將C＃中的二維數組轉換為javascript的Json字符串

Question

因此，在調試javascript代碼並最終使其他所有功能正常工作時，我遇到了意外問題。 我從控制器獲得的數據已被拉平。

[HttpPost]
    public JsonResult CalendarCellClasses(DateTime date)
    {
        DateTime firstOfMonth = date.AddDays(-(date.Day - 1));
        DateTime lastOfMonth = firstOfMonth.AddMonths(1).AddDays(-1);

        int additionalDaysBefore = firstOfMonth.DayOfWeek == DayOfWeek.Sunday ? 0 : (int)firstOfMonth.DayOfWeek - 1;
        int additionalDaysAfter = lastOfMonth.DayOfWeek == DayOfWeek.Sunday ? 0 : 7 - (int)lastOfMonth.DayOfWeek;
        int daysInMonth = lastOfMonth.Day;
        int totalDays = additionalDaysBefore + additionalDaysAfter + daysInMonth;
        int numWeeks = totalDays/7;

        DateTime firstDayInSeries = firstOfMonth.AddDays(-additionalDaysBefore);
        DateTime lastDayInSeries = lastOfMonth.AddDays(additionalDaysAfter);

        DateTime current = firstDayInSeries;

        string[,] dates = new string[numWeeks,7];

        for (int week = 0; week < numWeeks; week++)
        {
            for (int day = 0; day < 7; day++)
            {
                dates[week, day] = TrafficData.GetTrafficDate(current).CSSClass;
                current = current.AddDays(1);
            }
        }

        return Json(dates);
    }

如您所見，我有一個字符串[，]，我想將其傳遞給調用此方法的javascript函數。

jQuery(document).ready(function() {
    var calendar = $('#Trafikkalender');
    var date = $('#selectedDate').val();
    var param = { date: date }
    var url = $('#calArrayPostUrl').data('url');
    $.post(url, param, function(data) {
        var body = calendar.find('tbody');

        //var rows = body.getElementsByTagName('tr');
        var rows = body.find('tr');

        for (var i = 0; i < rows.length; i++) {
            //var cols = rows[i].getElementsByTagName('td');
            var cols = $(rows[i]).find('td');
            for (var j = 0; j < cols.length; j++) {
                var col = $(cols[j]);
                col.addClass(data[i][j]);
            }
        }
    });
});

但是根據調試器的數據，它是一個包含35個元素的數組，它們似乎是按一維數組排序的。 返回Json字符串時我做錯什么了嗎？還是2dim數組在javascript中不是問題？

Answer 1

json或多或少是一個文本文件。 我認為，您可以將數組編譯為json。

只需在開始處添加行，然后執行

while allarrayentries
{
 enter new line into json;
}

然后添加最后幾行，以完成json。

Answer 2

使用類模型代替數組來傳遞數據。

public class Dates{
   public DateTime current { get; set; }
   public int number { get; set; }
}

在Javascript中，您可以像這樣訪問它：

for (var i = 0; i < rows.length; i++) {
    var cols = $(rows[i]).find('td');
    for (var j = 0; j < cols.length; j++) {
        var col = $(cols[j]);
        col.addClass(data[i]["current"]);
        col.addClass(data[i]["number"]);
    }
}

Answer 3

使用JSON.NET而不是內置的序列化程序進行序列化。 似乎內置的數組對這些數組的序列化很差（並且JSON.NET也更快）。

這是一個基本的實現：

public class JsonNetResult : JsonResult
{
    public override void ExecuteResult(ControllerContext context)
    {
        if (context == null)
            throw new ArgumentNullException("context");
        if (this.JsonRequestBehavior == JsonRequestBehavior.DenyGet && string.Equals(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase))
            throw new InvalidOperationException("JSON GET is not allowed");

        HttpResponseBase response = context.HttpContext.Response;
        response.ContentType = string.IsNullOrEmpty(this.ContentType) ? "application/json" : this.ContentType;

        if (this.ContentEncoding != null)
            response.ContentEncoding = this.ContentEncoding;
        if (this.Data == null)
            return;

        if (Data != null)
        {
            var writer = new JsonTextWriter(response.Output);

            var serializer = JsonSerializer.Create(new JsonSerializerSettings
            {
                ReferenceLoopHandling = ReferenceLoopHandling.Ignore,
                DateFormatHandling = DateFormatHandling.IsoDateFormat
            });

            serializer.Serialize(writer, Data);

            writer.Flush();
        }
    }
}

並重寫控制器的Json函數：

    protected override JsonResult Json(object data, string contentType, Encoding contentEncoding, JsonRequestBehavior behavior)
    {

        return new JsonNetResult
        {
            Data = data,
            ContentType = contentType,
            ContentEncoding = contentEncoding,
            JsonRequestBehavior = behavior
        };
    }

    protected override JsonResult Json(object data, string contentType, Encoding contentEncoding)
    {
        return new JsonNetResult
        {
            Data = data,
            ContentType = contentType,
            ContentEncoding = contentEncoding
        };
    }

了解更多： http : //www.newtonsoft.com/json