截取指定长度字符串

 

方法1:

 

//截取指定长度字符串 public static string getStr(string s,int l) { string temp=s; if(Regex.Replace(temp,"[/u4e00-/u9fa5]","zz",RegexOptions.IgnoreCase).Length<=l) { return temp; } for(int i=temp.Length; i>=0; i--) { temp=temp.Substring(0,i); if(Regex.Replace(temp,"[/u4e00-/u9fa5]","zz",RegexOptions.IgnoreCase).Length<=l-3) { return temp+"..."; } } return ""; }

 

方法2:

/// <summary> /// 将指定字符串按指定长度进行剪切， /// </summary> /// <param name= "oldStr "> 需要截断的字符串 </param> /// <param name= "maxLength "> 字符串的最大长度 </param> /// <param name= "endWith "> 超过长度的后缀 </param> /// <returns> 如果超过长度，返回截断后的新字符串加上后缀，否则，返回原字符串 </returns> public static string StringTruncat(string oldStr, int maxLength, string endWith) { if (string.IsNullOrEmpty(oldStr)) // throw new NullReferenceException( "原字符串不能为空 "); return oldStr + endWith; if (maxLength < 1) throw new Exception("返回的字符串长度必须大于[0] "); if (oldStr.Length > maxLength) { string strTmp = oldStr.Substring(0, maxLength); if (string.IsNullOrEmpty(endWith)) return strTmp; else return strTmp + endWith; } return oldStr; }

方法3:

        private static byte [] SourceStr_Bytes;
        private static byte [] CutStr_Bytes = new byte[42];
        private static byte [] CutStr_Bytes1;
        private static int Bytes_Count = 0;
        private static string CutedStr;
        private static Encoding myEncoding = Encoding .GetEncoding("GB2312" );

/// <summary> /// 截取字符长度，以位长截取，并指定截取位长.不足指字位长,则不截取.原样返回. /// </summary> /// <param name="SourceString">被截取源字符串</param> /// <param name="CutLeng">指定要截取位长长度</param> /// <returns>返回截取后字符串</returns> public static string CutStringByByBytes(string SourceString,int CutLeng) { CutStr_Bytes1 = new byte[CutLeng]; SourceStr_Bytes = myEncoding.GetBytes(SourceString); Bytes_Count = SourceStr_Bytes.Length; if (Bytes_Count > CutLeng) { Array.Copy(SourceStr_Bytes, 0, CutStr_Bytes1, 0, CutLeng); CutedStr = myEncoding.GetString(CutStr_Bytes1); CutedStr = CutedStr.Substring(0, CutedStr.Length - 1) + "..."; return CutedStr; } return SourceString; }

 

http://hi.baidu.com/linrao/blog/item/82b9178a921cf11ac8fc7ada.html