Mandrill API to Google Sheets Script - google-apps-script

Question

I want to call the Mandrill API in order to have tagged template data populate in my google sheet labeled test, but I can't seem to get it working (spent hours I am a beginner). Below is my code
function mandrill() {
var from = "https://mandrillapp.com/api/1.0/tags/list.json";
var key = "*************";
var params = {
"return" :[
"time",
"sent",
"opens",
"clicks",
],
"sheet": "test"
}}
I followed the official docs but I can't seem to get it working. Any help at all would be greatly appreciated.
https://mandrillapp.com/api/docs/tags.JSON.html#method=all-time-series

Answer 1

The Apps Script reference specifies how to use the fetch() method to retrieve your request from Mandrill.
https://developers.google.com/apps-script/reference/url-fetch/url-fetch-app
Please use
UrlFetchApp.fetch(url, options);
You need to JSON.stringify() the JSON request for the Mandrill /urls/list.json and pass it as a payload into the options of the fetch request. According to the documentation you provided, the correct JSON request is:
{
"key": "example key"
}
Subsequently, you can proceed your response and use it to populate the sheet. Please refer to:
https://developers.google.com/apps-script/reference/url-fetch/http-response
Then you need to parse the answer to JavaScript object, you need to parse it before accessing the content with the method getContentText(). You can then access the entries of interest and populate your sheet with it:
In summary, your code to retrieve the response from Mandrill should look like this:
function mandrill() {
var from = "https://mandrillapp.com/api/1.0/tags/list.json";
var params = {
"key": "*************"
};
var payload = JSON.stringify(params);
var options = {
'method': 'post',
'payload': payload,
'contentType' : 'application/json'
};
var response = UrlFetchApp.fetch(url, options);
var data = JSON.parse(response).getContentText());
var ss=SpreadsheetApp.getActiveSpreadsheet().getSheetByName("test");
sheet.getRange("A1").setValue(data['sent'])
}