Louis Adekoya
Posts: 0
Joined: Sun Nov 17, 2013 10:51 pm

You did not supply enough values to fill path parameters ( @ 66 : 12 ) -> throw e;

I have the following server code.

code
var responseBody = {},
requestParams = {},
paramKeys = request.keys();
for (var key = 0; key < paramKeys&#46;length; key++) {
requestParams[paramKeys[key]] = request&#46;get(paramKeys[key]);
}

var vCampaign = 1122334;
var url = "https:&#47;&#47;api&#46;sendgrid&#46;com/v3/campaigns/{campaign_id}/schedules/now&quot

var XHRResponse = XHR2&#46;send("POST", url, {
"headers": {
"Content-Type": "application/json",
"Authorization": "Bearer xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
},
"parameters": {
"campaign_id:" vCampaign
},
"body":[{}]
});

Apperyio&#46;response&#46;success(XHRResponse&#46;body, "application/json");
/code

When I run this I get the error: "You did not supply enough values to fill path parameters ( @ 66 : 12 ) - throw e;"

If I replace {campaign_id} in the url with a hard value, it works, but I want to pass it as a parameter. What am I doing wrong.

Serhii Kulibaba
Posts: 150
Joined: Tue Aug 27, 2013 1:47 pm

You did not supply enough values to fill path parameters ( @ 66 : 12 ) -> throw e;

Hello,

Appery.io Server code doesn't parse request URL with request parameters.

So please use URL of the service:
pre"https:&#47;&#47;api&#46;sendgrid&#46;com/v3/campaigns/"+requestParams['campaign_id'] + "/schedules/now";/pre

instead of:
pre"https:&#47;&#47;api&#46;sendgrid&#46;com/v3/campaigns/{campaign_id}/schedules/now";/pre

Louis Adekoya
Posts: 0
Joined: Sun Nov 17, 2013 10:51 pm

You did not supply enough values to fill path parameters ( @ 66 : 12 ) -> throw e;

Perfect Sergiy! It's easy when you know how :-) Many thanks.

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