[英]Get current logged in user in JSON Format from views.py
我正在尝试构建一个关注和取消关注按钮,我只想显示该按钮对于登录的任何其他用户,用户不应该能够关注自己。
如何从我的views.py 将当前登录的用户以JSON 格式返回到我的JS 文件中? 我在 JavaScript 文件中而不是在模板中构建按钮,我试图以 JSON 格式返回但没有成功
视图.py
def display_profile(request, profile):
try:
profile_to_display = User.objects.get(username=profile)
profile_to_display_id = User.objects.get(pk=profile_to_display.id)
except User.DoesNotExist:
return JsonResponse({"error": "Profile not found."}, status=404)
# Return profile contents
if request.method == "GET":
return JsonResponse(profile_to_display.profile.serialize(), safe=False)
else:
return JsonResponse({
"error": "GET or PUT request required."
}, status=400)
模型.py
from django.contrib.auth.models import AbstractUser
from django.db import models
class User(AbstractUser):
pass
class NewPost(models.Model):
poster = models.ForeignKey("User", on_delete=models.PROTECT, related_name="posts_posted")
description = models.TextField()
date_added = models.DateTimeField(auto_now_add=True)
likes = models.IntegerField(default=0)
def serialize(self):
return {
"id": self.id,
"poster": self.poster.username,
"description": self.description,
"date_added": self.date_added.strftime("%b %d %Y, %I:%M %p"),
"likes": self.likes
}
class Profile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
following = models.ManyToManyField(User, blank=True, related_name="following")
followers = models.ManyToManyField(User, blank=True, related_name="followers")
def serialize(self):
return {
"profileID": self.user.id,
"following": int(self.following.all().count()),
"followers": int(self.followers.all().count()),
}
index.js
function load_user_info(user_clicked_on){
document.querySelector('#page-view').style.display = 'none';
document.querySelector('#posts-view').style.display = 'none';
document.querySelector('#show-posts').style.display = 'none';
document.querySelector('#load-profile').style.display = 'block';
fetch(`/profile/${user_clicked_on}`)
.then(response => response.json())
.then(profile => {
const profile_element = document.createElement('div');
const followers = document.createElement('div');
const following = document.createElement('div');
const follow_button = document.createElement('button');
followers.innerHTML = 'Followers: ' + profile.followers;
following.innerHTML = 'Following: ' + profile.following;
if (profile.profileID == ?? ){
follow_button.innerHTML = 'Sameuser';
}else{
follow_button.innerHTML = 'Not same user';
}
profile_element.appendChild(followers);
profile_element.appendChild(following);
profile_element.appendChild(follow_button);
profile_element.classList.add('profile_element');
document.querySelector('#user-profile').appendChild(profile_element);
});
document.querySelector('#user-profile').innerHTML = `<h3>${user_clicked_on.charAt(0).toUpperCase() + user_clicked_on.slice(1)} Profile</h3>`;
}
我认为您需要的只是登录用户的详细信息吗? 可能您可以通过使用从请求 object 中获取它
{{request.user}}
这将返回记录的用户实例
{{request.user.username}} {{request.user.email}}
这将返回用户名和 email
我不知道您是否可以直接在 javascript 中使用它,但是您可以使用隐藏字段将其保存在模板中
<input type="hidden" value="{{request.user.username}}" id='user_detail'>
并使用 id 选择器(getElementById 或 jquery)抓取它
$('#user_detail').val()
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