从字符串数组中提取标记位置-Java

Question

我只是想问是否有一种简单的方法可以从 java 中的字符串数组中提取字符串。 例如，如果我有输入：

String searchtext = "The one thing";
String source = "the one Thing in life is to not do in java";
String annote = "det num nn pp nn cop to neg vv pp nn";

我想要输出（我不想使用正则表达式，因为我的搜索文本会有所不同）

det num nn

这段代码能用吗？？？？

String searchtext = "The one thing";
String source = "the one Thing in life is to not do in java";
String annote = "det num nn pp nn cop to neg vv pp nn";
String[] annotelist = annote.split(" ");

List<String> sourcelist = Array.asList(sourcetext.split(" ")); 
search_startpt = searchlist.indexof(search[0]);

String[] searchannote = annotelist[search_startpt];
for (int j=1; j<sourcelist.length(); j++) 
  searchanote[j] = annotelist[sear_startpt+j];

System.out.println(StringUtils.join(searchannoate, " "));

原来，我试过下面的代码：

import org.apache.commons.lang.StringUtils;

String searchtext = "The one thing";
String[] search  = searchtext.split(" ");
String source = "the one Thing in life is to not do in java";
String[] sourcelist  = source.split(" ");
String annote = "det num nn pp nn cop to neg vv pp nn";
String[] annotelist = annote.split(" ");

int search_startpt = 0;

for (int i=0; i<sourcelist.length(); i++) {
  if (sourcelist[i].equalsIgnoreCase(search[0])) {
    for (int j=1; j<search.length(); j++) {
      if (sourcelist[i+j].equalsIgnoreCase(search[j]) ==0) break;
      if (sourcelist[i+search.length()].equalsIgnoreCase(search[search.length()-1])) search_startpt = i;
    }
  }
}

String[] searchannote = annotelist[search_startpt];

for (int j=1; j<sourcelist.length(); j++) 
  searchanote[j] = annotelist[sear_startpt+j];

System.out.println(StringUtils.join(searchannoate, " "));

Answer 1

将字符串之间的所有== 替换为.equals()。示例：

if (sourcelist[i] == search[0]) {

变成

if (sourcelist[i].equals(search[0])) {

原因是当你split()一个String时，所有创建的String对象都是新的和不同的，即使它们的内容是一样的。 == 操作符测试两个引用是否指向同一个对象，而.equals() 测试两个对象是否具有相同的内容。