как конвертироватьв JSON в Python с помощью встроенного оператора json.dumps

Question

как мне конвертировать в формат json,

я получаю сообщение об ошибке "не сериализуемо в JSON"

ниже моя программа

from urllib2 import urlopen as uReq
import re
from bs4 import BeautifulSoup, Comment
import requests
import json
my_url='https://uae.dubizzle.com/en/property-for-rent/residential/apartmentflat/?filters=(neighborhoods.ids=123)&page=1'

uClient=uReq(my_url)
page_html= uClient.read()
page_soup=BeautifulSoup(page_html, 'html.parser')
comments = page_soup.findAll(text=lambda text:isinstance(text, Comment))
[comment.extract() for comment in comments]
json_output= page_soup.find_all("script",type="application/ld+json",string=re.compile("SingleFamilyResidence")) #find_all("script", "application/ld+json")
#comments = json_output.findAll(text=lambda text:isinstance(text, Comment))
#[comment.extract() for comment in comments]
#json_output.find_all(text="<script type=""application/ld+json"">").replaceWith("")
#print json_output
jsonD = json.dumps(json_output)
uClient.close()

[{"@context ":" http://schema.org","@type":"SingleFamilyResidence","name":"Spacious 2BHK в аренду на Дамаск-стрит Al Qusais "," url ":" https://dubai.dubizzle.com/property-for-rent/residential/apartmentflat/2018/4/29/spacious-two-bed-room-available-for-rent-i-2/","address":{"@type":"PostalAddress","addressLocality":"Dubai","addressRegion":"Dubai"},"":{"@type":"Product","name":"Spacious 2BHK в аренду на Дамаск-стрит Al Qusais "," url ":" https://dubai.dubizzle.com/property-for-rent/residential/apartmentflat/2018/4/29/spacious-two-bed-room-available-for-rent-i-2/","offers":{"@type":"Offer","price":49000,"priceCurrency":"AED"}},"floorSize":1400,"numberOfRooms":2,"image":" * https://dbzlpvfeeds -a.akamaihd.net / images / user_images/2018/04/29/80881784_CP_photo.jpeg","geo": enj"@type":"GeoCoordinates","latitude":55.3923,"longitude":25.2893}}, {"@context": "http://schema.org","@type":"SingleFamilyResidence","name":"Fully Меблированная 2-комнатная квартира -Al Qusais "," url ":" https://dubai.dubizzle.com/property-for-rent/residential/apartmentflat/2017/10/9/fully-furnished-brand-new-2-bed-room-flat--2/","address":{"@type":"PostalAddress","addressLocality":"Dubai","addressRegion":"Dubai"},"":{"@type":"Product","name":"Fully Меблированная 2-комнатная квартира -Al Qusais "," url ":" https://dubai.dubizzle.com/property-for-rent/residential/apartmentflat/2017/10/9/fully-furnished-brand-new-2-bed-room-flat--2/","offers":{"@type":"Offer","price":70000,"priceCurrency":"AED"}},"floorSize":1400,"numberOfRooms":2,"image":" https://dbzlpvfeeds -a.akamaihd.net / images / user_images / 2018/09 /05 / 84371522_CP_photo.jpeg », "гео": { "@ типа": "GeoCoordinates", "Широта": 55.3959, "долгота": 25,2959}}]

Answer 1

сначала преобразовать bs4.element.ResultSet в строку после этого изменения в json

json_data = json.dumps(str(json_output))

Answer 2

Привет Добавил еще одну оболочку BeautifulSoup и получил ожидаемый json с помощью

, сначала получая текст и используя метод .get_text (), а затем используя json.loads

спасибо, интеллигенция.

from urllib2 import urlopen as uReq
import re
from bs4 import BeautifulSoup, Comment
import requests
import json
my_url='https://uae.dubizzle.com/en/property-for-rent/residential/apartmentflat/?filters=(neighborhoods.ids=123)&page=1'

uClient=uReq(my_url)
page_html= uClient.read()
page_soup=BeautifulSoup(page_html, 'lxml')# 'html.parser')
json_output= BeautifulSoup(str(page_soup.find_all("script",type="application/ld+json",string=re.compile("SingleFamilyResidence"))), 'lxml')#find_all("script", "application/ld+json")
json_text=json_output.get_text()
json_data = json.loads(json_text)
print json_data
uClient.close()