Question

我有3个字符串。第一组字符串是：

"1.0536"  
"2.1"  
"2"

第二个是：

"Round"  
"Square"  
"Hex"

最后一个是：

"6061-T6"  
"T351"  
"ASF.3.4.5"

我需要将三个字符串组合在一起，每个字符串之间的间距相同。我不能使用\ t进行制表，因为在我组合字符串后，我将它们发送到Access数据库。

当我将字符串组合起来时，它们看起来像：

"1.0536 Round 6061-T6"  
"2.1 Square T351"  
"2 Hex ASF.3.4.5"

我真的希望它们在每个字符串之间具有相同的间距：

“1.0536回合6061-T6”
“2.1坊T351”
“2 Hex ASF.3.4.5”

如何使用C＃执行此操作？

Answer 1

您可以使用advanced features的string.Format：

string.Format("{0,-10}{1,-10}{2}", ...)

您可以通过撰写str.PadRight(10)

来做同样的事情

Answer 2

如果您知道每列的最大长度，请执行以下操作：

String result = String.Format("{0} {1} {2}", strCol1.PadRight(10), strCol2.PadRight(9), strCol3.PadRight(9));

Answer 3

为了让生活更轻松，实用方法：

用法

var data = new[] {
    new[] { "ID", "NAME", "DESCRIPTION" },
    new[] { "1", "Frank Foo", "lorem ipsum sic dolor" },
    new[] { "2", "Brandon Bar", "amet forthrightly" },
    new[] { "3", "B. Baz", "Yeehah!" }
};

var tabbedData = EvenColumns(20, data);
var tabbedData2 = string.Join("\n", EvenColumns(20, false, data)); // alternate line separator, alignment

结果

ID                  NAME                DESCRIPTION
1                   Frank Foo           lorem ipsum sic dolor
2                   Brandon Bar         amet forthrightly
3                   B. Baz              Yeehah!

ID                NAME         DESCRIPTION
1           Frank Foolorem ipsum sic dolor
2         Brandon Bar   amet forthrightly
3              B. Baz             Yeehah!

代码

public string EvenColumns(int desiredWidth, IEnumerable<IEnumerable<string>> lists) {
    return string.Join(Environment.NewLine, EvenColumns(desiredWidth, true, lists));
}

public IEnumerable<string> EvenColumns(int desiredWidth, bool rightOrLeft, IEnumerable<IEnumerable<string>> lists) {
    return lists.Select(o => EvenColumns(desiredWidth, rightOrLeft, o.ToArray()));
}

public string EvenColumns(int desiredWidth, bool rightOrLeftAlignment, string[] list, bool fitToItems = false) {
    // right alignment needs "-X" 'width' vs left alignment which is just "X" in the `string.Format` format string
    int columnWidth = (rightOrLeftAlignment ? -1 : 1) *
                        // fit to actual items? this could screw up "evenness" if
                        // one column is longer than the others
                        // and you use this with multiple rows
                        (fitToItems
                            ? Math.Max(desiredWidth, list.Select(o => o.Length).Max())
                            : desiredWidth
                        );

    // make columns for all but the "last" (or first) one
    string format = string.Concat(Enumerable.Range(rightOrLeftAlignment ? 0 : 1, list.Length-1).Select( i => string.Format("{{{0},{1}}}", i, columnWidth) ));

    // then add the "last" one without Alignment
    if(rightOrLeftAlignment) {
        format += "{" + (list.Length-1) + "}";
    }
    else {
        format = "{0}" + format;
    }

    return string.Format(format, list);
}

特定问题

// for fun, assume multidimensional declaration rather than jagged
var data = new[,] {
    { "1.0536", "2.1", "2" },
    { "Round", "Square", "Hex" },
    { "6061-T6", "T351", "ASF.3.4.5" },
};

var tabbedData = EvenColumns(20, Transpose(ToJaggedArray(data)));

Transpose：

public T[][] Transpose<T>(T[][] original) {
    // flip dimensions
    var h = original.Length;
    var w = original[0].Length;

    var result = new T[h][];
    for (var r = 0; r < h; r++) {
        result[r] = new T[w];
        for (var c = 0; c < w; c++)
        {
            result[r][c] = original[c][r];
        }
    }
    return result;
}

多维数组（source）：

public T[][] ToJaggedArray<T>(T[,] multiArray) {
    // via https://stackoverflow.com/questions/3010219/jagged-arrays-multidimensional-arrays-conversion-in-asp-net
    var h = multiArray.GetLength(0);
    var w = multiArray.GetLength(1);

    var result = new T[h][];
    for (var r = 0; r < h; r++) {
        result[r] = new T[w];
        for (var c = 0; c < w; c++) {
            result[r][c] = multiArray[r, c];
        }
    }
    return result;
}

Answer 4

我知道这个问题早已得到解答，但是从C＃6.0开始还有一种新的方法

string[] one = new string[] { "1.0536", "2.1", "2" };
string[] two = new string[] { "Round", "Square", "Hex" };
string[] three = new string[] { "1.0536 Round 6061-T6", "2.1 Square T351", "2 Hex ASF.3.4.5" };

for (int i = 0; i < 3; i++) Console.WriteLine($"{one[i],-10}{two[i],-10}{three[i],-10}");

$"{one[i],-10}{two[i],-10}{three[i],-10}"是string.format的新替代品。我发现它在我的许多项目中非常有用。以下是c＃6.0中有关字符串插值的更多信息的链接： https://docs.microsoft.com/en-us/dotnet/csharp/tutorials/string-interpolation

Answer 5

使用String.Format("{0,10}", myString)
其中10是你想要的字符数

Answer 6

为了更加动态地执行此操作，您可以执行以下操作:(提前编写硬编码！）

    int padding = 3;
    int maxCol0width = "Hello World!".Length;
    int maxCol1width = "90.345".Length;
    int maxCol2width = "value".Length;

    string fmt0 = "{0,-" + (maxCol0width + padding) + "}";
    string fmt1 = "{1,-" + (maxCol1width + padding) + "}";
    string fmt2 = "{2,-" + (maxCol2width + padding) + "}";

    string fmt = fmt0 + fmt1 + fmt2;

    Console.WriteLine(fmt, "Hello World!", 90.345, "value");
    Console.WriteLine(fmt, "Hi!", 1.2, "X");
    Console.WriteLine(fmt, "Another", 100, "ZZZ");

您当然需要通过循环每个列的值来确定您的最大字宽。此外，格式字符串的创建可以显着清理和缩短。

Answer 7

另请注意，您需要使用非比例字体进行显示，否则您的列仍然无法正常排列。你在哪里显示这些数据？获得表格输出可能有更好的方法。

格式化值之间具有相同间距的C＃字符串

7 个答案:

用法

结果

代码

特定问题