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text.indexOf(match);

看看字符串 javadoc

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最佳答案

这样做的方法有:

返回此字符串中指定子字符串的第一个匹配项(或者最后一次)[从指定索引开始向前搜索(或者倒退)]的索引。

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"

小开

你可以通过在 while-loop 中赋值来获得一个文件中的所有匹配，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);
}
}

小开

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);
}
}

小开

int match_position=text.indexOf(match);

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查找单个索引

正如其他人所说，使用 text.indexOf(match)找到一个单一的匹配。

String text = "0123456789hello0123456789";
String match = "hello";
int position = text.indexOf(match); // position = 10

查找多个索引

由于 @ StephenC 的评论的代码可维护性和我自己在理解 @ polygenelubricants 的回答方面的困难，我想找到另一种方法来获得一个文本字符串中匹配的所有索引。下面的代码(从这个答案中修改)执行此操作:

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);
}

小开

如果要扫描搜索字符串的 n 个匹配项，我建议使用 Rel = “ nofollow”> 正则表达式 。他们有一个陡峭的学习曲线，但他们会节省你的时间，当涉及到复杂的搜索。

小开

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;


}
}
}
}
}
}

小开

//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);


}
}

小开

这使用正则表达式工作。

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));
}

产出:

在索引2-5找到“爱”

一般规则:

正则表达式从左到右搜索，一旦使用了匹配字符，就不能重用。

小开

对于多次出现和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);
}




}


}

小开

    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);

小开

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;
}

小开

     class Main{
public static int string(String str, String str1){
for (int i = 0; i <= str.length() - str1.length(); i++){
int j;
for (j = 0; j < str1.length(); j++) {
if (str1.charAt(j) != str.charAt(i + j)) {
break;
}
}
if (j == str1.length()) {
return i;
}}
return -1;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the string");
String str=sc.nextLine();
System.out.println("Enter the Substring");
String str1=sc.nextLine();
System.out.println("The position of the Substring is "+string(str, str1));
}
}

Java: method to get position of a match in a String?

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