WITH ?page how can i make ajax request code example

Example: make ajax calls with jQuery

// GET Request
    $.ajax({
        url: "example.php?firstParam=Hello&secondParam=World", //you can also pass get parameters
        dataType: 'json',	//dataType you expect in the response from the server
        timeout: 2000
    }).done(function (data, textStatus, jqXHR) {
        //your code here
    }).fail(function (jqXHR, textStatus, errorThrown) {
        console.log("jqXHR:" + jqXHR);
        console.log("TestStatus: " + textStatus);
        console.log("ErrorThrown: " + errorThrown);
    });

//POST Request
    var formData = {name: "John", surname: "Doe", age: "31"}; //Array 
    $.ajax({
        url: "example.php",
        type: "POST", // data type (can be get, post, put, delete)
        data: formData, // data in json format
       	timeout: 2000,	//Is useful ONLY if async=true. If async=false it is useless
        async: false, // enable or disable async (optional, but suggested as false if you need to populate data afterwards)
        success: function (data, textStatus, jqXHR) {
            //your code here
        },
        error: function (jqXHR, textStatus, errorThrown) {
            console.log("jqXHR:" + jqXHR);
            console.log("TestStatus: " + textStatus);
            console.log("ErrorThrown: " + errorThrown);
        }
    });


//Alternatively, the old aproach is
    $.ajax({
        url: "api.php?action=getCategories",
        dataType: 'json',
        timeout: 2000,
        success: function (result, textStatus, jqXHR) {   //jqXHR = jQuery XMLHttpRequest
            /*You could put your code here but this way of doing it is obsolete. Better to use .done()*/
        },
        error: function (jqXHR, textStatus, errorThrown) {
            console.log("jqXHR:" + jqXHR);
            console.log("TestStatus: " + textStatus);
            console.log("ErrorThrown: " + errorThrown);
        }
    });