我正在尝试将字符串截断到单个以逗号分隔的单词字符串中特定单词的第一次出现。例如：
deleteLastOccurrence("foo,bar,dog,cat,dog,bird","dog")
应该返回
"foo,bar,dog"
public String deleteLastOccurrence(String original, String target){
String[] arr = original.split(",");
arr = Arrays.copyOfRange(arr, Arrays.asList(arr).indexOf(target), original.length()-1);
path = StringUtils.join(pathArray,",");
}
使用正则表达式替换：
public static String deleteLastOccurrence(String original, String target){
return original.replaceAll("(,)?\\b" + target + "\\b.*", "$1" + target);
}
\b表示“单词边界”）。deleteAfterFirstOccurrence()，因为您当前的名称具有误导性：与您想要的无关的是“最后一次出现”。public static void main(String[] args) {
// Test for target in middle:
System.out.println(deleteLastOccurrence("foo,bar,dog,cat,dog,bird,dog", "dog"));
// Test for target at start:
System.out.println(deleteLastOccurrence("dog,bar,dog,cat,dog,bird,dog", "dog"));
// Test for target at end:
System.out.println(deleteLastOccurrence("foo,bar,cat,bird,dog", "dog"));
}
输出：
foo,bar,dog
dog
foo,bar,cat,bird,dog
.*。 - Bohemian更新：仔细查看问题后，我意识到我写的是方法名，而不是OP想要的结果。所以，它只是去掉了最后一次出现，而不是在它之后修剪。哦，好吧！:)
根据您的风格，您可能认为这并不更简单。但是，这是一个有趣的问题。我认为这段代码更清晰一些。
public class ReplaceLast {
public String deleteLastOccurrence(String fromThis, String word){
int wordLength = word.length();
if(fromThis.startsWith(word + ",")){
return fromThis.substring(wordLength + 1);
}
if(fromThis.endsWith("," + word)){
return fromThis.substring(0, fromThis.length() - wordLength - 1);
}
int index = fromThis.lastIndexOf("," + word + ",");
if(index == -1){
return fromThis;
}
return fromThis.substring(0, index) + fromThis.substring(index+word.length() + 1);
}
@Test
public void testNotThere() {
String actual = deleteLastOccurrence("foo,bar,dog,cat,dog,bird","moose");
assertEquals("foo,bar,dog,cat,dog,bird", actual);
}
@Test
public void testMiddle() {
String actual = deleteLastOccurrence("foo,bar,dog,cat,dog,bird","dog");
assertEquals("foo,bar,dog,cat,bird", actual);
}
@Test
public void testFirst() {
String actual = deleteLastOccurrence("foo,bar,dog,cat,dog,bird","foo");
assertEquals("bar,dog,cat,dog,bird", actual);
}
@Test
public void testLast() {
String actual = deleteLastOccurrence("foo,bar,dog,cat,dog,bird","bird");
assertEquals("foo,bar,dog,cat,dog", actual);
}
@Test
public void testSubword() {
String actual = deleteLastOccurrence("foo,bar,dog,cat,dog,bird","bir");
assertEquals("foo,bar,dog,cat,dog,bird", actual);
}
}
我尝试解决截取字符串中第一个特定单词的问题，而我并不关心原方法的名称（deleteLastOccurrence），因为我认为这个名称会误导人。
对于我来说，匹配单个单词而不是子单词的技巧是在句子前后添加两个逗号，然后使用带有逗号的单词进行检查。
例如，",dog," 将与 ",foo,bar,dog,cat,dog,bird," 进行匹配以检查其是否存在。
package gicappa;
public class So {
public static String trimSentenceOnFirstOccurrenceOf(String sentence, String word) {
if (word.isEmpty()) return sentence;
if (!addCommasAround(sentence).contains(addCommasAround(word))) return sentence;
return trimAddedCommasOf(substringOfSentenceUntilEndOfWord(addCommasAround(sentence), addCommasAround(word)));
}
public static String substringOfSentenceUntilEndOfWord(String string, String word) {
return string.substring(0, string.indexOf(word) + word.length());
}
public static String trimAddedCommasOf(String string) {return string.substring(1,string.length()-1);}
public static String addCommasAround(String s) {return "," + s + ","; }
}
如果您想进行一些测试，我用于TDD的方法如下：
package gicappa;
import org.junit.Test;
import static gicappa.So.trimSentenceOnFirstOccurrenceOf;
import static org.hamcrest.core.Is.is;
import static org.hamcrest.core.IsEqual.equalTo;
import static org.junit.Assert.assertThat;
public class SoTest {
@Test
public void it_returns_the_same_sentence_for_empty_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", ""), is(equalTo("foo,bar,dog,cat,dog,bird")));
}
@Test
public void it_returns_the_same_sentence_for_not_contained_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", "s"), is(equalTo("foo,bar,dog,cat,dog,bird")));
}
@Test
public void it_returns_the_first_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", "foo"), is(equalTo("foo")));
}
@Test
public void it_returns_the_same_sentence_if_is_matched_the_last_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", "bird"), is(equalTo("foo,bar,dog,cat,dog,bird")));
}
@Test
public void it_trims_after_the_end_of_the_first_matched_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", "dog"), is(equalTo("foo,bar,dog")));
}
@Test
public void it_does_not_trim_for_a_subword_of_a_contained_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("foo,bar,dog,cat,dog,bird", "do"), is(equalTo("foo,bar,dog,cat,dog,bird")));
}
@Test
public void it_does_not_trim_for_a_subword_of_an_already_contained_word() {
assertThat(trimSentenceOnFirstOccurrenceOf("dog,foozzo,foo,cat,dog,bird", "foo"), is(equalTo("dog,foozzo,foo")));
}
}
package gicappa;
public class Sentence {
private String s;
public Sentence(String sentence) {
this.s = sentence;
}
public String trimOnFirstOccurrenceOf(String word) {
if (word.isEmpty() || csvSentenceContainsWord(word)) return s;
return substringSentenceToEndOf(word);
}
private String substringSentenceToEndOf(String word) {
return addCommasTo(s).substring(1, addCommasTo(s).indexOf(addCommasTo(word)) + addCommasTo(word).length()-1);
}
private boolean csvSentenceContainsWord(String word) {
return !addCommasTo(s).contains(addCommasTo(word));
}
public static String addCommasTo(String s) {return "," + s + ",";}
}
使用方式如下：
new Sentence("dog,foozzo,foo,cat,dog,bird").trimOnFirstOccurrenceOf("foo"), is(equalTo("dog,foozzo,foo"))
public String deleteLastOccurrence(String original, String target){
return original.replace("(^|,)" + target + "(,|$)", "");
}
replace 函数只替换一个字符，而 replaceAll 函数则替换所有匹配项 :) - Wolphpublic String trimTo(String in, String matchNoCommas) {
if (in.startsWith(matchNoCommas + ",")) // special check here...
return matchNoCommas;
int idx = in.indexOf("," + matchNoCommas+ ",");
if (idx < 0)
return in;
return in.substring(0, idx + matchNoCommas.length()+1);
}
提供与@Bohemian的正则表达式版本相同的结果。由您决定哪个更易于理解。
也许我错了，但这样做不行吗？
public trimCommaSeparatedListToIncludeFirstOccurrenceOfWord(String listOfWords, String wordToMatch) {
int startOfFirstOccurrenceOfWordToMatch = listOfWords.indexOf(wordToMatch);
int endOfFirstOccurrenceOfWordToMatch = startOfFirstOccurrenceOfWordToMatch + wordToMatch.length() - 1;
return listOfWords.substring(0, endOfFirstOccurrenceOfWordToMatch);
}
现在这可能不是原帖作者想要的，但我认为这就是原帖作者所要求的。例如：f("doggy,cat,bird", "dog")将返回"dog"。
对于全词匹配，我会像其他人建议的那样使用正则表达式。
gonzoc0ding，在阅读所有回复后，我认为你的方法更简单更清晰，除了这一点需要修正：
public String deleteLastOccurrence(String original, String target){
String[] arr = original.split(",");
arr = Arrays.copyOfRange(arr,0, Arrays.asList(arr).indexOf(target));
path = StringUtils.join(arr,",");
}
但也许我还没有理解您的要求...