How do I write to console.log() from ajax calls two levels deep?

Question

Please Note: I am new to using ajax and have no knowledge of error handling between different pages.

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I am calling an ajax call from within an ajax call; and would like to obtain a console.log() return from the 2nd page, which is two levels deep. Link to related question -- but to my understanding it is only for one level deep, not two levels.

The return I am getting with this code is a portion of the second page's HTML code, and not my console.log() call from within that writefile page. How can I get this to work?

function update_static_HTML(url,filename,cdn){
    console.log('GET PAGE CONTEN FOR URL = '+url);
    console.log('cdn = '+cdn);  
    $.ajax({
        url: url,
        dataType: "html",
        success: function(data) {
            console.log('success - 1st');
            var save_to_file_url = 'https://www.example.com/writefile?name='+filename+'&cdn='+cdn;
            $.ajax({
                type: 'POST',
                url: save_to_file_url,
                dataType: "html",
                data: {
                    content: data 
                },
                success: function( result ) {
                    console.log('success - 2nd');


                    console.log( result ); // << returns page content, not console.log value

                },
                error:function(error){
                    console.log('ajax Error ' + JSON.stringify(error));
                } 
            });
        },
        error:function(error){
            console.log('ajax Error ' + JSON.stringify(error));
        } 
    }); 
    return false;
} 

UPDATE:

I now understand I need to somehow handle it with some call-back using jqXHR.

I am not clear how to implement the suggested code:

// Assign handlers immediately after making the request,
// and remember the jqxhr object for this request
var jqxhr = $.get( "example.php", function() {
  alert( "success" );
})
  .done(function() {
    alert( "second success" );
  })
  .fail(function() {
    alert( "error" );
  })
  .always(function() {
    alert( "finished" );
  });
 
// Perform other work here ...
 
// Set another completion function for the request above
jqxhr.always(function() {
  alert( "second finished" );
});

What section of it needs to go onto what page?

Answer 1

As temporary workaround I am now writing debug notes to an id withing the returned page, and taking the debug info from there. Note that this is not a answer to my question but a way to get still use console.log to help me debug my code.

    ...
    success: function( result ) {
        var info = $(result).find('#info').html();
        console.log(info);
    },
    ...