I have a model like this:
class MyClass(models.Model):
typea = models.CharField(max_length=128)
typeb = models.CharField(max_length=128)
If for example, the resulting json from an API is like this:
{
"count": 75,
"results": [
{
"typea": "This tipe",
"typeb": "A B type",
"jetsons": [],
"data": [
"https://myurl.com/api/data/2/",
],
"created": "2014-12-15T12:31:42.547000Z",
"edited": "2017-04-19T10:56:06.685592Z",
},
I need to parse this result and save typea and typeb into the database, I'm kind of confused on how to do this.
I mean, there is the JSONField on Django, but I don't think that will work for me, since I need to save some specific nested string of the json dict.
Any example or idea on how to achieve this?
I mean, my confusion is on how to parse this and "extract" the data I need for my specific fields.
Thank You
You can always do an import json and use json.load(json_file_handle) to create a dictionary and extract the values you need. All you need is to open the .json file (you can use with open("file.json", "r") as json_file_handle) and load the data.
Related
In purpose to delete multiple data using function
Product.objects.in_bulk([pk1,pk2,pk3,...]).delete()
I'm trying to grab pk value from json as list for in_bulk function param.
my json :
[
{
"Pk": 1,
"Product": "testing"
},
{
"Pk": 2,
"Product": "testing"
}
]
But i don't know how to achieve this in django (i'm new in django from .NET backgroud). Should i iterate each object in json in order to get each pk or there's smartest way to get list pk value ?
You delete this with:
Product.objects.filter(pk__in=list_of_pks).delete()
So if you have a JSON blob, you can work with:
from json import loads as jloads
my_json_blob = '[ {"Pk": 1, …}, … ]'
data = jloads(my_json_blob)
Product.objects.filter(pk__in=[d['Pk'] for d in data]).delete()
In my django application with database engine djongo, I'm trying to return a JSON response by retrieving from my database. But, I'm receiving JSON array instead of JSON object. Currently, there is only one record in my database. Please see the code below.
model.py
class bloodDonors(models.Model):
location=models.CharField(max_length=20)
name=models.CharField(max_length=20)
phone=models.IntegerField()
address=models.TextField()
bloodGroup=models.CharField(max_length=5)
type=models.CharField(max_length=20)
def __str__(self):
return self.name
views.py
class bloodDonersView(viewsets.ModelViewSet):
queryset = bloodDonors.objects.all()
serializer_class = bloodDonorsSerializer
JSON Reponse:
[
{
"id": 3,
"location": "Delhi",
"name": "Tony",
"phone": 888,
"address": "South street",
"bloodGroup": "B+",
"type": "blood-donation"
}
]
But, actually I needed the response as given below:
{
"id": 3,
"location": "Delhi",
"name": "Tony",
"phone": 888,
"address": "South street",
"bloodGroup": "B+",
"type": "blood-donation"
}
It sounds like you are using the 'list' endpoint that retrieves all objects matching the queryset, but you want to be using the detail retrieve endpoint, where you just retrieve one object. The good news is that because you are using a Model viewset, you get a detail retrieve GET endpoint for free. All you have to do is pass in the pk of the object you want as part of the url and it will return only the single object you requested.
So if your url is currently:
/api/v1/blood-donors
you'd want to use the url:
/api/v1/blood-donors/3
to get just the single object you want (with pk=3). The GET endpoint with no /{PK} at the end is always going to return a list, even if there's only one object in your database, and if you add the pk to the url you will always only get one object back, the object that matches the pk you pass in.
Update to answer your question in the comment. The new method would look something like this:
class bloodDonersView(viewsets.ModelViewSet):
queryset = bloodDonors.objects.all()
serializer_class = bloodDonorsSerializer
#action(detail=False, methods=['get',], url_path='first')
def get_first_object(self, request):
object = bloodDonors.objects.first()
serializer = bloodDonorsSerializer(object, many=False)
return response.Response(serializer.data)
the url you would use is
api/v1/blood-donors/first
and you need to check the documentation about where to import response and the action decorator.
I am trying to dynamically build some json based on data I retrieve from a database. Up until the opening '[' is the "root" I guess you could say. The next parts with name and value are dynamic and will be based on the number of results I get from the db. I query the db and then the idea was to iterate through the result adding to the json. Can I use jsonBuilder for the root section and then loop with jsonSlurper to add each additional section? Most of the examples I have seen deal with a root and then a one time "slurp" and then joining the two so wasn't sure if I should try a different method for looping and appending multiple sections.
Any tips would be greatly appreciated. Thanks.
{
"hostname": "$hostname",
"path": "$path",
"extPath": "$extPath",
"appName": "$appName",
"update": {"parameter": [
{
"name": "$name",
"value": "$value"
},
{
"name": "$name",
"value": "$value"
}
]}
}
EDIT: So what I ended up doing was just using StringBuilder to create the initial block and then append the subsequent sections. Maybe not the most graceful way to do it, but it works!
//Create the json string
StringBuilder json = new StringBuilder("""{
"hostname": "$hostname",
"path": "$path",
"extPath": "$extPath",
"appName": "$appName",
"update": {"parameter": ["""
)
//Append
sql.eachRow("""<query>""",
{ params ->
json.append("""{ "name": "params.name", "value": "params.value" },""");
}
)
//Add closing json tags
json.append("""]}}""")
If I got your explanation correctly and if the data is not very big (it can live in memory), I'd build a Map object (which is very easy to work with in groovy) and convert it to JSON afterwards. Something like this:
def data = [
hostname: hostname,
path: path,
extPath: extPath,
appName: appName,
update: [parameter: []]
]
sql.eachRow(sqlStr) { row ->
data.update.parameter << [name: row.name, value: row.value]
}
println JsonOutput.toJson(data)
If you're using Grails and Groovy you can utilize grails.converters.JSON.
First, define a JSON named config:
JSON.createNamedConfig('person') {
it.registerObjectMarshaller(Person) {
Person person ->
def output = [:]
output['name'] = person.name
output['address'] = person.address
output['age'] = person.age
output
}
}
This will result in a statically defined named configuration for the Object type of person. Now, you can simply call:
JSON.use('person') {
Person.findAll() as JSON
}
This will return every person in the database with their name, address and age all in one JSON request. I don't know if you're using grails as well in this situation though, for pure Groovy go with another answer here.
i have a big json object with a list of "tickets". schema looks like below
{
"Artist": "Artist1",
"Tickets": [
{
"Id": 1,
"Attr2Array": [
{
"Att41": 1,
"Att42": "A",
"Att43": null
},
{
"Att41": 1,
"Att42": "A",
"Att43": null
},
],
.
.
.
(more properties)
"Price": "20",
"Description": "I m a ticket"
},
{
"Id": 4,
"Attr2Array": [
{
"Att41": 1,
"Att42": "A",
"Att43": null
},
{
"Att41": 1,
"Att42": "A",
"Att43": null
},
],
.
.
.
.
(more properties)
"Price": "30",
"Description": "I m a ticket"
}
]
}
each item in the list has around 25-30 properties (some simple types, and others complex array as nested objects)
i have to read the object from an api endpoint and extract only "ID" and "Description" but they need to be sorted by "Price" which is an int for example
In what order shall i proceed with this data manipulation
Shall i use the json object, deserialised it into another object with just those 2 properties (which i need) and THEN perform sort "asc" on the "Price"?
Please note that after i have the sorted list i will have to convert it back to a json list because the front end consumes a json after all.
What i dont like about this approach is the cycle of serialisation and deserialisation that happens
or
I perform a sort on the json object first (using for example a binary/bubble sort) and then use the object to create a strongly typed (deserialised) object with just those 2 properties and then serialise it back to pass to the front end
I dont know how performant the bubble sort will be and if at all i will get any gain in performance for large chunks of data processing.
I also need to keep in mind that this implementation can take into account other properties like "availabilitydate" because at a later date, this front end could add one more filter like "availabilitdate" asc
any help is much appreciated
thanks
You can deserialize your JSON string (or file) using the Microsoft System.Web.Extensions and JavaScriptSerializer.DeserializeObject.
First, you must have classes associated to your JSON. To create classes, select your JSON sample data and, in Visual Studio, go to Edit / Paste Special / Paste JSON As Classes.
Next, use this sample to deserialize a JSON string to typed objects, and to sort all Tickets by Price property using Linq.
String json = System.IO.File.ReadAllText(#"C:\Data.json");
var root = new System.Web.Script.Serialization.JavaScriptSerializer().Deserialize<Rootobject>(json);
var sortedTickets = root.Tickets.OrderBy(t => t.Price);
I have a simple custom data structure which I use to map the results from the database:
case class Filter(id: Int, table: String, name: String, Type: String, structure: String)
The resulting object type is List[Filter] and if converted to JSON, it should look something like this:
[
{
"id": 1,
"table": "table1",
"name": "name1",
"Type": "type1",
"structure": "structure1"
},
{
"id": 2,
"table": "table2",
"name": "name2",
"Type": "type2",
"structure": "structure2"
}
]
Now when I try to serialize my object into JSON
val result: String = Json.toJson(filters)
I am getting something like
No Json deserializer found for type List[Filter]. Try to implement an implicit Writes or Format for this type.
How do I solve this seemingly simple problem without writing some ridiculous amount of boilerplate?
My stack is Play 2.2.1, Scala 2.10.3, Java 8 64bit
Short answer:
Just add:
implicit val filterWrites = Json.writes[Filter]
Longer answer:
If you look at the definition of Json.toJson, you will see that its complete signature is:
def toJson[T](o: T)(implicit tjs: Writes[T]): JsValue = tjs.writes(o)
Writes[T] knows how to take a T and transform it to a JsValue. You will need to have an implicit Writes[Filter] around that knows how to serialize your Filter instance. The good news is that Play's JSON library comes with a macro that can instantiate those Writes[_] for you, so you don't have to write boring code that transforms your case class's fields into JSON values. To invoke this macro and have its value picked up by implicit search add the line above to your scope.