比较 Django 中 2 个不同模型对象的 views.py 中的对象

Question

我正在尝试比较 2 个对象（Web 开发、市场营销等职位发布的技能与登录用户的相同技能），如果匹配，将显示该职位发布。

目标是显示与用户匹配的多个职位发布。当前，jobmatch.html 不显示任何职位。

感谢您抽出宝贵时间！！

views.py

from apps.job.models import Job
from apps.userprofile.models import User_profile

def match_jobs(request):
    match_jobs = {}
    for job in Job.objects.all():
        if job.product_management == User_profile.product_management:
            match_jobs['product management'] = job
        elif job.web_development == User_profile.web_development:
            match_jobs['web development'] = job
        elif job.user_experience == User_profile.user_experience:
            match_jobs['user experience'] = job
        elif job.marketing == User_profile.marketing:
            match_jobs['marketing'] = job
        elif job.finance == User_profile.finance:
            match_jobs['finance'] = job

    return render(request, 'jobmatch.html', match_jobs)

Job.models.py

class Job(models.Model):
    title = models.CharField(max_length=255)    
    location = models.CharField(max_length=255, blank=True, null=True)
    description = models.TextField()
    requirements = models.TextField(blank=True, null=True)
    product_management = models.BooleanField(default=False)
    web_development = models.BooleanField(default=False)
    user_experience = models.BooleanField(default=False)
    marketing = models.BooleanField(default=False)
    finance = models.BooleanField(default=False)

User.models.py

class User_profile(models.Model):
    user = models.OneToOneField(User, related_name='userprofile', on_delete=models.CASCADE)
    is_employer = models.BooleanField(default=False)
    resume = models.ImageField(default='default.jpg', upload_to='resume')
    full_name = models.CharField(max_length=255, default='Enter full name')
    relevant_background = models.TextField()
    product_management = models.BooleanField(default=False)
    web_development = models.BooleanField(default=False)
    user_experience = models.BooleanField(default=False)
    marketing = models.BooleanField(default=False)
    finance = models.BooleanField(default=False)

jobmatch.html

            {% for job in match_jobs %}
                <div class="colum is-4">
                    <div class="card">
                        <div class="card-content">
                            <div class="media">
                                <div class="media-content">
                                    <p class="title is-4">{{ job.title }}</p>
                                    <p class="subtitle is-6">{{ job.created_at }}</p>
                                </div>
                            </div>

                            <div class="content">
                                {{ job.location }}

                                <br>

                                <a href="{% url 'job_detail' job.id %}">View</a>
                            </div>
                        </div>
                    </div>
                </div>
            {% endfor %}

Answer 1

您现在可以通过字典键进行迭代。您必须执行以下操作：{% for job in match_jobs.values %} 或 {% for job_identifier, job in match_jobs.items %}

Answer 2

您可以创建一个 Skill 模型来创建您想要的任何技能，并将它们与使用 ManyToManyField 的工作和用户相关联。这将使添加、更新或删除任何技能变得容易。

class Skill(models.Model):
    name = models.CharField(max_length=250)

class Job(models.Model):
    title = models.CharField(max_length=255)    
    location = models.CharField(max_length=255, blank=True, null=True)
    description = models.TextField()
    requirements = models.TextField(blank=True, null=True)
    skills = models.ManyToManyField(Skill, blank=True, related_name='as_job_skill')

class User_profile(models.Model):
    user = models.OneToOneField(User, related_name='userprofile', on_delete=models.CASCADE)
    is_employer = models.BooleanField(default=False)
    resume = models.ImageField(default='default.jpg', upload_to='resume')
    full_name = models.CharField(max_length=255, default='Enter full name')
    relevant_background = models.TextField()
    skills = models.ManyToManyField(Skill, blank=True, related_name='as_user_skill')

您还可以非常方便地为用户找到匹配的职位。

def match_jobs(request):
    user_profile = request.user.userprofile
    matched_jobs = Job.objects.filter(skills__in=user_profile.skills.all()).distinct()
    return render(request, 'jobmatch.html', matched_jobs)

Answer 3

您需要重新设计模型。

将所有技能存储在单独的 table 中。

class Skill(models.Model):
  name = models.CharField(max_length=50) # example web development

现在您可以通过多对多关系link您的工作和用户配置文件模型与技能模型。

class Job(models.Model):
  ......
  skills = models.ManyToManyField(Skill)
 
class UserProfile(models.Model):
.........
  skills = models.ManyToManyField(Skill)

现在你可以用这个找到匹配的工作。

 user_skills = request.user.userprofile.skills.all()
 matched_jobs = Job.objects.filter(skills__in=user_skills)

比较 Django 中 2 个不同模型对象的 views.py 中的对象

compare objects in views.py of 2 different models objects in Django

html

python

django

compare

django-views