Question

//警报没有处理错误消息，我想只插入少数4个问题，之后它应该不起作用。

＆＃13;

success: function(msg, String, jqXHR) {
    window.location = 'home.html';


    $("#result").html(msg, String, jqXHR)
    alert("Data Uploaded: ");

    if (msg.success == 'Error') {

      alert("Please close the previous issue");

      window.location = 'home.html';
    }

＆＃13;

// here（msg.success ==＆＃39;错误＆＃39;）无法正常工作我想显示此消息 - ＆＃34;请关闭上一期＆＃34;对此。

Answer 1

希望这适合你。

$(document).ready(function(){
$('#your_form').submit(function(event) {
     event.preventDefault();
     dataString = $("#your_form").serialize();
     $.ajax({
        type: "post",
        url: "post.php",
        dataType:"json",
        data: dataString,
        success: function (response) {
            if(response.status === "success") {
                // do something with response.message or whatever other data on success
                alert("Data Uploaded: ");
                window.location='home.html';
            } else if(response.status === "error") {
                alert("Please close the previous issue");

                 window.location='home.html';
            }
        }
    })
    return false;
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

警报不能处理错误消息

1 个答案: