Question

我在文本文件“test”中有以下字符串

Table Name.type
Market Drinks.tea

我wana拆分字符串，以便得到以下输出

ObjectName = Table    AttributeName = Name    Attribute Type =type
ObjectName = Market   AttributeName = Drinks  Attribute Type =tea

这是我的代码

            string[] lines = File.ReadAllLines(@"d:\test.txt");
            int i = 0;
            var items = from line in lines
                        where i++ != 0
                        select new{
                            objectName = line.Split(new char[] {' '})[0],
                            attrName = line.Split(new char[]{'.'})[1],
                            attrType = line.Split(new char[] { ' ' })[2]
                        };
            foreach (var item in items)
            {
                Console.WriteLine("ObjectName = {0}, AttributeName = {1}, Attribute Type = {2}",
                    item.objectName, item.attrName, item.attrType);
            }

我越过界限异常了。

PS：文本文件中字符串末尾没有空格我只想测试一个字符！

Answer 1

您不需要new char[] { ... }周围因为String.Split（）使用params

要修复索引越界，select的最后部分应变为：

        attrType = line.Split(' ', '.' )[2]

修改

感谢@Kobi，let只允许你进行一次拆分，当你有很多行和/或列时，这是一个很大的改进。

var items = from line in lines
            where i++ != 0
            let words = line.Split(' ', '.')
            select new
            {
                objectName = words[0],
                attrName = words[1],
                attrType = words[2]
            };

旧答案

您可以对所有3个部分使用相同的分割，使其更容易阅读：

       select new{
             objectName = line.Split(' ', '.' )[0],
             attrName   = line.Split(' ', '.' )[1],
             attrType   = line.Split(' ', '.' )[2]
        };

Answer 2

使用更健壮的正则表达式：

        static void Main()
    {

        const string PATTERN = @"^([\w]+)\s+([\w]+)\.(\w+)";
        const string TEXT = "Table Name.type\r\nMarket Drinks.tea";

        foreach (Match match in Regex.Matches(TEXT, PATTERN, RegexOptions.Multiline))
        {
            Console.WriteLine("ObjectName = {0}   AttributeName = {1}  Attribute Type ={2}",
                match.Groups[1].Value, match.Groups[2].Value, match.Groups[3].Value);
        }

    }

输出：

ObjectName = Table   AttributeName = Name  Attribute Type =type
ObjectName = Market   AttributeName = Drinks  Attribute Type =tea

Answer 3

在分割部分，您应该这样做（前提是您确定输入的格式正确）：

attrName = line.Split(' ')[1].Split('.')[0],
attrType = line.Split(' ')[1].Split('.')[1]

Answer 4

界限在这一行 - attrType = line.Split(new char[] { ' ' })[2]

你的attrType应为= line.Split(new char[] { '.' } )[1];

attrName应为= line.Split(new char[] {' '})[1].Split(new char[] {'.'})[0]

正如Henk Holterman所说，你不需要在分裂中使用新的char []，所以你的行将是 -

attrType = line.Split('.')[1];
attrName = line.Split(' ')[1].Split('.')[0];

Answer 5

这里的回复是所需输出的正确答案

objectName = line.Split(' ')[0],
attrName = line.Split(' ')[1].Split('.')[0],
attrType = line.Split('.')[1]

Answer 6

如果你想以这种方式做，只需使用Regex它更灵活

const string pattern = @"(?<objectName>\w+)\s(?<attrName>\w+)\.(?<attrType>\w+)";

string[] lines = File.ReadAllLines(@"e:\a.txt");
var items = from line in lines
            select new
            {
                objectName = Regex.Match(line, pattern).Groups["objectName"].Value,
                attrName = Regex.Match(line, pattern).Groups["attrName"].Value,
                attrType = Regex.Match(line, pattern).Groups["attrType"].Value
            };

foreach (var item in items.ToList())
{
    Console.WriteLine("ObjectName = {0}, AttributeName = {1}, Attribute Type = {2}",
        item.objectName, item.attrName, item.attrType);
}

从文本文件中拆分字符串

6 个答案:

修改