invalid literal for int() with base 10: 'John Doe'

Question

i have a model patient and another one exam, now i want to retrieve patient's exams based on the name, but i am stuccoed with the error invalid literal for int() with base 10: 'John Doe'

ValueError at /labtech/John Doe/see_exam
invalid literal for int() with base 10: 'John Doe'
Request Method: GET
Request URL:    http://homasoft.com:8000/labtech/John%20Doe/see_exam
Django Version: 1.8
Exception Type: ValueError
Exception Value:    
invalid literal for int() with base 10: 'John Doe'
Exception Location: /Library/Python/2.7/site-        packages/django/db/models/fields/__init__.py in get_prep_value, line 985
Python Executable:  /usr/bin/python
Python Version: 2.7.5
Python Path:    
['/Users/mymacbookpro/Documents/Virtualenvs/hospital/src',
'/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg',
'/Users/mymacbookpro/Documents/Virtualenvs/hospital/src',
'/Users/mymacbookpro/Documents/Virtualenvs/hospital/src/django-social-auth',
'/Users/mymacbookpro/Documents/Virtualenvs/hospital/src/django-socialprofile',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python27.zip',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/plat- darwin',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/plat-mac',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/plat-mac/lib-scriptpackages',
'/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/lib-tk',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/lib-old',
'/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/lib-dynload',
 '/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/PyObjC',
'/Library/Python/2.7/site-packages']
Server time:    Mon, 1 Jun 2015 12:27:27 +0000

here is my view

def see_exam(request, name):
    exam = Test.objects.filter(patient__user__exact=name)

    context = {
        'exam' : exam
    }

    return render(request, 'account/labtechs/see_results.html', context)

model exam

class Test(models.Model):
    exam = models.ForeignKey(Exam)
    patient = models.ForeignKey(Patient)
    date = models.DateField()
    result = models.TextField(default="negative")
    done_by = models.CharField(max_length=120)
    def __unicode__(self):
        return self.exam.name

model for patient

class Patient(models.Model):
    """docstring for Patient"""
    user = models.OneToOneField(MyUser)
    date_of_birth = models.DateTimeField(blank=True, null=True)
    age = models.IntegerField(default=1)
    sex = models.OneToOneField(Sex)
    religion = models.CharField(max_length=120, blank=True, null=True)
    village = models.CharField(max_length=120)
    status = models.OneToOneField(Status, blank=True, null=True)
    relative = models.CharField(max_length=120)
    phone = models.IntegerField(default=1)
    allergies = models.TextField(default="", blank=True, null=True)
    defficiencies = models.TextField(default="", blank=True, null=True)
    created_at = models.DateTimeField(auto_now=True)
    updated_at = models.DateTimeField(auto_now_add=True)
    def __unicode__(self):
        return "%s %s" %(self.user.first_name, self.user.last_name)

http://stackoverflow.com/questions/16742432/python-error-valueerror-invalid-literal-for-int-with-base-10-stop — chandu, Jun 01 '15 at 12:39

score 0 · Answer 1 · answered Jun 01 '15 at 12:49

0

Change the query

exam = Test.objects.filter(patient__user__exact=name)

to

exam = Test.objects.filter(patient__user__name__exact=name)
#------------------------------------------^

Note the user__name added. Here I'm assuming your MyUser class has attribute name that you want to compare.

In your query you are trying to compare string name with object user (or rather internally id of that object) which is not compatible.

answered Jun 01 '15 at 12:49

Rohan

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no, there is attr called name, there is first and last name and there is a function get_full_name() – King Yohel Jun 09 '15 at 02:02
@user134186, then you need to split the provided name into first and last name and compare them as suggested. Or do `startswith` and `endswith` filter on them – Rohan Jun 09 '15 at 04:21

invalid literal for int() with base 10: 'John Doe'

1 Answers1