I need to get salaries so I copied PHP Curl code for “Get the Salary/Wage per Employee per Annum”, it works fine on rapid but it does not works on my side and getting error “Internal Server Error”, please guide.
Thanks Asif
I’m using USA zipcode as a region to query house-hold income. https://rapidapi.com/idealspot-inc-idealspot-inc-default/api/idealspot-geodata/endpoints
I set the JSON strng-encoded location = { “type”:“region”, “regiontype”: “zipcode”, “region_id”: “78702” }
import requests
url = “https://idealspot-geodata.p.rapidapi.com/api/v1/data/insights/household-income/query”
querystring = {“version”:“v2”,“location”:"%7B %22type%22%3A%22region%22%2C %22regiontype%22%3A %22zipcode%22%2C %22region_id%22%3A %2278702%22 %7D"}
headers = { ‘x-rapidapi-host’: “idealspot-geodata.p.rapidapi.com”, ‘x-rapidapi-key’: “YOUR API KEY” }
response = requests.request(“GET”, url, headers=headers, params=querystring)
print(response.text)
Cheers,
아래에 의견을 추가하고 토론에 참여하세요.
I’m using USA zipcode as a region to query house-hold income.
https://rapidapi.com/idealspot-inc-idealspot-inc-default/api/idealspot-geodata/endpoints
I set the JSON strng-encoded location = {
“type”:“region”,
“regiontype”: “zipcode”,
“region_id”: “78702”
}
import requests
url = “https://idealspot-geodata.p.rapidapi.com/api/v1/data/insights/household-income/query”
querystring = {“version”:“v2”,“location”:"%7B %22type%22%3A%22region%22%2C %22regiontype%22%3A %22zipcode%22%2C %22region_id%22%3A %2278702%22 %7D"}
headers = {
‘x-rapidapi-host’: “idealspot-geodata.p.rapidapi.com”,
‘x-rapidapi-key’: “YOUR API KEY”
}
response = requests.request(“GET”, url, headers=headers, params=querystring)
print(response.text)
Cheers,