如何在Java中将子字符串与普通空白进行比较

如何在Java中将子字符串与普通空白进行比较,java,string,compare,substring,whitespace,Java,String,Compare,Substring,Whitespace,很抱歉,如果这看起来像一个愚蠢的问题,但我不知道如何将Java中用户输入的子字符串与空白进行比较。在我的程序中,我使用 Input.getString() 这不常见,但Input.class用于接收用户的输入。这是我的节目: public class Project1 { static String E [] = { "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q"

很抱歉,如果这看起来像一个愚蠢的问题,但我不知道如何将Java中用户输入的子字符串与空白进行比较。在我的程序中,我使用

      Input.getString()
这不常见,但Input.class用于接收用户的输入。这是我的节目:

public class Project1
{
static String E [] = { "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", " " };
static String M [] = { ".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--..", ".----", "..---", "...--", "....-", ".....", "-....", "--...", "---..", "----.", "-----", "|" };

public String input[];
public String output[];
static String TranslationType;
static boolean Bool;
static String TextToTranslate;
static int i;
static int SectionOfString;
static StringBuffer emcStringBuffer = new StringBuffer();
static StringBuffer mceStringBuffer = new StringBuffer();
static String emc = "English-MorseCode";
static String mce = "MorseCode-English";
static StringBuffer MorseCodeLetter = new StringBuffer();
static String emcString;
static String mceString;
static String space = " ";

public static void main( String [] args )
{
    if ( E.length != M.length )
    {
        System.out.println( "Error" );
    }
    TranslationType = Input.getString( "Please choose a method of translation (\"English-MorseCode\" or \"MorseCode-English\") " );
    if ( TranslationType.equals( "English-MorseCode" ) )
    {
        Bool = true;
        System.out.println( TranslationType );
    }
    else if( TranslationType.equals( "MorseCode-English" ) )
    {
        Bool = false;
        System.out.println( TranslationType );
    }

    while ( !TranslationType.equals( emc ) && !TranslationType.equals( mce ) )
    {
        System.out.println( "Error: \"" + TranslationType + "\" is not a valid response.");
        TranslationType = Input.getString( "Please choose a method of translation (\"English-MorseCode\" or \"MorseCode-English\") " );
        if ( TranslationType.equals( "English-MorseCode" ) )
        {
            Bool = true;
            System.out.println( TranslationType );
        }
        else if( TranslationType.equals( "MorseCode-English" ) )
        {
            Bool = false;
            System.out.println( TranslationType );
        }
    }
    TextToTranslate = Input.getString( "What would you like to translate?" );
    TextToTranslate = TextToTranslate.toUpperCase();
    System.out.println( TextToTranslate );
    if ( Bool == true )
    {
        for ( SectionOfString = 0; SectionOfString < TextToTranslate.length(); SectionOfString++ )
        {
            for ( i = 0; i < E.length; i++ )
            {
                if ( TextToTranslate.substring( SectionOfString, SectionOfString + 1 ).equals( E[i] ) )
                {
                    emcStringBuffer.append( M[i] + " " );
                }
            }
        }
        emcString = emcStringBuffer.toString();
        System.out.println( emcString );
    }
    else if ( Bool == false )
    {
        for ( SectionOfString = 0; SectionOfString < TextToTranslate.length(); SectionOfString++ )
        {
            if ( TextToTranslate.substring( SectionOfString, SectionOfString + 1 ).equals( space ) )
            {
                for ( i = 0; i < M.length; i++ )
                {
                    if ( MorseCodeLetter.equals( M[i] ) )
                    {
                        mceStringBuffer.append( E[i] );
                        System.out.println( MorseCodeLetter + ";" + mceStringBuffer );
                    }
                }
            }
            else
            {
                if ( TextToTranslate.substring( SectionOfString, SectionOfString + 1 ).equals( M[36] ) )
                {
                    mceStringBuffer.append( E[36] );
                }
                else
                {
                    MorseCodeLetter.append( TextToTranslate.substring( SectionOfString, SectionOfString + 1 ) );
                }
            }
        }
        mceString = mceStringBuffer.toString();
        System.out.println( mceString );
    }
    System.out.println( "Done" );
}
}
然后程序会认为…-。这是一个词。我不确定这是否是翻译结果未打印到终端(或windows上的命令提示符)的原因

此外,我以前尝试过将空间作为静态字符,我只是尝试过将空间作为最终字符,在if构造中使用以下内容,我遇到了相同的问题:

TextToTranslate.charAt( SectionOfString ) == space
我不知道为什么会出现这个问题,我试图找到一个类似的问题,但没有找到,所以如果有人知道如何解决这个问题,我将不胜感激


注意:我在Java编程论坛上问了同样的问题(用户:Indybones33)

我没有看透代码示例的复杂性,但这里有一个简单的空间比较,以防万一这是有帮助的

String str = "x y";
String strSpace = " ";

if (strSpace.equals(str.substring(1,2))) {
    System.out.println("Bingo");
}
这导致摩尔斯电码字母不能分开。例如 如果TextToTranslate的输入为

. .-. | .-. .
..-.|.-。然后程序会认为…-。这是一个词。 我不确定这是否是翻译结果不正确的原因 打印到终端(或windows上的命令提示符)

你知道这个吗?还是你在假设?我认为你完全错了

我认为真正的问题在于:

if ( MorseCodeLetter.equals( M[i] ) 
这应该总是返回
false
,因为
morsecodeler
属于
StringBuffer
类型,
M[i]
属于
String
类型

您只需执行以下操作即可解决此问题:

if ( MorseCodeLetter.toString().equals( M[i] ) )
作为旁注,使用
StringBuffer
不必要地过于复杂。您可以简单地使用字符串连接(
String MorseCodeLetter=“”;MorseCodeLetter+=letter;
)。另外,通常
StringBuffer
用于多线程程序,而
StringBuilder
是首选


旁注2:所有的符号和命名约定都不是标准的Java约定。这意味着其他Java程序员很难轻松阅读您的代码。只有
名称应为大写。变量名(如代码中的
MorseCodeLetter
)应为小写。

有趣的问题,但您的解码过程似乎与维基百科对摩尔斯电码间距的定义不符。。。那是从哪里来的?这相当于七个点吗?建议:在
if
中使用
TextToTranslate.charAt(SectionOfString)=空格
,这要求
空格
最终字符空格=“”改为。感谢您的建议,但修改后出现了相同的问题。谢谢。这解决了问题的一部分,所以现在程序的一部分被打印到终端:MorseCode English…。|。EF F完成
if ( MorseCodeLetter.toString().equals( M[i] ) )