String match = "hello";
String text = "0123456789hello0123456789";
int position = getPosition(match, text); // should be 10, is there such a method?
问问题
491919 次
13 回答
276
执行此操作的方法系列是：
返回此字符串中第一次（或最后一次）出现指定子字符串的索引[从指定索引开始向前（或向后）搜索]。
String text = "0123hello9012hello8901hello7890";
String word = "hello";
System.out.println(text.indexOf(word)); // prints "4"
System.out.println(text.lastIndexOf(word)); // prints "22"
// find all occurrences forward
for (int i = -1; (i = text.indexOf(word, i + 1)) != -1; i++) {
System.out.println(i);
} // prints "4", "13", "22"
// find all occurrences backward
for (int i = text.length(); (i = text.lastIndexOf(word, i - 1)) != -1; i++) {
System.out.println(i);
} // prints "22", "13", "4"
于 2010-04-11T02:21:34.767 回答
48
这使用正则表达式。
String text = "I love you so much";
String wordToFind = "love";
Pattern word = Pattern.compile(wordToFind);
Matcher match = word.matcher(text);
while (match.find()) {
System.out.println("Found love at index "+ match.start() +" - "+ (match.end()-1));
}
输出 ：
在索引 2 - 5 处找到“爱”
一般规则 ：
- 正则表达式从左到右搜索，一旦匹配字符被使用过，就不能重复使用。
于 2015-11-07T14:00:49.357 回答
20
text.indexOf(match);
请参阅字符串 javadoc
于 2010-04-11T01:46:01.250 回答
12
查找单个索引
正如其他人所说，用于text.indexOf(match)查找单个匹配项。
String text = "0123456789hello0123456789";
String match = "hello";
int position = text.indexOf(match); // position = 10
查找多个索引
由于@StephenC 对代码可维护性的评论以及我自己难以理解@polygenelubricants 的答案，我想找到另一种方法来获取文本字符串中匹配项的所有索引。以下代码（从this answer修改）这样做：
String text = "0123hello9012hello8901hello7890";
String match = "hello";
int index = text.indexOf(match);
int matchLength = match.length();
while (index >= 0) { // indexOf returns -1 if no match found
System.out.println(index);
index = text.indexOf(match, index + matchLength);
}
于 2015-06-17T00:28:44.327 回答
2
您可以通过在while循环中简单地分配来获取文件中的所有匹配项，很酷：
$ javac MatchTest.java
$ java MatchTest
1
16
31
46
$ cat MatchTest.java
import java.util.*;
import java.io.*;
public class MatchTest {
public static void main(String[] args){
String match = "hello";
String text = "hello0123456789hello0123456789hello1234567890hello3423243423232";
int i =0;
while((i=(text.indexOf(match,i)+1))>0)
System.out.println(i);
}
}
于 2010-04-11T02:21:47.667 回答
2
int match_position=text.indexOf(match);
于 2014-03-03T10:24:41.817 回答
1
import java.util.StringTokenizer;
public class Occourence {
public static void main(String[] args) {
String key=null,str ="my name noorus my name noorus";
int i=0,tot=0;
StringTokenizer st=new StringTokenizer(str," ");
while(st.hasMoreTokens())
{
tot=tot+1;
key = st.nextToken();
while((i=(str.indexOf(key,i)+1))>0)
{
System.out.println("position of "+key+" "+"is "+(i-1));
}
}
System.out.println("total words present in string "+tot);
}
}
于 2014-02-09T20:34:59.563 回答
1
我有一些大代码，但工作得很好......
class strDemo
{
public static void main(String args[])
{
String s1=new String("The Ghost of The Arabean Sea");
String s2=new String ("The");
String s6=new String ("ehT");
StringBuffer s3;
StringBuffer s4=new StringBuffer(s1);
StringBuffer s5=new StringBuffer(s2);
char c1[]=new char[30];
char c2[]=new char[5];
char c3[]=new char[5];
s1.getChars(0,28,c1,0);
s2.getChars(0,3,c2,0);
s6.getChars(0,3,c3,0); s3=s4.reverse();
int pf=0,pl=0;
char c5[]=new char[30];
s3.getChars(0,28,c5,0);
for(int i=0;i<(s1.length()-s2.length());i++)
{
int j=0;
if(pf<=1)
{
while (c1[i+j]==c2[j] && j<=s2.length())
{
j++;
System.out.println(s2.length()+" "+j);
if(j>=s2.length())
{
System.out.println("first match of(The) :->"+i);
}
pf=pf+1;
}
}
}
for(int i=0;i<(s3.length()-s6.length()+1);i++)
{
int j=0;
if(pl<=1)
{
while (c5[i+j]==c3[j] && j<=s6.length())
{
j++;
System.out.println(s6.length()+" "+j);
if(j>=s6.length())
{
System.out.println((s3.length()-i-3));
pl=pl+1;
}
}
}
}
}
}
于 2015-07-08T07:58:39.113 回答
1
//finding a particular word any where inthe string and printing its index and occurence
class IndOc
{
public static void main(String[] args)
{
String s="this is hyderabad city and this is";
System.out.println("the given string is ");
System.out.println("----------"+s);
char ch[]=s.toCharArray();
System.out.println(" ----word is found at ");
int j=0,noc=0;
for(int i=0;i<ch.length;i++)
{
j=i;
if(ch[i]=='i' && ch[j+1]=='s')
{
System.out.println(" index "+i);
noc++;
}
}
System.out.println("----- no of occurences are "+noc);
}
}
于 2015-09-01T07:18:56.553 回答
1
String match = "hello";
String text = "0123456789hello0123456789hello";
int j = 0;
String indxOfmatch = "";
for (int i = -1; i < text.length()+1; i++) {
j = text.indexOf("hello", i);
if (i>=j && j > -1) {
indxOfmatch += text.indexOf("hello", i)+" ";
}
}
System.out.println(indxOfmatch);
于 2018-04-13T16:32:09.567 回答
0
如果您要扫描搜索字符串的“n”个匹配项，我建议您使用正则表达式。他们有一个陡峭的学习曲线，但在复杂的搜索方面，他们会为你节省几个小时。
于 2015-06-22T14:26:55.127 回答
0
对于多次出现和在字符串中找到的字符？？是或否
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class SubStringtest {
public static void main(String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter the string");
String str=br.readLine();
System.out.println("enter the character which you want");
CharSequence ch=br.readLine();
boolean bool=str.contains(ch);
System.out.println("the character found is " +bool);
int position=str.indexOf(ch.toString());
while(position>=0){
System.out.println("the index no of character is " +position);
position=str.indexOf(ch.toString(),position+1);
}
}
}
于 2018-01-04T06:30:41.957 回答
0
public int NumberWordsInText(String FullText_, String WordToFind_, int[] positions_)
{
int iii1=0;
int iii2=0;
int iii3=0;
while((iii1=(FullText_.indexOf(WordToFind_,iii1)+1))>0){iii2=iii2+1;}
// iii2 is the number of the occurences
if(iii2>0) {
positions_ = new int[iii2];
while ((iii1 = (FullText_.indexOf(WordToFind_, iii1) + 1)) > 0) {
positions_[iii3] = iii1-1;
iii3 = iii3 + 1;
System.out.println("position=" + positions_[iii3 - 1]);
}
}
return iii2;
}
于 2020-05-29T08:29:29.617 回答