将文本文件读入字符串变量的最快方法是什么?
我理解它可以通过几种方式完成,比如读取单个字节,然后将它们转换为字符串。我在寻找一种编码最少的方法。
将文本文件读入字符串变量的最快方法是什么?
我理解它可以通过几种方式完成,比如读取单个字节,然后将它们转换为字符串。我在寻找一种编码最少的方法。
当前回答
你可以这样用
public static string ReadFileAndFetchStringInSingleLine(string file)
{
StringBuilder sb;
try
{
sb = new StringBuilder();
using (FileStream fs = File.Open(file, FileMode.Open))
{
using (BufferedStream bs = new BufferedStream(fs))
{
using (StreamReader sr = new StreamReader(bs))
{
string str;
while ((str = sr.ReadLine()) != null)
{
sb.Append(str);
}
}
}
}
return sb.ToString();
}
catch (Exception ex)
{
return "";
}
}
希望这对你有所帮助。
其他回答
用最少的c#代码最快的方法可能就是这个:
string readText = System.IO.File.ReadAllText(path);
string contents = System.IO.File.ReadAllText(path)
这是MSDN文档
我在一个ReadAllText和StreamBuffer之间做了一个比较,一个2Mb的csv,它的差异似乎是相当小的,但ReadAllText似乎从完成函数所需的时间占上风。
string text = File.ReadAllText("Path");一个字符串变量中包含了所有文本。如果你需要每一行单独,你可以使用这个:
string[] lines = File.ReadAllLines("Path");
看一下File.ReadAllText()方法
一些重要备注:
This method opens a file, reads each line of the file, and then adds each line as an element of a string. It then closes the file. A line is defined as a sequence of characters followed by a carriage return ('\r'), a line feed ('\n'), or a carriage return immediately followed by a line feed. The resulting string does not contain the terminating carriage return and/or line feed. This method attempts to automatically detect the encoding of a file based on the presence of byte order marks. Encoding formats UTF-8 and UTF-32 (both big-endian and little-endian) can be detected. Use the ReadAllText(String, Encoding) method overload when reading files that might contain imported text, because unrecognized characters may not be read correctly. The file handle is guaranteed to be closed by this method, even if exceptions are raised