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如何从jqueryajax调用向UI反馈代码隐藏过程的当前进度

jquery web-services asp.net c#

Tim Meers · 技术社区 · 16 年前

我有一个窗体将要执行一个失败的长时间运行的进程。安装过程中不仅需要很长时间,而且需要很多小步骤。我尽量避免在我的整个应用程序中使用内置的msajax,但如果这是一种更简单的方法,我会在本页中使用它。

<head runat="server">
    <title></title>
    <script type="text/javascript">
        $(function() {
            $(this).find("#submitForm").click(function() {
                RunCodeBehind();
            });
        });

        function RunCodeBehind() {
            $.ajax({
                error: function(msg) { alert(msg) }, 
                type: "POST",
                url: 'Other.aspx/RunLongRunningProcess',
                data: "{}",
                contentType: "application/json; charset=utf-8",
                dataType: "json",
                success: function(responses) {
                    if (responses.d != "") {
                        //Display process in UI
                        var divEvents = document.getElementById("events");
                        divEvents.innerText = divEvents.innerText + "\n" + data;
                    }
                    else {
                        //no response, update as failed
                        var divEvents = document.getElementById("events");
                        divEvents.innerText = divEvents.innerText + "No response from code behind";
                    }
                }
            });
        }

    </script>
</head>
<body>
    <form id="form1" runat="server">
    <div>
        <a id="submitForm">Start Process</a>
        <br />
        <div id="events"></div>
    </div>
    </form>
</body>

代码隐藏:

[WebMethod]
public static string RunLongRunningProcess()
{
    string returnValue;
    var sqlQuery = "SELECT COUNT(*) FROM Users;"; //generic sql query
    returnValue = ExecuteQuery(sqlQuery);

    //feedback cout total to UI, continue process

    var sqlQueryInsert = @"INSERT INTO Log (UserCount) 
                            SELECT COUNT(*) FROM Users;"; //generic sql insert
    returnValue = ExecuteNonQuery(sqlQueryInsert);
//feedback insert step to UI
    var sqlQuery = "SELECT user + ' - ' + name + ' - ' + favday FROM Users;"; //generic sql query
    returnValue = ExecuteQuery(sqlQuery);
    //feedback selection to UI

    return returnValue;
}

有没有关于如何让它通过一次调用多次反馈到UI的建议?