Django：更新創建新的 object 而不是更新現有的 django？

Question

我正在使用 django 和 ajax 創建評論更新 function，並且我正在嘗試在不刷新頁面的情況下更新評論，現在當我點擊編輯框以進行編輯時，我正在編輯之前文本（評論）而不是將新編輯保存到舊編輯； 它繼續創建一個新的評論，我該如何防止它。

視圖.py

## edit comment
@csrf_exempt
def edit_comment(request):
    if request.method == "POST":
        id = request.POST.get("cid")
        comment = Comment.objects.get(pk=id)
        comment_data = {"id": comment.id, "comment": comment.comment, "video": comment.video.id}
        return JsonResponse(comment_data)

## Create comment
def ajaxComment(request):
    if request.method == "POST":
        pk = request.POST.get("id")
        video = Video.objects.get(id=pk)
        user = request.user
        comment = request.POST.get("comment")
        new_comment = Comment(user=user, video=video, comment=comment)
        new_comment.save()

        print(id)
        print(user)
        print(comment)
        print(video)

        response = "Comment Posted"
        return HttpResponse(response)

ajax代碼

// Edit
$('.comment-wrapper').on("click", ".btn-edit", function(){
    console.log("Edit Button Cliked");
    let id = $(this).attr("data-cid");

    console.log(id);

    mydata = {cid:id}

    $.ajax({
        url: "{% url 'edit-comment' %}",
        method:"POST",
        data:mydata,

        success: function(data){
            console.log(data);
            $("#id").val(data.id)
            $("#id").val(data.video)
            $("#comment").val(data.comment)
            // $(".new_comment").val(data.video)

            console.log(data.id);


        },
    })
})

$(document).on("submit", "#comment_form", function(e){
        e.preventDefault();
        let _comment = $("#comment").val()

        console.log("send button clicked")
        $.ajax({
            type: "POST",
            url: "{% url 'save-comment' %}",
            data:{
                id: $("#id").val(),
                comment: _comment,
                csrfmiddlewaretoken: $('input[name=csrfmiddlewaretoken]').val(),
            },
            success: function(){
                // Some code
            }

索引.html

<button data-cid="{{ c.id }}" class="btn-edit">Edit</button>

什么可能是最好的解決方法？

Answer 1

我假設您發布到ajaxComment是新的還是編輯的。 如果它是一個編輯，只需傳遞一個帶有評論 PK 的額外 POST：

def ajaxComment(request):
    if request.method == "POST":
        pk = request.POST.get("id")
        video = Video.objects.get(id=pk)
        user = request.user

        comment = request.POST.get("comment")

        if pk:
            # is an edit
            o = Comment.objects.get(pk=pk)
            o.comment = comment
            # you could chop these two out
            o.video = video
            o.user = user
            o.save()
            response = "Comment Edited"
        else:
            # is new
            new_comment = Comment(user=user, video=video, comment=comment)
            new_comment.save()
            response = "Comment Created"

        print(id)
        print(user)
        print(comment)
        print(video)

        return HttpResponse(response)