Question

\编辑sb.Append中的代码获取异常（＆＃34;允许前缀：＆＃34;）行 \类型＆＃39; System.ArgumentNullException＆＃39;发生在System.Core.dll中但未在用户代码中处理

public partial class IndexingPolicy：IEquatable

{
  public IndexingPolicy(List<string> allowPrefixes = default(List<string>), List<string> denyPrefixes = default(List<string>), bool? disableIndexing = default(bool?))

    {

        this.AllowPrefixes = allowPrefixes;

        this.DenyPrefixes = denyPrefixes;

        this.DisableIndexing = disableIndexing;
    }


    [DataMember(Name="allowPrefixes", EmitDefaultValue=false)]

    public List<string> AllowPrefixes { get; set; }


    [DataMember(Name="denyPrefixes", EmitDefaultValue=false)]

    public List<string> DenyPrefixes { get; set; }


    [DataMember(Name="disableIndexing", EmitDefaultValue=false)]

    public bool? DisableIndexing { get; set; }


   ").Append(DisableIndexing).Append("\n");

    public override string ToString()
    {
        var sb = new StringBuilder();

        sb.Append("class IndexingPolicy {\n");

        sb.Append(" AllowPrefixes: ").Append(
          string.Join(",", AllowPrefixes.ToList())
        ).Append("\n");


        sb.Append(" DenyPrefixes: ").Append(
          string.Join(",", DenyPrefixes.ToList())
        ).Append("\n");

        sb.Append("  DisableIndexing: ").Append(DisableIndexing).Append("\n");

        sb.Append("}\n");

        return sb.ToString();
    }

Answer 1

问题是默认的ToString（）函数只描述了类（在本例中为png 1 [System.String]`）。要获取字符串，您必须以某种方式覆盖字符串，或将列表内容转换为您要查找的字符串。

我喜欢在这种情况下使用string.Join（）。

System.Collections.Generic.List

...作为一个旁注，我发现使用带有string.Join的数组比使用stringbuilder更“干净”，但效果也不错。

public override string ToString()
{   
  var sb = new StringBuilder();    
  sb.Append("class IndexingPolicy {\n");    
  sb.Append(" AllowPrefixes: ").Append(
    string.Join(",", AllowPrefixes.ToList())
  ).Append("\n");

  sb.Append(" DenyPrefixes: ").Append(
    string.Join(",", DenyPrefixes.ToList())
  ).Append("\n");

  sb.Append(" DisableIndexing: ").Append(
    string.Join(","< DisableIndexing.ToList())
  ).Append("\n");

  sb.Append("}\n");

  return sb.ToString();
}

...如果使用加法使用扩展方法：

public override string ToString()
{   
  var response = new List<string>() {
     "class IndexingPolicy {",
    $" AllowPrefixes: {string.Join(",", AllowPrefixes.ToList())}",   
    $" DenyPrefixes: {string.Join(",", DenyPrefixes.ToList())}",    
    $" DisableIndexing: {string.Join(",", DisableIndexing.ToList())}",
     "}",
     ""

  return string.Join(Environment.NewLine, response);
}

它变得更漂亮：

public static string Join(this IEnumerable @this, string connector)
  => string.Join(connector, @this.ToList());

我进入输出System.Collections.Generic.List而不是数据

1 个答案: