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I was trying to fetch text content from table which works well but along with result it print unwanted codes
my code is here
searchitem = searchme.objects.filter(face = after) .values_list ("tale" , flat = True)
the contents are text
the result I receive is "querySet Prabhakaran seachitem"
but I only want o get result "Prabhakaran"
model is this
class searchme ( models.Model):
face = models.TextField()
tale = models.TextField ()
From the official django documentation :
A common need is to get a specific field value of a certain model instance. To achieve that, use values_list() followed by a get() call:
So use:
searchme.objects.values_list('tale', flat=True).get(face=after)
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i have soup response text with multiple group and sub groups.
i want to get automatic all groups and their values .
how can i do it ?
In the end, I want to get the title and the value for each group. The best thing for me is for each group to have its values separately.
OrderedDict([('#id',
'boic'),
('mc:id',
'boic'),
('mc:ocb-conditions',
OrderedDict([('mc:rule-deactivated',
'true'),
('mc:international',
'true')])),
('mc:cb-actions',
OrderedDict([('mc:allow',
'false')]))])
My goal is to get to a state where I get the following output:
'#id','boic'
'mc:id','boic'
'mc:ocb-conditions'
'mc:rule-deactivated','true'
'mc:international', 'true'
'mc:cb-actions'
'mc:allow','false'
i try to use
' '.join(BeautifulSoup(soup_response, "html.parser").findAll(text=True))
and got all values But I'm missing the titles of the values.
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Input: abc#xyz.com
Output: xyz
my code is below
email = "123#helloworld.com"
new_email = email.split('#')[1]
domain = new_email.split('.')[0]
Please help me with other methods to get the part of it?
UPDATE (Based on a comment, see below)
If you have input like 123#helloworld.python.com or 123#helloworld.yahoo. And you want to extract only helloworld. You can use
result = re.search(r'#([^\.]+)\.', email).group(1)
BEFORE UPDATE
You can use re module.
import re
email = email = '123#helloworld.com'
result = re.search(r'#(.+)\.com$', email)
if result:
result = result.group(1)
print(result)
Output:
helloworld
If you are sure that you will always have something in between # and .com. You can remove if statement.
import re
email = email = '123#helloworld.com'
result = re.search(r'#(.+)\.com$', email).group(1)
print(result)
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I want to print firstname of employees who have both work number and mobile number. Below is my json body. I am facing difficulty in getting inside phoneNumbers attribute. My final output should be: "Adhi as Adhi has both work and mobile numbers".
I am not able to iterate the inner dictionary of phoneNumbers attribute.Can you please help me on this.
This is my python code
for i in Data['users']:
for j in i['phoneNumbers']:
for i in range(len(j)):
if j['type']=="work" and j['type']=="mobile":
print("Firstname",i['firstName'])
You can loop over the users and check if the work and mobile number are present:
for user in Data['users']:
has_mobile_number = False
has_work_number = False
for phonenumber in user['phoneNumbers']:
if phonenumber['type'] == 'work':
has_work_number = True
if phonenumber['type'] == 'mobile':
has_mobile_number = True
if has_work_number and has_mobile_number:
print('Firstname', user['firstName'])
Also, I recommend not using i and j when not talking about indexes. In you code, i is a dict representing a user and j is a dict representing a phone. I replaced them with user and phonenumber for more clarity in the code above.
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I have a problem with django pagination. In my table I have 13,618 records but do pagination, I do not return results.
>>> from api.models import Post
>>> posts = Post.objects.all()
>>> posts.count()
13618
>>> posts = Post.objects.all()[10:10]
>>> posts.count()
0
The problem is in your slicing:
posts = Post.objects.all()[10:10]
You're asking for the 10th item to the 9th (10-1) item, which is an empty list. The same would happen if you did this:
ls = [1,2,3]
ls[1:1] # => []
It looks like you want 10 items starting from the 10th, in which case you should do:
posts = Post.objects.all()[10:20]
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Mutliple fields in the where clause of a QuerySet
qs_new = model_obj.objects.all()
You can use filter()to define a WHERE clause for your query:
qs_new = model_obj.objects.filter(...)
The QuerySet documentation describes various ways in which you can combine conditions and filters.
qs_new = model_obj.objects.filter(col_name='value', col_name2='value2')
filters_con = { 'a': 1, 'b': 2, 'c': 3 }
model_name.objects.filter(**filters_con)
it is working fine for me.