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Answers for "observe an ajax call in jquery"

Javascript
5

ajax call

$.ajax({
        url: "Url",
        dataType: "json",
        type: "Post",
        async: true,
        data: {"Key":value,"Key2":value2},
        success: function (data) {
                
        },
        error: function (xhr, exception, thrownError) {
            var msg = "";
            if (xhr.status === 0) {
                msg = "Not connect.\n Verify Network.";
            } else if (xhr.status == 404) {
                msg = "Requested page not found. [404]";
            } else if (xhr.status == 500) {
                msg = "Internal Server Error [500].";
            } else if (exception === "parsererror") {
                msg = "Requested JSON parse failed.";
            } else if (exception === "timeout") {
                msg = "Time out error.";
            } else if (exception === "abort") {
                msg = "Ajax request aborted.";
            } else {
                msg = "Error:" + xhr.status + " " + xhr.responseText;
            }
            if (callbackError) {
                callbackError(msg);
            }
           
        }
    });
Posted by: Guest on January-06-2021
1

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!" );
  });
Posted by: Guest on September-21-2021

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