Answers for "jquery ajax html response"

ajax exmaple\

$.ajax({
    url:'your url',
    type: 'POST',  // http method
    data: { myData: 'This is my data.' },  // data to submit
    success: function (data, status, xhr) { // after success your get data
        $('p').append('status: ' + status + ', data: ' + data);
    },
    error: function (jqXhr, textStatus, errorMessage) { // if any error come then 
            $('p').append('Error' + errorMessage);
    }
});

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            /*You could put your code here but this way of doing it is obsolete.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);
        }
    });

jquery ajax request

// Using the core $.ajax() method
$.ajax({
 
    // The URL for the request
    url: "post.php",
 
    // The data to send (will be converted to a query string)
    data: {
        id: 123
    },
 
    // Whether this is a POST or GET request
    type: "GET",
 
    // The type of data we expect back
    dataType : "json",
})
  // Code to run if the request succeeds (is done);
  // The response is passed to the function
  .done(function( json ) {
     $( "<h1>" ).text( json.title ).appendTo( "body" );
     $( "<div class=\"content\">").html( json.html ).appendTo( "body" );
  })
  // Code to run if the request fails; the raw request and
  // status codes are passed to the function
  .fail(function( xhr, status, errorThrown ) {
    alert( "Sorry, there was a problem!" );
    console.log( "Error: " + errorThrown );
    console.log( "Status: " + status );
    console.dir( xhr );
  })
  // Code to run regardless of success or failure;
  .always(function( xhr, status ) {
    alert( "The request is complete!" );
  });

js ajax receive html

$.ajax({
  	type: 'POST',
  	url: "<Your URL>",
	contentType: 'application/json; charset=utf-8'
  	// Set your dataType to either 'html' or 'text'.
	// Keep in mind: dataType is for receiving,
	// contentType is for sending
  	dataType: 'html',
  	data: { example: 1, id: "0x100"},
	// Note: the data above is used in sending,
	// data below is a variables that stores received data
  	success: function (data){
     	// Suppose you have an html element, where you want to append 
     	// the response:
     	$('#<Your html element id>').html(data);
  		// .html(data) overwrites existing data
		// Use .append(data) to add response without overwriting!
  	}
});