Question

我的应用（示例）使用html表单来获取用户输入。然后我使用此输入来搜索数据库表中的匹配项。然后，我想打印匹配数据库列中每一行的结果。

Models.py

...

class Method(models.Model):
    method_name = models.CharField(max_length=50)
    method_description = models.TextField(max_length=100, blank=True, null=True)

    def __unicode__(self): #python 3.3 __str__
        return self.method_name

    class Meta:
        ordering = ['method_name']

class Project(models.Model):
    project_name = models.CharField(max_length=50)
    project_description = models.TextField(max_length=100)
    contact_person = models.ForeignKey(Contact)

    def __unicode__(self):
        return self.project_name

    class Meta:
        ordering = ['project_name']




class Sample(models.Model):
    sample_ID = models.CharField(max_length=20)
    sample_name = models.CharField(max_length=30)
    date = models.DateField()      
    project = models.ManyToManyField(Project)
    method = models.ForeignKey(Method)


    def __unicode__(self):
        return self.sample_ID

    class Meta:
        ordering = ['sample_ID']

views.py

def sample_lookup(request):
    form = SampleLookupForm(request.POST or None)
    id = ''
    results = []
    errors = False
    catresults = ()
    categories = ['Sample Name', 'Date', 'Project', 'Method']

    if form.is_valid():
        id = form.cleaned_data['Sample_ID'].upper()
        # Iterate through all sample IDs and check.
        # Not efficient for large query sets. 
        for s in Sample.objects.all():
            if s.sample_ID == id:
                results = [s.sample_name, s.extraction_date, s.project, s.method]
                catresults = zip(categories, results)
                errors = False
                break
        if not results:
            errors = True
    context = {
    "errors": errors, "catresults": catresults,
    "q": id, "form": form,
    }
    return render(request, 'sample_lookup.html', context)

（我知道这种观点非常丑陋但现在仍有效。）

在此之后，我使用

在html中打印出来

{% for result in catresults %}
<!-- Field = result.0, value = result.1 -->

查询本身就是样本ID。我在Sample的所有列中搜索匹配的ID。当/如果匹配时，我打印该列的每个字段。对于CharField，DateField和ForeignKey字段，我得到一个匹配该样本信息的漂亮字符串。

但是，对于ManyToManyField，我得到samples.Project.None。

Field         Value
Sample_Name   Sample_A
Date          2.2.15
Project       samples.Project.None
Method        A real method

所以看来s.project给出了samples.Project.None。我有一种感觉，这是因为这是一个ManyToMany字段。如何从该字段获得所有结果？

我尝试过s.project.project_name，但是没有用。

django = 1.8 python = 2.7

谢谢

显示查询集的ManyToManyField结果

0 个答案: