keyword and results, respectively.

Answer 6:

For me, this error occurred in my django project because of the following:

  1. I inserted a new hyperlink in my home.html present in templates folder of my project as below:

    <input type="button" value="About" onclick="location.href='{% url 'about' %}'">
    1. In views.py, I had the following definitions of count and about:

    def count(request):
    fulltext = request.GET['fulltext']
    wordlist = fulltext.split()
    worddict = {}
    for word in wordlist:
    if word in worddict:
    worddict[word] += 1
    else:
    worddict[word] = 1
    worddict = sorted(worddict.items(), key = operator.itemgetter(1),reverse=True)
    return render(request,'count.html', 'fulltext':fulltext,'count':len(wordlist),'worddict'::worddict})

    def about(request):
    return render(request,"about.html")

    1. In urls.py, I had the following url patterns:

    urlpatterns = [
    path('admin/', admin.site.urls),
    path('',views.homepage,name="home"),
    path('eggs',views.eggs),
    path('count/',views.count,name="count"),
    path('about/',views.count,name="about"),
    ]

As can be seen in no. 3 above,in the last url pattern, I was incorrectly calling views.count whereas I needed to call views.about.
This line fulltext = request.GET['fulltext'] in count function (which was mistakenly called because of wrong entry in urlpatterns) of views.py threw the multivaluedictkeyerror exception.

Then I changed the last url pattern in urls.py to the correct one i.e. path('about/',views.about,name="about"), and everything worked fine.

Apparently, in general a newbie programmer in django can make the mistake I made of wrongly calling another view function for a url, which might be expecting different set of parameters or passing different set of objects in its render call, rather than the intended behavior.

Hope this helps some newbie programmer to django.

Answer 7:

First check if the request object have the ‘is_private’ key parameter.
Most of the case’s this MultiValueDictKeyError occurred for missing
key in the dictionary-like request object. Because dictionary is an
unordered key, value pair “associative memories” or “associative
arrays”

In another word. request.GET or request.POST is a dictionary-like
object containing all request parameters. This is specific to Django.

The method get() returns a value for the given key if key is in the
dictionary. If key is not available then returns default value None.

You can handle this error by putting :

is_private = request.POST.get('is_private', False);

References