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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.
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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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data = [{'name':'Albert','rel':'Head','unique_number': 101},
{'name':'Sheen','rel':'Head','unique_number': 201},
{'name':'Peter','rel':'Son','unique_number': 101},
{'name':'Chloe','rel':'Daughter','unique_number': 101}]
can you help me out in getting data like this? filtered on unique_number
updated_data = [
{'house_head':'Albert','members':['Peter','Chloe']},
{'house_head':'Sheen','members':[]}
]
The following should work:
numbers=set([i['unique_number'] for i in data])
dict={i:{'Head':'', 'members':[]} for i in numbers}
for i in data:
if i['rel']=='Head':
dict[i['unique_number']]['Head']=i['name']
else:
dict[i['unique_number']]['members'].append(i['name'])
new_data=[{'house_head':dict[i]['Head'], 'members':dict[i]['members']} for i in dict]
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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 want to output of sql as row wise, dont want output to be inline.
This is the code and output in the image :--
So, how can i view my result of a query row wise, not inline.
for eg like this :-
Ursula La Multa | 4790940
Rudolf von Treppenwitz | 3593205
Markoff Chaney | 2395470
Anonymous Contributor | 1197735
MOREOVER, HOW CAN I GET RID OF THAT KEYWORD 'DECIMAL' FROM [Decimal('4790940')] AS SHOWN IN OUTPUT
for result in results:
print ("%s\t | %s" % (result[0], result[1]))
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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]