Mixing
How to search String array through char array?
1
I want to build a simple dictionary search program without using dictionary library. I want to search the string array if the string in array is equal to the char array. It should print the string and its index number. If there are 2 strings that are matching the char then it should print the first string and leave the after string
.e.g String array["fine","rest","door","shine"] char character ['t','r','e','s']. the answer should be "rest" at index 1. and if the String rest is repeat then it should only print first one.
I tried to compare the string array and char array but its returning all the words of string array that matches char array.
String strArray=new String[4];
char chrArray=new char[4];
String value="";
char compare;
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
for (int i = 0; i < strArray.length; i++) {
if(strArray[i].length()==chrArray.length){
if(""+strArray[i]!=value){
value="";
}
for (int j = 0; j < strArray[i].length(); j++) {
for (int k = 0; k < chrArray.length; k++) {
if(strArray[i].charAt(j)==chrArray[k]){
value=value+strArray[i].charAt(j);
}
}
}
}
}
System.out.println(value);
The output should be the string from array that is equal to char array.
java arrays
asked Jan 1 at 11:18
AliAli
83
add a comment |
1
I want to build a simple dictionary search program without using dictionary library. I want to search the string array if the string in array is equal to the char array. It should print the string and its index number. If there are 2 strings that are matching the char then it should print the first string and leave the after string
.e.g String array["fine","rest","door","shine"] char character ['t','r','e','s']. the answer should be "rest" at index 1. and if the String rest is repeat then it should only print first one.
I tried to compare the string array and char array but its returning all the words of string array that matches char array.
String strArray=new String[4];
char chrArray=new char[4];
String value="";
char compare;
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
for (int i = 0; i < strArray.length; i++) {
if(strArray[i].length()==chrArray.length){
if(""+strArray[i]!=value){
value="";
}
for (int j = 0; j < strArray[i].length(); j++) {
for (int k = 0; k < chrArray.length; k++) {
if(strArray[i].charAt(j)==chrArray[k]){
value=value+strArray[i].charAt(j);
}
}
}
}
}
System.out.println(value);
The output should be the string from array that is equal to char array.
java arrays
asked Jan 1 at 11:18
AliAli
83
"if(""+strArray[i]!=value){" isn't the correct way to compare strings. The+is redundant (unless you expectstrArray[i]to be null), and then you'd need to useequals(eitherstrArray[i].equals(value), orObjects.equals(strArray[i], value)if you think it might be null).
– Andy Turner
Jan 1 at 11:20
Format code correctly
– beso9595
Jan 1 at 11:22
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29
add a comment |
1
1
1
I want to build a simple dictionary search program without using dictionary library. I want to search the string array if the string in array is equal to the char array. It should print the string and its index number. If there are 2 strings that are matching the char then it should print the first string and leave the after string
.e.g String array["fine","rest","door","shine"] char character ['t','r','e','s']. the answer should be "rest" at index 1. and if the String rest is repeat then it should only print first one.
I tried to compare the string array and char array but its returning all the words of string array that matches char array.
String strArray=new String[4];
char chrArray=new char[4];
String value="";
char compare;
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
for (int i = 0; i < strArray.length; i++) {
if(strArray[i].length()==chrArray.length){
if(""+strArray[i]!=value){
value="";
}
for (int j = 0; j < strArray[i].length(); j++) {
for (int k = 0; k < chrArray.length; k++) {
if(strArray[i].charAt(j)==chrArray[k]){
value=value+strArray[i].charAt(j);
}
}
}
}
}
System.out.println(value);
The output should be the string from array that is equal to char array.
java arrays
asked Jan 1 at 11:18
AliAli
83
I want to build a simple dictionary search program without using dictionary library. I want to search the string array if the string in array is equal to the char array. It should print the string and its index number. If there are 2 strings that are matching the char then it should print the first string and leave the after string
.e.g String array["fine","rest","door","shine"] char character ['t','r','e','s']. the answer should be "rest" at index 1. and if the String rest is repeat then it should only print first one.
I tried to compare the string array and char array but its returning all the words of string array that matches char array.
String strArray=new String[4];
char chrArray=new char[4];
String value="";
char compare;
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
for (int i = 0; i < strArray.length; i++) {
if(strArray[i].length()==chrArray.length){
if(""+strArray[i]!=value){
value="";
}
for (int j = 0; j < strArray[i].length(); j++) {
for (int k = 0; k < chrArray.length; k++) {
if(strArray[i].charAt(j)==chrArray[k]){
value=value+strArray[i].charAt(j);
}
}
}
}
}
System.out.println(value);
The output should be the string from array that is equal to char array.
java arrays
java arrays
asked Jan 1 at 11:18
AliAli
83
asked Jan 1 at 11:18
AliAli
83
edited Jan 1 at 11:35
Ali
asked Jan 1 at 11:18
AliAli
83
asked Jan 1 at 11:18
AliAli
83
asked Jan 1 at 11:18
AliAli
83
83
"if(""+strArray[i]!=value){" isn't the correct way to compare strings. The+is redundant (unless you expectstrArray[i]to be null), and then you'd need to useequals(eitherstrArray[i].equals(value), orObjects.equals(strArray[i], value)if you think it might be null).
– Andy Turner
Jan 1 at 11:20
Format code correctly
– beso9595
Jan 1 at 11:22
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29
add a comment |
"if(""+strArray[i]!=value){" isn't the correct way to compare strings. The+is redundant (unless you expectstrArray[i]to be null), and then you'd need to useequals(eitherstrArray[i].equals(value), orObjects.equals(strArray[i], value)if you think it might be null).
– Andy Turner
Jan 1 at 11:20
Format code correctly
– beso9595
Jan 1 at 11:22
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29
"
if(""+strArray[i]!=value){" isn't the correct way to compare strings. The + is redundant (unless you expect strArray[i] to be null), and then you'd need to use equals (either strArray[i].equals(value), or Objects.equals(strArray[i], value) if you think it might be null).– Andy Turner
Jan 1 at 11:20
"
if(""+strArray[i]!=value){" isn't the correct way to compare strings. The + is redundant (unless you expect strArray[i] to be null), and then you'd need to use equals (either strArray[i].equals(value), or Objects.equals(strArray[i], value) if you think it might be null).– Andy Turner
Jan 1 at 11:20
Format code correctly
– beso9595
Jan 1 at 11:22
Format code correctly
– beso9595
Jan 1 at 11:22
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29
add a comment |
2 Answers
2
active
oldest
votes
6
You can sort char array and then compare it using Arrays.equal. By sorting char array, there will be no need to use 2 for loops.
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String args) {
String strArray=new String[4];
char chrArray=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
add a comment |
0
You could also create a Map<Character, Integer> to store counts for both the words and the characters, and compare each word character counts with the character counts. If both are equal, print it out.
Demo:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String words = {"fine", "rest", "door", "shine"};
private static final char chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
Output:
rest, at index = 1
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
You can sort char array and then compare it using Arrays.equal. By sorting char array, there will be no need to use 2 for loops.
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String args) {
String strArray=new String[4];
char chrArray=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
add a comment |
6
You can sort char array and then compare it using Arrays.equal. By sorting char array, there will be no need to use 2 for loops.
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String args) {
String strArray=new String[4];
char chrArray=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
add a comment |
6
6
6
You can sort char array and then compare it using Arrays.equal. By sorting char array, there will be no need to use 2 for loops.
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String args) {
String strArray=new String[4];
char chrArray=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
You can sort char array and then compare it using Arrays.equal. By sorting char array, there will be no need to use 2 for loops.
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String args) {
String strArray=new String[4];
char chrArray=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
answered Jan 1 at 11:39
Mukesh prajapatiMukesh prajapati
865418
865418
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
add a comment |
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Thanks u Mukesh for your code ,i was trying this from 2 days.God Bless You.
– Ali
Jan 1 at 12:23
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
Happy to help. Welcome @Ali
– Mukesh prajapati
Jan 1 at 12:27
add a comment |
0
You could also create a Map<Character, Integer> to store counts for both the words and the characters, and compare each word character counts with the character counts. If both are equal, print it out.
Demo:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String words = {"fine", "rest", "door", "shine"};
private static final char chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
Output:
rest, at index = 1
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
add a comment |
0
You could also create a Map<Character, Integer> to store counts for both the words and the characters, and compare each word character counts with the character counts. If both are equal, print it out.
Demo:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String words = {"fine", "rest", "door", "shine"};
private static final char chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
Output:
rest, at index = 1
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
add a comment |
0
0
0
You could also create a Map<Character, Integer> to store counts for both the words and the characters, and compare each word character counts with the character counts. If both are equal, print it out.
Demo:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String words = {"fine", "rest", "door", "shine"};
private static final char chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
Output:
rest, at index = 1
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
You could also create a Map<Character, Integer> to store counts for both the words and the characters, and compare each word character counts with the character counts. If both are equal, print it out.
Demo:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String words = {"fine", "rest", "door", "shine"};
private static final char chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
Output:
rest, at index = 1
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
edited Jan 1 at 12:51
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
answered Jan 1 at 12:23
RoadRunnerRoadRunner
11.3k31341
11.3k31341
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"
if(""+strArray[i]!=value){" isn't the correct way to compare strings. The+is redundant (unless you expectstrArray[i]to be null), and then you'd need to useequals(eitherstrArray[i].equals(value), orObjects.equals(strArray[i], value)if you think it might be null).– Andy Turner
Jan 1 at 11:20
Format code correctly
– beso9595
Jan 1 at 11:22
I m a beginner and i don't know if my code is correct but if some one just write this program i will be very thankful.The code i want is described above.
– Ali
Jan 1 at 11:29