如何在找到单词时断开字符串

Question

我的String对象中有一个数据，如下面的

Scanner Inlist 1,2,3
 Resolved scan set NotEqual to  Non Scan Set
 Area of intrest equal to Total Intrest
 Initial responder Inlist enter values

现在当我读到每一行时，如果我找到了单词（Inlist，NotEqual，Inlist），那么它需要打破这一行并需要读取下一行。

Output would be:

Scanner 
Resolved scan set
Area of intrest
Initial responder

直到现在我所尝试的是

String filterstringobj=promtchild.toString();
StringTokenizer str=new StringTokenizer(filterstringobj,"");
while(str.hasMoreTokens())
{
    String Inlistremove=str.nextToken("InList");
    if(Inlistremove.length()!=0)
    {                       
         System.out.println(Inlistremove);
         if(Inlistremove.equalsIgnoreCase("InList") && 
            Inlistremove.equalsIgnoreCase("NotEqual") && 
            Inlistremove.equalsIgnoreCase("Equal")
           )
         {
            System.out.println(Inlistremove);
         }
    }
}

Answer 1

你的逻辑有一个很大的缺陷：

看着你if我看到了

if(Inlistremove.equalsIgnoreCase("InList")&&Inlistremove.equalsIgnoreCase("NotEqual")&&...

Inlistremove如何等同于"InList" AND 同时等于"NotEqual"？您在寻找 OR 吗？那将是||

Answer 2

使用此行：

StringTokenizer str=new StringTokenizer(filterstringobj," "); 

而不是

StringTokenizer str=new StringTokenizer(filterstringobj,"");

修改
好的，然后观看以下演示代码：

import java.util.StringTokenizer;
class  WordsFromString
{
    public static void main(String st[])
    {
        String data = "Scanner Inlist 1,2,3\n"+
                      "Resolved scan set NotEqual to  Non Scan Set\n"+
                      "Area of intrest equal to Total Intrest\n"+
                      "Initial responder Inlist enter values";
        StringTokenizer tokenizer = new StringTokenizer(data,"\n",true);
        StringBuilder output = new StringBuilder();
        while (tokenizer.hasMoreElements())
        {
            String sLine = tokenizer.nextToken();
            StringTokenizer tokenizerWord = new StringTokenizer(sLine," ",true);
            while (tokenizerWord.hasMoreElements())
            {
                String word = tokenizerWord.nextToken();
                if ("Inlist".equals(word) || "NotEqual".equals(word) || "Inlist".equals(word) || "equal".equals(word))
                {
                    break;
                }
                else
                {
                    output.append(word);
                }
            }
        }
        System.out.println(output.toString());
    }

}

Answer 3

非常灵活，只有一行：

public static String parseLine(String line){
    return line.replaceAll("(?i)(inlist|notequal|equal).*", "");
}

public static void main(String[] a){
    System.out.println(parseLine("Resolved scan set NotEqual to  Non Scan Set"));
    System.out.println(parseLine("Area of intrest equal to Total Intrest"));
    System.out.println(parseLine("Initial responder Inlist enter values"));
}

将打印：

  

已解决的扫描集

     

兴趣区

     

Total Intrest Initial Responder

Answer 4

您不需要使用StringTokenizer。 请注意an old question知道原因吗？

您可以使用正则表达式来匹配不需要的字符串部分，而不是StringTokenizer，而是用空字符串替换它们。

Answer 5

代码中有三个问题：

StringTokenizer str=new StringTokenizer(filterstringobj,""); 

应该是

StringTokenizer str=new StringTokenizer(filterstringobj," "); 

其次是：

if(Inlistremove.equalsIgnoreCase("InList") && 
            Inlistremove.equalsIgnoreCase("NotEqual") && 
            Inlistremove.equalsIgnoreCase("Equal")
           )

应该是：

if(Inlistremove.equalsIgnoreCase("InList") ||
            Inlistremove.equalsIgnoreCase("NotEqual") || 
            Inlistremove.equalsIgnoreCase("Equal")
           )

第三个是，你将如何移动到字符串数组的下一行？您必须稍微更改代码以包含移动到另一个要解析的字符串。

为此，我建议你创建一个函数：

public static void Parse(String s){ 
   String filterstringobj=s;
   StringTokenizer str=new StringTokenizer(filterstringobj," ");
   while(str.hasMoreTokens())
   {
       String Inlistremove=str.nextToken("InList");
       if(Inlistremove.length()!=0)
       {                       
           System.out.println(Inlistremove);
            if(Inlistremove.equalsIgnoreCase("InList") ||
               Inlistremove.equalsIgnoreCase("NotEqual") ||
               Inlistremove.equalsIgnoreCase("Equal")
            )
           {
               System.out.println(Inlistremove);
               return;
           }
      }
   }
}

并在main()方法中执行此操作：

public static void main(String[] args)
{
    String[] array = new String[3];
    array[0] = "Resolved scan set NotEqual to  Non Scan Set";
    array[1] = "Area of intrest equal to Total Intrest";
    array[2] = "Initial responder Inlist enter values";
    for(int i = 0; i < 3; i++) {
          Parse(array[i]);
    }
}