entityMap|blocks|key|89geq|text|如果您可以保证这些值将按上述顺序排列，那么应该可以使用下面这样的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|50ad6|public+static+void+main(String[]+args)+{
++++String+my+=+"StackOverflow+PF+Apple+FT+Laptop+HW";
++++String[]+words+=+my.split("+");
++++for+(i+=+0;+i+<+words.length;+i%2B%2B)+{
++++++++if+(i+%25+2+==+0)+{
++++++++++++System.out.print(words(i)+%2B+"+is+a+");
++++++++}+else+{
++++++++++++System.out.println(getTranslation(words(i)));
++++++++}
++++}
}

private+String+getTranslation(String+code)+{
++++if+("PF".equals(code))+{
++++++++return+"Platform";
++++}
++++//etc...
}|code-block|syntax|javascript|a05t4|本质上，这将做的是将字符串拆分成所有的单词。因为单词是“配对”在一起的，所以它们将以2个一组的形式出现。这意味着您可以检查单词的索引是偶数还是奇数。如果它是偶数，那么你就知道它是第一个配对的单词，这意味着你可以在它后面附加"is+a“字符串。如果它是奇数，那么您需要附加转换后的值。|1fggt^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|K|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|L|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

If you can guarantee that the values will be in the above order, something like this should work

<pre><code>public static void main(String[] args) {
 String my = "StackOverflow PF Apple FT Laptop HW";
 String[] words = my.split(" ");
 for (i = 0; i &lt; words.length; i++) {
 if (i % 2 == 0) {
 System.out.print(words(i) + " is a ");
 } else {
 System.out.println(getTranslation(words(i)));
 }
 }
}

private String getTranslation(String code) {
 if ("PF".equals(code)) {
 return "Platform";
 }
 //etc...
}
</code></pre>

Essentially, what this will do is split the string into all of the words. Since the words are "paired" together, they will come in groups of 2. This means that you can check if the index for the word is even or odd. If it's even, then you know it's the 1st paired word meaning you can append the "is a " string to it. If it's odd, then you want to append the translated value.

entityMap|blocks|key|e3o2p|text|使用正则表达式并将孔文本拆分为两个单词一组...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9iu60|然后用空格拆分数组中的每个元素，就完成了！|6kv2v|示例：|7t168|public+static+void+main(String[]+args)+throws+ParseException+{
String+inputTxt+=+"StackOverFlow+PF+Apple+FT+Laptop+HW.";
String[]+elements+=+inputTxt.split("(?<!\\G\\w%2B)\\s");
System.out.println(Arrays.toString(elements));
System.out.println(elements[0].split("+")[0]+%2B+"+is+a+Platform");
System.out.println(elements[1].split("+")[0]+%2B+"+is+a+Fruit");
System.out.println(elements[2].split("+")[0]+%2B+"+is+a+Hardware");
}|code-block|syntax|javascript|9onr4^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|N|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|O|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|P|9|@]|A|@]|B|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

Use regex amd split the hole text in groups of 2 words...
then split every element of the array by spaces and you are done!
<h1>Example:</h1>
<pre><code>public static void main(String[] args) throws ParseException {
String inputTxt = &quot;StackOverFlow PF Apple FT Laptop HW.&quot;;
String[] elements = inputTxt.split(&quot;(?&lt;!\\G\\w+)\\s&quot;);
System.out.println(Arrays.toString(elements));
System.out.println(elements[0].split(&quot; &quot;)[0] + &quot; is a Platform&quot;);
System.out.println(elements[1].split(&quot; &quot;)[0] + &quot; is a Fruit&quot;);
System.out.println(elements[2].split(&quot; &quot;)[0] + &quot; is a Hardware&quot;);
}
</code></pre>

entityMap|blocks|key|9c5pb|text|考虑到你问题中的有限规格，没有理由分开。你可以像这样替换你的占位符：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|d2kme|String+my+=+"StackOverFlow+PF+Apple+FT+Laptop+HW.";
my+=+my.replaceAll("PF[\\s.]?",+"++is+a+Platform.\n");
my+=+my.replaceAll("FT[\\s.]?",+"++is+a+Fruit.\n");
my+=+my.replaceAll("HW[\\s.]?",+"+is+a+hardware.\n");
System.out.print(my);|code-block|syntax|javascript|escje|输出：|uanp|StackOverFlow+++is+a+Platform.
Apple+++is+a+Fruit.
Laptop++is+a+hardware.|8lf09^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|N|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

Given the limited specification in your question, there's no reason to split. You can just replace your placeholders like this:

<pre><code>String my = "StackOverFlow PF Apple FT Laptop HW.";
my = my.replaceAll("PF[\\s.]?", " is a Platform.\n");
my = my.replaceAll("FT[\\s.]?", " is a Fruit.\n");
my = my.replaceAll("HW[\\s.]?", " is a hardware.\n");
System.out.print(my);
</code></pre>

Output:

<pre><code>StackOverFlow is a Platform.
Apple is a Fruit.
Laptop is a hardware.
</code></pre>

entityMap|blocks|key|dfkmh|text|public+class+StackOverflow+{

++++public+static+void+main(String[]+args)+{
++++++++//+Fields
++++++++String+myString+=+"StackOverFlow+PF+Apple+FT+Laptop+HW";

++++++++String[]+tokens+=+myString.split("+");

++++++++for+(int+i+=+0;+i+<+tokens.length;+i%2B%2B)+{
++++++++++++System.out.print(tokens[i]);
++++++++++++//+Every+other+token
++++++++++++if+(i+%25+2+==+0)+{
++++++++++++++++System.out.print("+is+a+"+%2B+convert(tokens[i+%2B+1]));
++++++++++++++++i%2B%2B;
++++++++++++}
++++++++++++System.out.println();
++++++++}

++++}

++++/**
+++++*+convert+method+turns+the+abbreviations+into+long+form
+++++*/
++++private+static+String+convert(String+s)+{
++++++++String+str;
++++++++switch+(s)+{
++++++++++++case+"PF":
++++++++++++++++str+=+"Platform";
++++++++++++++++break;
++++++++++++case+"FT":
++++++++++++++++str+=+"Fruit";
++++++++++++++++break;
++++++++++++case+"HW":
++++++++++++++++str+=+"Hardware";
++++++++++++++++break;
++++++++++++default:
++++++++++++++++str+=+"Unknown";
++++++++++++++++break;
++++++++}
++++++++return+str;
++++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|fcl82|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|G|9|@]|A|@]|B|$C|D]]|$2|E|4|-4|6|F|8|H|9|@]|A|@]|B|$]]]]

<pre><code>public class StackOverflow {

 public static void main(String[] args) {
 // Fields
 String myString = "StackOverFlow PF Apple FT Laptop HW";

 String[] tokens = myString.split(" ");

 for (int i = 0; i &lt; tokens.length; i++) {
 System.out.print(tokens[i]);
 // Every other token
 if (i % 2 == 0) {
 System.out.print(" is a " + convert(tokens[i + 1]));
 i++;
 }
 System.out.println();
 }

 }

 /**
 * convert method turns the abbreviations into long form
 */
 private static String convert(String s) {
 String str;
 switch (s) {
 case "PF":
 str = "Platform";
 break;
 case "FT":
 str = "Fruit";
 break;
 case "HW":
 str = "Hardware";
 break;
 default:
 str = "Unknown";
 break;
 }
 return str;
 }

}
</code></pre>

Is there any way to split the words from a string in <code>java</code>?.

<pre><code>String my ="StackOverFlow PF Apple FT Laptop HW."
</code></pre>

PF = Platform, FT = Fruit, HW = Hardware.

The expected output should be

<pre><code>StackOverFlow is a Platform.
Apple is a Fruit.
Laptop is a hardware.
</code></pre>

I am doing in this way:

<pre><code>String[] words = my.split(" ");
 for(int u=0 ; u&lt;words.length ; u++){
 System.out.println( words(u));
 }
</code></pre>

Separate specific words from a string in java

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 在java中，有没有办法将单词从字符串中拆分出来？ String my ="StackOverFlow PF Apple FT Laptop HW." PF =平台，FT =水果，HW =硬件。 预期输出应为 StackOverFlow  is a Platform.Apple  is a Fruit.Laptop ...

问在java中将特定的单词与字符串分开。
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问在java中将特定的单词与字符串分开。EN