Question

我在尝试将Enum转换为jQGrid的JSON字符串时遇到问题。我之前使用的格式（进行手动转换）是：

{{0: '-', 1: 'Active', 2: 'Deactive', 3: 'Pending'}}

    public static string GetStatuses(bool addDefault = false)
    {
        var statusesEnum = Enum.GetValues(typeof(StatusEnum));
        string statuses = "{value: {0: '-', ";

        foreach (StatusEnum status in statusesEnum)
            statuses += String.Format("{0}: '{1}', ", (byte)status, Enum.GetName(typeof(StatusEnum), status));

        return statuses.Substring(0, statuses.Length - 2) + "}}";
    }

所以我需要避免使用这种方法，因为我认为这不是最好的方法，我想使用JSON.NET库对其进行序列化。所以我写了这个：

public class StatusJSON
{
    public byte ID { get; set; }
    public string Name { get; set; }

    public StatusJSON() { }

    public StatusJSON(byte id, string name)
    {
        ID = id;
        Name = name;
    }
}

public class JSONUtils
{
    /// <summary>
    /// Get all the posible statuses of selected <paramref name="type"/> in JSON
    /// </summary>
    /// <param name="type">Type of the status</param>
    /// <param name="addDefault">Check if add a default / NULL status</param>
    /// <returns>A string JSON with the statuses</returns>
    public static string GetStatuses(Type type, bool addDefault = false)
    {
        var statusesEnum = Enum.GetValues(type);
        List<StatusJSON> statuses = new List<StatusJSON>();

        if (addDefault)
            statuses.Add(new StatusJSON(0, "-"));

        foreach (var statusEnum in statusesEnum)
            statuses.Add(new StatusJSON((byte)statusEnum, Enum.GetName(type, statusEnum)));

        return JsonConvert.SerializeObject(statuses);
    }
}

您可以将其用作：string statuses = JSONUtils.GetStatuses(typeof(StatusEnum), addDefault);。问题是这会返回一个字符串，如：

[{"ID":0,"Name":"-"},{"ID":1,"Name":"Active"},{"ID":2,"Name":"Deactive"},{"ID":3,"Name":"Pending"}]

库中是否有任何方法可以获得类似我需要的字符串？感谢

Answer 1

我最终做的是重复使用我的旧代码。所以现在我有了这个：

public class Statutes
{
    public byte ID { get; set; }
    public string Name { get; set; }

    public Statutes() { }

    public Statutes(byte id, string name)
    {
        ID = id;
        Name = name;
    }

    /// <summary>
    /// Get all the posible statuses of selected <paramref name="type"/>
    /// </summary>
    /// <param name="type">Type of the status</param>
    /// <param name="addDefault">Check if add a default / NULL status</param>
    /// <returns>A list with the statuses</returns>
    public static List<Statutes> SelectAll(Type type, bool addDefault = false)
    {
        var statusesEnum = Enum.GetValues(type);
        List<Statutes> statuses = new List<Statutes>();

        if (addDefault)
            statuses.Add(new Statutes(0, "-"));

        foreach (var statusEnum in statusesEnum)
            statuses.Add(new Statutes((byte)statusEnum, Enum.GetName(type, statusEnum)));

        return statuses;
    }
}

public class JSONUtils
{
    /// <summary>
    /// Get all the posible statuses of selected <paramref name="type"/> in JSON
    /// </summary>
    /// <param name="type">Type of the status</param>
    /// <param name="addDefault">Check if add a default / NULL status</param>
    /// <returns>A string JSON for jQGrid with the statuses</returns>
    public static string GetStatusesJQGrid(Type type, bool addDefault = false)
    {
        var statuses = Statutes.SelectAll(type, addDefault);
        string result = "{value: {";

        foreach (Statutes status in statuses)
            result += String.Format("{0}: '{1}', ", status.ID, status.Name);

        return result.Substring(0, result.Length - 2) + "}}";
    }
}

您可以将其用作：string statuses = JSONUtils.GetStatusesJQGrid(typeof(StatusEnum), true);

在我找到使用JSON.NET的更好方法之前，我认为这是一个很好的代码片段，可以为使用jQGrid的人重复使用。这对选择选项有效：

colModel: {name: 'status_id', label: 'Status', edittype: 'select', sortable: true, search: true, stype:'select', editoptions: " + statuses + @", searchoptions: {sopt: ['eq', 'ne']}}

使用JSON.NET枚举jQGrid的JSON对象

1 个答案: