c# - Visual Studio 2015 在 Text Visualizer 上截断长字符串

Question

更新:经过一些研究，我发现这是 Visual Studio 2015 和 Text Visualizer 的问题，您可以查看 here .

要重现它，打开 QuickWatch (Shift+F9) 并在搜索框中输入 new string(' ', 32769)。之后单击放大镜玻璃，您应该会在字符串中间看到一个“...”...

所以，改变我的问题，有没有办法解决这个问题并使其不被截断，这样我就可以在没有变通方法的情况下进行复制？

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我有这段代码:

        JArray bundlesJson = (JArray)JObject.Parse(System.IO.File.ReadAllText(Server.MapPath("~/itens.json")))["bundles"]; // file "itens.json" can be found at http://pastebin.com/7K15yAVd
        System.Text.RegularExpressions.Regex rgx = new System.Text.RegularExpressions.Regex(@"(?<sNome>.*?)\/(?<sClasse>.*?)\/(?<sDescricao>.*?)(?:\/(?<sExtras>.*?)\n|\n|$)");
        var text = System.IO.File.ReadAllText(Server.MapPath("~/TextFile1.txt")); // TextFile1.txt is too big to put here, so i've uploaded it here: http://pastebin.com/AtxbYPXc

        var matches = rgx.Matches(text);
        var lstItens = new List<Item>();
        var oJson = new JArray();

        foreach (System.Text.RegularExpressions.Match match in matches)
        {
            var item = new Item()
            {
                sClasse = match.Groups["sClasse"].Value.Trim(),
                sDescricao = match.Groups["sDescricao"].Value.Trim(),
                sNome = match.Groups["sNome"].Value.Trim(),
                sExtras = match.Groups["sExtras"].Value.Trim(),
            };
            item.PreencherListaBundles(bundlesJson.ToString());

            lstItens.Add(item);
        }
        var result = JsonConvert.SerializeObject(lstItens, Formatting.Indented);
        var backResult = JsonConvert.DeserializeObject<List<Item>>(result);

lstItens 列表正确包含所有项目(501 项)，但 result 字符串在搜索 "Nome"时仅返回 48 个对象:，这是必填字段。为什么会这样？

为了举例说明错误，如果您搜索 "Nome":"Fiddlehead Fern，请在 result var 中查找项目 lstItens[166] " 你可以看到该项目不存在......

奇怪的是 backResult.Count 将显示 501 个结果，并且似乎包含所有项目，但在 result var 生成的 json 中进行简单搜索必填字段“Nome”将产生 48 个结果，如图所示:

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Item.cs:

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public class Item
{
    [JsonProperty(PropertyName = "Nome")]
    public string sNome { get; set; }

    [JsonProperty(PropertyName = "Classe")]
    public string sClasse { get; set; }

    [JsonProperty(PropertyName = "Descricao")]
    public string sDescricao { get; set; }

    [JsonProperty(PropertyName = "Extras")]
    public string sExtras { get; set; }

    [JsonProperty(PropertyName = "Bundles")]
    public List<Bundle> lstBundles { get; set; }

    public void PreencherListaBundles(string jsonBundles)
    {
        List<Bundle> lstBundle = JsonConvert.DeserializeObject<List<Bundle>>(jsonBundles.ToString());

        this.lstBundles = new List<Bundle>();

        lstBundle.ForEach(x =>
        {
            if (!lstBundles.Select(y => y.sNome).Contains(x.sNome) && x.lstItens.Select(y => y.sNome).Contains(sNome))
            {
                lstBundles.Add(new Bundle() { sLocal = x.sLocal, sNome = x.sNome, sRecompensa = x.sRecompensa, lstItens = x.lstItens });
            }
        });
    }
}

---

Bundle.cs

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public class Bundle
{
    [JsonProperty(PropertyName = "bundle")]
    public string sNome { get; set; }

    [JsonProperty(PropertyName = "location")]
    public string sLocal { get; set; }

    [JsonProperty(PropertyName = "reward")]
    public string sRecompensa { get; set; }

    [JsonProperty(PropertyName = "items")]
    public List<BundleItem> lstItens { get; set; }

    public class BundleItem
    {
        [JsonProperty(PropertyName = "name")]
        public string sNome { get; set; }

        [JsonProperty(PropertyName = "description")]
        public string sDescricao { get; set; }

        [JsonProperty(PropertyName = "quantity")]
        public int nQuantidade { get; set; }

        [JsonProperty(PropertyName = "quality")]
        public string sQualidade { get; set; }
    }
}

编辑: 看起来某些机器上没有出现该错误，就像您在 Gerard Sexton 的回答中看到的那样，但是当我运行他运行的相同代码时，我仍然得到 48 个结果。可以在此讨论中找到更多详细信息:https://chat.stackoverflow.com/rooms/106307/discussion-between-gerard-sexton-and-gabriel-duarte

Answer 1

试试这个:

var result = JsonConvert.SerializeObject(lstItens, Formatting.Indented);
//Testing deserialization
var backResult = JsonConvert.DeserializeObject<List<Item>>(result);

结果变量是一个包含格式正确的字符串的字符串。