jquery ajax错误帮助 [英] jquery ajax error help

问题描述

$("#form1").submit(function(e){
    e.preventDefault();
    });
    $("#submit").click(function(e){
    dataString = $("#form1").serialize();
    var new = $("input#new").val();
    dataString = "new=" + new;
    $.ajax({
    type: "POST",
    url: "Servlet",
    data: dataString,
    dataType: "json",
    success: function( data, textStatus, jqXHR) {
    //our country code was correct so we have some information to display
    if(data.success){
    $("#ajaxResponse").html("");
    $("#ajaxResponse").append("Country Code: " + data.countryInfo.code + "
    ");
    }
    //display error message
    else {
    $("#ajaxResponse").html("<div>Category Name is invalid!</div>");
    }
    },
    error: function(jqXHR, textStatus, errorThrown){
    console.log("Something really bad happened " + textStatus);
    $("#ajaxResponse").html(jqXHR.responseText);
    },
    //capture the request before it was sent to server
    beforeSend: function(jqXHR, settings){
    //adding some Dummy data to the request
    settings.data += "&dummyData=whatever";
    //disable the button until we get the response
    $('#submit').attr("disabled", true);
    },
    //this is called after the response or error functions are finsihed
    //so that we can take some action
    complete: function(jqXHR, textStatus){
    //enable the button
    $('#submit').attr("disabled", false);
    }
    });
    });

推荐答案

（ ＃form1）。submit（ function （e）{
e.preventDefault（）;
}）;

("#form1").submit(function(e){ e.preventDefault(); });

（ ＃submit）。 click（ function （e）{
dataString =

("#submit").click(function(e){ dataString =

（ ＃form1）。serialize（）;
var new =

("#form1").serialize(); var new =

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