I am storing output of each column in a list for the below query
select table,
columns,
flag
from {{ model }}
for each table I need to compare the flag. how do i get flag in list for the corresponding table
{% for n in table %}
{%- if flag[n] == 'False' %}
I tried as above its not working its returning null value
It's important to understand that jinja is a templating language. In dbt, we use jinja to automate the creation of sql that then gets executed in our data warehouse. Accordingly, data from your data warehouse does not (typically) flow through your jinja code.
flag is a reference to a column in your database, so to evaluate the contents of flag, you need to write sql to do that job.
The only exception is if you are using run_query or a similar macro to pull the data from your database into your jinja context at model-compile time. But I don't see any reference to that in your code (and frankly that's a complicating detail that's only relevant if you are trying to write SQL dynamically). If you are just trying to compare the value from a record in your database as part of your model, just do it in sql.
I have a query set like the below.
<QuerySet[{'user':'xyz','id':12,'home':'qwe','mobile':1234},
{'user':'xyz','id':12,'home':'qwe','mobile':4321},
{'user':'abc','id':13,'home':'def','mobile':1233},
{'user':'abc','id':13,'home':'def','mobile':1555},]>
This QuerySet is returned by django using
users.objects.all()
For Each query in the query set, I'm drawing a user table in the front-end which shows the details of users. If the same 'user' registers with two 'mobile' numbers, it is showed as two rows in the table instead of one. My goal is to display two numbers in the same row instead of creating two rows.
I have thought of two possible solutions which are below:
Solution 1: I have thought of merging the two queries into one if the 'user' value matches in both the queries. For this we need to make lot of checks using conditional statements which works slowly when lot of users are there.
Solution 2: I have searched on Google and came up with Group By Django, but it is not working. I have tried below
query = users.objects.all().query
query.group_by = ['mobile']
results = QuerySet(query=query, model=users)
Please provide a way so that two queries can be clubbed into one based on 'user' and after clubbing 'mobile' should contain two values.
Edit: I will pass this query set to a template via view where the table is being drawn. At present, for each query, one row is the in table. Since the above query set has four queries, we will have four rows. But I want to display the information in just two rows since the 'id' and 'user' are same. Code for template is below:
{% if users %}
{% for user in users %}
{% for k,v in required_keys.items %}
<td>{{ user | get_item:k }}
{% endfor %}
{% endfor %}
{% endif %}
Required key is an dictionary which contains the keys which are used to display in table. Example:
Required_keys = {
'User Name':user,
'Contact':mobile,
'Address':home,
}
The get item is a function which gets the value when key is passed.
def get_item(dictionery,key):
return dictionery.get(key)
Edit 2: I have tried for a solution and it's not a full solution but it partially solves the problem. The solution perfectly works with one user. However if there are many users, the solution doesn't work. See below for example:
# input query set to the function which partially solves the problem
<QuerySet[{'user':'xyz','id':12,'home':'qwe','mobile':1234},
{'user':'xyz','id':12,'home':'qwe','mobile':4321},]>
# Now our team has written a function which gives the following output
<QuerySet[{'user':'xyz','id':12,'home':'qwe','mobile':1234,4321},]>
But if there are more than one user, the output is same as input, it doesn't club the queries. See the function below:
def merge_values(cls,values):
import itertools
grouped_results = itertools.groupby(values,key=lambda x:x['id'])
merged_values = []
for k,g in grouped_results:
groups=list(g)
merged_value = {}
for group in groups:
for key, val in group.iteritems():
if not merged_value.get(key):
merged_value[key] = val
elif val != merged_value[key]:
if isinstance(merged_value[key], list):
if val not in merged_value[key]:
merged_value[key].append(val)
else:
old_val = merged_value[key]
merged_value[key] = [old_value, val]
merged_values.append(merged_value)
return merged_values
The values parameter for the function is the whole query set. But this function
works if there is only one user in query set as mentioned above. But if there
are multiple users, it fails.
Edit 3: I have found why above function won't work for multiple users(don't know whether this is correct). The input query set to the function (for one user) is
<QuerySet[{'user':'xyz','id':12,'home':'qwe','mobile':1234},
{'user':'xyz','id':12,'home':'qwe','mobile':4321},]>
Since the two user queries are one after another, they are clubbed. But if there are multiple users, the query set passed to the function is
<QuerySet[{'user':'xyz','id':12,'home':'qwe','mobile':1234},
{'user':'xyz','id':13,'home':'qwe','mobile':4321},
{'user':'abc','id':12,'home':'def','mobile':1233},
{'user':'abc','id':13,'home':'def','mobile':1555},]>
In the above query set, the user with same id are not one after another, so the function failed to get desired output.
You can use the function which you have written, only change is required is need to order the values based on id.
So before taking the values from the objects, order_by('id'), then take your required values and pass it to your function to merge the values.
Eg.
query_set = Mymodel.objects.filter(name='Required').order_by('id')
values = query_set.values('id', 'field1', 'field2')
merged_values = merge_values(values)
As a start, what would be helpful is to organise your models better. You have user information duplicated in your user model. If a you know that a user has multiple mobile numbers, it may be better to create a separate model for that:
class Mobile(models.Model)
number = models.IntegerField()
user = models.ForeignKey(User)
Then, when you are make a query and you want to get all the mobile numbers, you would first get all the users using users = User.objects.all() in your view, then in your template:
{% for user in users %}
{% for mobile in user.mobile_set.all %}
<td>{{ user }}</td>
<td>{{ mobile.number }}</td>
{% endfor %}
{% endfor %}
As a side note, I feel it's better practice to name your models as a singular (User instead of Users)
The following:
{% assign devices = site.data.equipment | where:"department","sound" | group_by:"servicelocation" %}
Is helping me build a list of sound devices from the equipment data list, grouped by location. The file resides in the _includes folder and is entitled equipment_list.html.
When I include this file in a _page, it displays exactly what I need, but I am attempting to take it one step further.
{% assign departmentname = page.name | remove:"equipment_list_" | remove:".md" %} yields the word sound based upon the _page/equipment_list_sound.md file.
{{ departmentname }} = "sound" as expected.
Can I insert this value into the where filter somehow to be able to re-use the equipment list page over and over again for different departments?
where:"department","{{departmentname}}" fails, and any variation I can imagine also fails. Is it possible?
You can use : {% assign devices = site.data.equipment | where:"department", departmentname | group_by:"servicelocation" %}
I have been playing with the paginator function in Bolt CMS, it is easy to use.
Now I need to know if there is a way to implement the pagination in the contenttype yaml.
I think, is it possible something like this?
entries:
name: Entries
singular_name: Entry
fields:
...
taxonomy: [ categories ]
allowpaging: true
I only have found that you need explicity write the allowpaging flag when you fetch the content via setcontent:
{% setcontent entries = "entries/latest/4" allowpaging %}
But what if you want to use the same template for displaying the related taxonomy records? The problem is that you always will be fetching the last 4 entries regardless the taxonomy.
If there's no way to do this, there would be a way to implementing it?
Paging will also be automatically set if you use listing records setting
listing_records: 10
But your template still needs a pager that will use this setting - the listing templates in the theme/base-2014 will work and can be used as an example
The docs have more information https://docs.bolt.cm/contenttypes-and-records#defining-contenttypes
In config.yml set listing_records: xx or the number of records you want to show
then set in your .twix template {% setcontent entries = "entries/latest/xx" allowpaging %}with the same number
and add at the end of .twix file put this code {{ pager('pages') }} to show pages
You can see the official bolt docs for further informations
https://docs.bolt.cm/3.1/templating/content-paging
Hopefully easy question. I have a query set of actions and each action has a time_estimate. I want to get the average time_estimate across all of the actions in the queryset such that this returns the average:
{{ actions.avg_time_estimate }}
I tried
actions.avg_time_estimate = actions.aggregate(Avg('time_estimate'))
but this is assigning a dictionary to actions.avg_time_estimate, meaning I get my average like this:
{{ actions.avg_time_estimate.effort__avg }}
What is the correct way to pull the average into the queryset? I also want to retain the actual list of actions in the query.
Try actions = actions.aggregate(avg_time_estimate=Avg('time_estimate'))