31 - Handling Form Errors

plucena24 · Dec 16, 2019 · 31429a1 · 31429a1
1 parent 05d048d
commit 31429a1
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diff --git a/db.sqlite3 b/db.sqlite3
diff --git a/templates/pages/home.html b/templates/pages/home.html
@@ -51,9 +51,17 @@ <h1>Welcome to Tweetme 2</h1>
             const ogHtml = tweetsContainerElement.innerHTML
             tweetsContainerElement.innerHTML = newTweetElement + ogHtml
             myForm.reset()
+        } else if (xhr.status === 400) {
+            const errorJson = xhr.response
+            console.log(errorJson)
+        } else if (xhr.status === 500) {
+            alert("There was a server error, please try again.")
         }
 
     }
+    xhr.onerror = function() {
+        alert("An error occurred. Please try again later.")
+    }
     xhr.send(myFormData)
 }
 const tweetCreateFormEl = document.getElementById("tweet-create-form")

diff --git a/tweets/views.py b/tweets/views.py
@@ -14,6 +14,7 @@ def home_view(request, *args, **kwargs):
     return render(request, "pages/home.html", context={}, status=200)
 
 def tweet_create_view(request, *args, **kwargs):
+    print(abc)
     form = TweetForm(request.POST or None)
     next_url = request.POST.get("next") or None
     if form.is_valid():
@@ -25,6 +26,9 @@ def tweet_create_view(request, *args, **kwargs):
         if next_url != None and is_safe_url(next_url, ALLOWED_HOSTS):
             return redirect(next_url)
         form = TweetForm()
+    if form.errors:
+        if request.is_ajax():
+            return JsonResponse(form.errors, status=400)
     return render(request, 'components/form.html', context={"form": form})