Django錯誤提交形式爲

Question

我有我想提交給我的服務器的Django形式。當提交按鈕被按下時，它發送一個post請求，並且兩個對象（一個stream_id和一個圖像）被髮送到view函數。我可以從調試頁面看到，對象分別包含POST和FILES對象中的stream_id和image。

然而，一旦視圖功能被擊中，我嘗試初始化形式如下：

uploadImageForm = UploadImageForm(request.POST, request.FILES) 

並獲得我拋出驗證錯誤。

[28/May/2013 18:46:32] DEBUG [ct:194] uploadImage - method is post, errors: 
<ul class="errorlist"><li>stream_id<ul class="errorlist"> 
<li>This field is required.</li></ul></li> 
<li>image<ul class="errorlist"><li>This field is required.</li></ul></li></ul> 
[28/May/2013 18:46:32] DEBUG [ct:195] cleaned data: {} 

任何想法爲什麼？

我的形式：

class UploadImageForm(forms.Form): 
    image = forms.ImageField() 
    image.widget.attrs["onchange"]="this.form.submit();" 
    stream_id = forms.CharField(max_length=100) 
    stream_id.widget.attrs["style"]="display:none;" 

    def __init__(self,stream_id,*args,**kwrds): 
     super(UploadImageForm,self).__init__(*args,**kwrds) 
     self.fields['stream_id'].widget.attrs["value"]= stream_id 

型號：

class Stream(models.Model): 
    tracked_user = models.ForeignKey(TrackedUser) 
    stream_id = models.CharField(max_length=255) 
    stream_hash = models.CharField(max_length=255) 
    name = models.CharField(max_length=60) 
    start_time = models.DateTimeField(default=datetime.datetime.now(pytz.utc)) 
    end_time = models.DateTimeField(default=None) 
    end_time.null = True 
    image = models.ImageField(upload_to="/") 
    image.null = True 

    def __unicode__(self): 
     return self.name 

代碼在views.py：

def uploadImage(request): 
    uploadImageForm = None 
    if (request.method == "POST"): 
     uploadImageForm = UploadImageForm(request.POST, request.FILES) 
     log.debug("uploadImage - method is post, errors: " + 
      str(uploadImageForm.errors)) 

有趣的是，結合的形式輸出如下：

<p> 
    <label for="id_image">Image:</label> 
    <input id="id_image" name="image" onchange="this.form.submit();" type="file" /> 
</p> 
<ul class="errorlist"> 
    <li>This field is required.</li> 
</ul> 

<p> 
    <label for="id_stream_id">Stream id:</label> 
    <input id="id_stream_id" maxlength="100" name="stream_id" 
     style="display:none;" type="text" value="&lt;QueryDict: 
     {u&#39;stream_id&#39;:[u&#39;d21256f37601d2800b0b9604f0e94e1e&#39;], 
     u&#39;csrfmiddlewaretoken&#39;: 
     [u&#39;F0fmAD0VAj0RHrM0GGfnaSb6vTNgj9ZJ&#39;]}&gt;" /> 
</p> 

Answer 1

默認情況下，Django表單中的所有字段都是必需的，這就是您在Stream ID和其他字段上收到錯誤的原因。在你不希望被需要，或者使任選被你的表單類所需的字段設置blank=True：

class Stream(models.Model): 
    tracked_user = models.ForeignKey(TrackedUser) 
    stream_id = models.CharField(max_length=255, blank=True) 
    stream_hash = models.CharField(max_length=255, blank=True) 
    name = models.CharField(max_length=60) 
    start_time = models.DateTimeField(default=datetime.datetime.now(pytz.utc)) 
    end_time = models.DateTimeField(default=None, null=True) 
    image = models.ImageField(upload_to="/") 

    def __unicode__(self): 
     return self.name 

我不太清楚你與ImageUploadForm做什麼，但你要按照當前寫入的方式獲取初始化錯誤。您也沒有在您的視圖中傳遞stream_id。

class UploadImageForm(forms.Form): 
    image = forms.ImageField() 
    image.widget.attrs["onchange"]="this.form.submit();" 
    stream_id = forms.CharField(max_length=100) 
    stream_id.widget.attrs["style"]="display:none;" 

    def __init__(self, *args, **kwargs): 
     stream_id = kwargs.pop('stream_id') 
     super(UploadImageForm,self).__init__(*args, **kwargs) 
     self.fields['stream_id'].widget.attrs["value"]= stream_id 

我想你的意思是從繼承的ModelForm ...

class UploadImageForm(forms.ModelForm): 
    class Meta: 
     model = Stream 

    def __init__(self, *args, **kwargs): 
     stream_id = kwargs.pop('stream_id') 

     super(UploadImageForm, self).__init__(*args, **kwargs) 

      # this is really bad practice. I would add this click handler 
      # unobtrusively 
      self.fields['image'].widget.attrs["onchange"] = "this.form.submit();" 
      stream_id = self.fields['stream_id'] 

      # not sure what you're trying to accomplish here... 
      self.fields['stream_id'].widget.attrs["style"] = "display:none;" 

      stream_id.initial = stream_id