Create a code snippet that is capable of interpreting and processing the data from this

Is there a way I can extract information from this API using javascript?

I tried the following approach but it doesn't seem to be working - 
loadJSON("https://bitcoinfees.earn.com/api/v1/fees/recommended", gotData, 'jsonp');

function gotData(data) {
  println(data);
}

Answer №1

loadJSON is not a built-in JavaScript function (it is part of the p5 library). To achieve similar functionality, you can utilize the fetch function in the following manner:

    fetch("https://bitcoinfees.earn.com/api/v1/fees/recommended") // Use the fetch function with the API URL as the parameter
    .then((response) => response.json()) // Convert the data to JSON format
    .then(function(apiData) {
        // Your code for processing the API data goes here
    console.log(apiData);
        console.log(apiData.fastestFee); // Access any value from the data object like this
    })
    .catch(function(error) {
        // Optional handling of errors if the server responds with an error
    console.log(error)
    });

Answer №2

Using pure javascript:

let request = new XMLHttpRequest();
request.onreadystatechange = function() {
    if (request.readyState == 4 && request.status == 200) {
       // Perform action once response is ready:
       document.getElementById("output").innerHTML = request.responseText;
       alert(JSON.stringify(request.responseText));
      // Display the value of fastestFee  
      let parsedData = JSON.parse(request.responseText);
      alert(parsedData.fastestFee);
    }
};
request.open("GET", "https://bitcoinfees.earn.com/api/v1/fees/recommended", true);
request.send();
<div id="output"> </div> 

Answer №3

Top search result! Understand it. Apply it.

    let address = "https://bitcoinfees.info/api/v1/fees/recommended";
    
    let request = $.ajax(address, {
      success: function(info) {
        console.log(info);
      },
      error: function() {
        console.log('Oops! Something went wrong.');  
      }
    });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>

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