Hi -
Im using the Reed jobs API and am making the request:
http://www.reed.co.uk/api/1.0/search?...
I have put my API key as a request parameter called username & set it as a header.
I get the below error, can anyone advise?
"status":"502 ",
"url":"http://www.reed.co.uk/api/1.0/search?...",
"response":"org.apache.http.client.ClientProtocolException"
Hello Kieran,
As far as I can get from http://www.reed.co.uk/api
"You will need to include your api key for all requests in a basic authentication http header as the username, leaving the password empty."
Please look here to find the way how it can be done:
https://getsatisfaction.com/apperyio/...
I have looked on the link and am still a little confused, I altered my param to:
NAME - Authorization
VALUE - username b164f851-5f34-46eb-ab0a-d0fda45acb84 =
Can you let me know what I am doing wrong?
oh and i have password as a separate param as empty
Im not sure if I am supposed to comment or put a new message - but here is both
I have looked on the link and am still a little confused, I altered my param to:
NAME - Authorization
VALUE - username b164f851-5f34-46eb-ab0a-d0fda45acb84 =
NAME - Password
VALUE -
Can you let me know what I am doing wrong?
Hello Kieran,
You need to make settings as in this response:
https://getsatisfaction.com/apperyio/...
I have now done this, and still the same error,
NAME Authorization
VALUE Basic b164f851-5f34-46eb-ab0a-d0fda45acb84=
I have tried this previously also, but with no luck.
Can you help?
Kieran,
Please also look here:
http://devcenter.appery.io/documentat...
You should copy your api key, then go to http://www.base64encode.org/
paste and add colon at the end, press encode. The resulting value you should use as a value for the Header Authorization like this
Basic ZDA5ZjhkZTMtYmNmZi00ZTM2LTlmMWYtXXXXXXliZTY5YWJhOg==