图片未上传到文件夹

时间:2019-09-25 11:40:39

标签: django django-models django-views

我要上传图片,但是它没有显示在媒体文件夹中。

这是HTML模板

<form class="form" action="/profile/edit/{{user.id}}" method="post" enctype="multipart/form-data" style="padding-top:20px;">
                        {% csrf_token %}
                        <input type="file" class="btn " name="pro_pic" accept=".png, .jpg, .jpeg">
                        <button type="submit" class="btn text-success" name="pr_pic" id="pic_submit"><i class="fas fa-check"></i> Submit</button>
                    </form>

Models.py

class UserDetails(models.Model):

    user_id = models.IntegerField(unique=True)
    bio = models.TextField(null = True)
    profession = models.CharField(max_length = 100, null = True)
    city = models.CharField(max_length = 100, null = True)
    country = models.CharField(max_length = 100, null = True)
    img = models.ImageField(upload_to='pic_user', null = True)

views.py

def edit (request, id = '') :


    if request.user.is_authenticated == True :

        if request.method == 'POST' :
        # MyModel.objects.filter(pk=some_value).update(field1='some value')

            if request.POST.get('pr_pic') == '' :

                post_image = request.FILES['pro_pic']
                # pic = UserDetails.objects.filter(user_id = id).update(img = post_image)
                # UserDetails.objects.filter(user_id = id).update(img = post_image)
                pic = UserDetails.objects.filter(user_id = id)
                pic.img = post_image
                pic.save()

无论何时创建用户,他的user_id都会在另一个表(UserDetails模型)中更新,并且字段为空。

但是在HTML ive中,有一个选项可以稍后上传图像(该ive在HTML中显示)。

我认为我的views.py错误,请帮助我更正

1 个答案:

答案 0 :(得分:0)

尝试:

import uuid
from PIL import Image

format_image, img_str = request.FILES['pro_pic'].split(';base64,')
file_content = ContentFile(base64.b64decode(img_str))

unique_filename = str(uuid.uuid4())
pic.img.save(unique_filename, file_content)

file_url = settings.MEDIA_FILES + '/tmp/' + unique_filename

# convert image into JPEG format and save
img = Image.open(file_url)
img.save(file_url, format='jpeg', optimize=True, quality=85)
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