如何将全部转换为一个字符串？

Question

我需要将此行转换为一个字符串，因为我的方法“DisplayMessage”只接受1个参数，所以我该怎么做？

_userOptions.DisplayMessage("\nFile Generated: " + 
    BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)) + 
    "\nTime Elapsed: {0} minute(s) {1} second(s)", 
    timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10 + "\n"); 

Answer 1

你的字符串表示您要拨打的static Format method on the String class，像这样：

_userOptions.DisplayMessage(string.Format(
    "\nFile Generated: {0}\nTime Elapsed: {1} minute(s) {2} second(s)\n", 
    BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)), 
    timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10)); 

然而，这会给你一个问题，因为你有更多的参数比你在字符串中有占位符。

此外，给出有关使用“\ n”作为新行分隔符的注释，除非您对该特定格式有特定需求（并且您似乎没有这样做，但并不表示您正在撰写数据传到外部系统的东西），最好使用Environment.NewLine，你可以这样使用它（注意，这仍然没有解决你拥有比占位符更多参数的事实：

_userOptions.DisplayMessage(string.Format(
    "{0}File Generated: {1}{0}Time Elapsed: {2} minute(s) {3} second(s){0}", 
    Environment.NewLine, 
    BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)), 
    timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10)); 

Answer 2

它看起来像你的DisplayMessage方法不允许字符串格式。尝试将整个内容（DisplayMessage括号内的所有内容）放入String.Format()方法中。这将使它成为一个字符串，并仍然允许传递多个参数。

Answer 3

试试这个：

string s= String.Format(
    "\nFile Generated: " + 
    BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)) + 
    "\nTime Elapsed: {0} minute(s) {1} second(s) {2} msec(s)\n", 
    timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10); 
_userOptions.DisplayMessage(s); 

Answer 4

下面查找。

string str = String.Format("\n" + "File Generated: " + BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)) + "\n" + "Time Elapsed: " + " {0} minute(s)" + " {1} second(s)", timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10 + "\n"); 

_userOptions.DiplayMessage(str); 

希望这有助于。

Answer 5

而不是{0}和{1}，只需直接使用你的论点：

_userOptions.DisplayMessage("\n" + "File Generated: " + BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)) + "\n" + "Time Elapsed:" + timeSpan.Minutes + "minute(s)" + timeSpan.Seconds + "second(s)" + timeSpan.Milliseconds/10 + "\n"); 

Answer 6

var msg = String.Format("\nFile Generated: {0}\nTime Elapsed: {1} minute(s) {2} second(s)\n", BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)), timeSpan.Minutes, timeSpan.Seconds); 

_userOptions.DisplayMessage(msg); 

这应该这样做...

Answer 7

我猜你想use String.Format。这需要一个字符串，并用参数索引替换{#}。

例子：

String.Format("Hi {0}, welcome to Stack Overflow!", "ale"); 

你可能想看看How should I concatenate strings?

Answer 8

这是在我看来，更优雅：

string message = string.Format("{0}File Generated: {1}{0}Time Elapsed: {2} minute(s) {3} second(s) {4} milliseconds{0}", 
    "\n", BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected)), timeSpan.Minutes, timeSpan.Seconds, timeSpan.Milliseconds/10); 
_userOptions.DisplayMessage(message); 

使用Format不需要在字符串上使用任何+运算符。

Answer 9

我认为，你可以使用StringBuilder

StringBuilder sb = new StringBuilder(); 
     sb.Append("\n"); 
     sb.Append("File Generated: "); 
     sb.Append(BinaryWriter.GetBinaryFileName(filePath, Convert.ToInt32(logSelected))); 
     sb.Append("\n"); 
     sb.Append("Time Elapsed: "); 
     sb.Append(timeSpan.Minutes); 
     sb.Append(" minute(s)"); 
     sb.Append(timeSpan.Seconds); 
     sb.Append(" second(s)"); 
     sb.Append(); 
     _userOptions.DisplayMessage(sb.ToString()); 

但我认为，你有一些bug：你有2个参数，但实际上只有3个