Develop Reference

Parse lines with different number of string words [duplicate]

This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 1 year ago.
I have a text file with the following format(the words at each line are seperated with tab):
stri1 stri2 stri3
stri4 stri5 stri6
stri7 stri8
stri9 stri0 stri5
As you can see i have some lines with only two words. I have a class to save the words of each line:
public class Entity{
private word1,word2,word3;
//constructor and getter/setter methods
}
I want to save the text values using the following code:
for(String i : filelines){
String[] line = i.split("\t");
if(line[2] == null){
listOfEntities.add(new Entity(line[0], line[1], null));
}
else{
listOfEntities.add(new Entity(line[0], line[1], line[2]));
}
When i try to execute this code i get an ArrayIndexOutOfBoundsException because some line have only 2 words. How can i handle this situation because i want also the null values in order to make some sql queries later.

Mod with length of array will return 0 if array exceed legth it will store zeroth index other than that you have to do multiple ifs as you have done for second index.
Another way I would suggest pass the whole array in Entity parameter and in Entity class run a for loop till array length which will run till available lines in array only.
for(String i : filelines){
String[] line = i.split("\t");
if(line[2 % line.length] == null){
listOfEntities.add(new Entity(line[0], line[1 % line.length], null));
}
else{
listOfEntities.add(new Entity(line[0], line[1 % line.length], line[2 % line.length]));
}

I have another option for you. You have to separate the options in the file with an element. This looks like this:
stri1-stri2-stri3
stri4-stri5-stri6
stri7-stri8
stri9-stri0-stri5
For this you have to change your entity class a little bit:
public static class Entity {
private String[] words;
public Entity(String[] words) {
this.words = words;
}
public String[] getWords() {
return words;
}
}
To read out the whole thing now, I wrote you a code.
List<Entity> array = new ArrayList<>();
Scanner scanner = new Scanner(new File("MyFile.txt"));
while (scanner.hasNext()) {
String line = scanner.next();
String[] options = line.split("-");
array.add(new Entity(options));
}
array.forEach(entity -> {
System.out.println("Entity Words > " + Arrays.toString(entity.getWords()));
});
I hope I could help you with this.

Check if any part of a string input is not a number

I couldnt find an answer for this in Java, so I'll ask here. I need to check if 3 parts of a string input contains a number (int).
The input will be HOURS:MINUTES:SECONDS (E.g. 10:40:50, which will be 10 hours, 40 minutes and 50 seconds). So far I am getting the values in String[] into an array by splitting it on :. I have parsed the strings into ints and I am using an if statement to check if all 3 parts is equal or larger than 0. The problem is that if I now use letters I will only just get an error, but I want to check if any of the 3 parts contains a character that is not 0-9, but dont know how.
First I thought something like this could work, but really dont.
String[] inputString = input.split(":");
if(inputString.length == 3) {
String[] alphabet = {"a","b","c"};
if(ArrayUtils.contains(alphabet,input)){
gives error message
}
int hoursInt = Integer.parseInt(inputString[0]);
int minutesInt = Integer.parseInt(inputString[1]);
int secondsInt = Integer.parseInt(inputString[2]);
else if(hoursInt >= 0 || minutesInt >= 0 || secondsInt >= 0) {
successfull
}
else {
gives error message
}
else {
gives error message
}
In the end I just want to check if any of the three parts contains a character, and if it doesnt, run something.

If you are sure you always have to parse a String of the form/pattern HH:mm:ss
(describing a time of day),
you can try to parse it to a LocalTime, which will only work if the parts HH, mm and ss are actually valid integers and valid time values.
Do it like this and maybe catch an Exception for a wrong input String:
public static void main(String[] arguments) {
String input = "10:40:50";
String wrongInput = "ab:cd:ef";
LocalTime time = LocalTime.parse(input);
System.out.println(time.format(DateTimeFormatter.ISO_LOCAL_TIME));
try {
LocalTime t = LocalTime.parse(wrongInput);
} catch (DateTimeParseException dtpE) {
System.err.println("Input not parseable...");
dtpE.printStackTrace();
}
}
The output of this minimal example is
10:40:50
Input not parseable...
java.time.format.DateTimeParseException: Text 'ab:cd:ef' could not be parsed at index 0
at java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1949)
at java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1851)
at java.time.LocalTime.parse(LocalTime.java:441)
at java.time.LocalTime.parse(LocalTime.java:426)
at de.os.prodefacto.StackoverflowDemo.main(StackoverflowDemo.java:120)

I would personally create my own helper methods for this, instead of using an external library such as Apache (unless you already plan on using the library elsewhere in the project).
Here is an example of what it could look like:
public static void main(String[] arguments) {
String time = "10:50:45";
String [] arr = time.split(":");
if (containsNumbers(arr)) {
System.out.println("Time contained a number!");
}
//You can put an else if you want something to happen when it is not a number
}
private static boolean containsNumbers(String[] arr) {
for (String s : arr) {
if (!isNumeric(s)) {
return false;
}
}
return true;
}
public static boolean isNumeric(String str) {
return str.matches("-?\\d+(.\\d+)?");
}
containsNumbers will take a String array as an input and use an enhanced for loop to iterate through all the String values, using the other helper method isNumeric that checks if the String is a number or not using regex.
This code has the benefit of not being dependent on Exceptions to handle any of the logic.
You can also modify this code to use a String as a parameter instead of an array, and let it handle the split inside of the method instead of outside.
Note that typically there are better ways to work with date and time, but I thought I would answer your literal question.
Example Runs:
String time = "sd:fe:gbdf";
returns false
String time = "as:12:sda";
returns false
String time = "10:50:45";
returns true

You can check the stream of characters.
If the filter does not detect a non-digit, return "Numeric"
Otherwise, return "Not Numeric"
String str = "922029202s9202920290220";
String result = str.chars()
.filter(c -> !Character.isDigit(c))
.findFirst().isEmpty() ? "Numeric"
: "Not Numeric";
System.out.println(result);

If you want to check with nested loop you can see this proposal:
Scanner scanner = new Scanner(System.in);
String [] inputString = scanner.nextLine().split(":");
for (int i = 0; i < inputString.length; i++) {
String current = inputString[i];
for (int k = 0; k < current.length(); k++) {
if (!Character.isDigit(current.charAt(k))) {
System.out.println("Error");
break;
}
}
}

you could use String.matches method :
String notANum= "ok";
String aNum= "7";
if(notANum.matches("^[0-9]+$") sop("no way!");
if(aNum.matches("^[0-9]+$") sop("yes of course!");
The code above would print :
yes of course
The method accepts a regex, the one in the above exemple is for integers.
EDIT
I would use this instead :
if(input.matches("^\d+:\d+:\d+$")) success;
else error
You don't have to split the string.

I tried to make your code better, take a look. You can use Java regex to validate numbers. also defined range for time so no 24:61:61 values is allowed.
public class Regex {
static boolean range(int timeval,int min,int max)
{
boolean status=false;
if(timeval>=min && timeval<max)
{status=true;}
return status;
}
public static void main(String[] args) {
String regex = "[0-9]{1,2}";
String input ="23:59:59";
String msg="please enter valid time ";
String[] inputString = input.split(":");
if(inputString[0].matches(regex) && inputString[1].matches(regex) && inputString[2].matches(regex) )
{
if(Regex.range(Integer.parseInt(inputString[0]), 00, 24) &&Regex.range(Integer.parseInt(inputString[1]), 00, 60) && Regex.range(Integer.parseInt(inputString[2]), 00, 60))
{msg="converted time = " + Integer.parseInt(inputString[0]) + " : " +Integer.parseInt(inputString[1])+ " : " +Integer.parseInt(inputString[2]) ;}
}
System.out.println(msg);
}
}

Program not noticing when the string is equal to a specific string [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 6 years ago.
I've been trying to create a program that censors a word but I was having difficulty with that so I tried going back to some of the fundamental code and testing it and I am coming across an odd result.
import java.util.Scanner;
public class TextCensor
{
public static void main(String[] args)
{
String input;
Scanner keyboard = new Scanner(System.in);
input = keyboard.nextLine();
int length = input.length() - 1;
if (length + 1 >= 3)
{
for (int i=0; i<(length - 1); i=i+1 )
{
char first = input.charAt(i);
char second = input.charAt(i+1);
char third = input.charAt(i+2);
String censorCheck = "" + first + second + third;
if (censorCheck == "tag")
{
System.out.println("success");
}
else
{
System.out.println(censorCheck);
}
}
}
else
{
System.out.println(input);
}
}
}
If I input the string "adtag" I will obtain the following output:
adt
dta
tag
yet "success" will never be printed despite the fact that I have printed a censorCheck that is equal to "tag".

String is an object. You have to compare objects by equals():
censorCheck.equalsIgnoreCase("tag")
Ignore case works fir upper letters as well.
Only for primitives you can use comparison by ==:
3 == 3

You are trying to check whether both instance of String is same or not instead of checking contents of both string.
You should try censorCheck.equals("tag") .

To compare whether contents of two string are equal or not in JAVA you should use the equals() method. You cannot compare the value of two string by the == operator . In your case use if (censorCheck.equals("tag")) and see if you get the desired result.

java find if the string contains 2 other strings

I have 2 strings "test" "bet" and another string a="tbtetse". I need to check if the "tbtetse" contains the other two strings.
I was thinking if I could find all the anagrams of string a and and then find the other two strings in those, but it doesn't work that way and also my anagram code is failing for a lengthy string.
Could you please help with any other ways to solve it?

Assuming you're trying to test whether the letters in a can be used to form an anagram of the test strings test and bet: I recommend making a dictionary (HashMap or whatever) of character counts from string a, indexed by character. Build a similar dictionary for the words you're testing. Then make sure that a has at least as many instances of each character from the test strings as they have.
Edit: Alcanzar suggests arrays of length 26 for holding the counts (one slot for each letter). Assuming you're dealing with only English letters, that is probably less of a hassle than dictionaries. If you don't know the number of allowed characters, the dictionary route is necessary.

Check below code, it may help you.
public class StringTest {
public static void main(String[] args) {
String str1 = "test";
String str2 = "bev";
String str3 = "tbtetse";
System.out.println(isStringPresent(str1, str2, str3));
}
private static boolean isStringPresent(String str1, String str2, String str3) {
if ((str1.length() + str2.length()) != str3.length()) {
return false;
} else {
String[] str1Arr = str1.split("");
String[] str2Arr = str2.split("");
for (String string : str1Arr) {
if (!str3.contains(string)) {
return false;
}
}
for (String string : str2Arr) {
if (!str3.contains(string)) {
return false;
}
}
}
return true;
}
}

basically you need to count characters in both sets and compare them
void fillInCharCounts(String word,int[] counts) {
for (int i = 0; i<word.length(); i++) {
char ch = word.charAt(i);
int index = ch - 'a';
counts[index]++;
}
}
int[] counts1 = new int[26];
int[] counts2 = new int[26];
fillInCharCounts("test",counts1);
fillInCharCounts("bet",counts1);
fillInCharCounts("tbtese",counts2);
boolean failed = false;
for (int i = 0; i<counts1.length; i++) {
if (counts1[i] > counts2[i]) {
failed = true;
}
}
if (failed) {
whatever
} else {
something else
}
If you are generalizing it, don't forget to call .toLowerCase() on the word before sending it in (or fix the counting method).

Pseudo code:
Make a copy of string "tbtetse".
Loop through each character in "test".
Do a indexOf() search for the character in your copied string and remove it if found.
If not found, fail.
Do the same for the string "bet".

class WordLetter {
char letter;
int nth; // Occurrence of that letter
...
}
One now can use Sets
Set<WordLetter>
// "test" = { t0 e0 s0 t1 }
Then testing reduces to set operations. If both words need to be present, a union can be tested. If both words must be formed from separate letters, a set of the concatenation can be tested.

Returning Words Within a String Array

I created a method to output a String. Using the split method and a for loop, I added each word in my sentence into a String array, replacxing the last two letters of each word with "ed". Now, my return statement should return each of the words. When I used System.out.print, it worked. When I use a return and call it in my main method, I get this output: "[Ljava.lang.String;#1b6235b"
The error seems so simple but I just don't know where I'm going worng. Any help would be appreciated.
Here is my method:
public String[] processInfo() {
String sentence = this.phrase;
String[] words = sentence.split(" ");
if (!this.phrase.equalsIgnoreCase("Fred")) {
for (int i = 0; i < words.length; i++) {
words[i] = words[i].substring(0, words[i].length() - 2).concat(
"ed ");
// System.out.print(words[i]);
}
}
return words;
}

You are printing arrays but arrays don't have a proper implementation of toString() method by default.
What you see is
"[Ljava.lang.String;#1b6235b"
This is [Ljava.lang.String; is the name for String[].class, the java.lang.Class representing the class of array of String followed by its hashCode.
In order to print the array you should use Arrays.toString(..)
System.out.println(Arrays.toString(myArray));
A good idea however, it returns my Strings in an Array format. My aim
is to return them back into sentence format. So for example, if my
input is, "Hey my name is Fred", it would output as, "Hed ed naed ed
Fred". Sorry, I forgot to add that it also seperates it with commas
when using Arrays.toString
Then you should modify your processInfo() returning a String or creating a new method that convert your String[] to a String.
Example :
//you use like this
String [] processInfoArray = processInfo();
System.out.println(myToString(processInfoArray));
// and in another part you code something like this
public static String myToString(String[] array){
if(array == null || array.length == 0)
return "";
StringBuilder sb = new StringBuilder();
for(int i=0;i<array.length-1;i++){
sb.append(array[i]).append(" ");
}
return sb.append(array[array.length -1]).toString();
}

As much as I can get from your question and comment is that your aim is to return them back into sentence format. So for example, if your input is, "Hey my name is Fred", it would output as, "Hed ed naed ed Fred".
In that case you should return a String, and not an array. I have modified your method a bit to do so. Let me know if you wanted something else.
public String processInfo() {
String sentence = this.phrase;
String[] words = sentence.split(" ");
if (!this.phrase.equalsIgnoreCase("Fred")) {
sentence = "";
for (int i = 0; i < words.length; i++) {
words[i] = words[i].substring(0, words[i].length() - 2).concat(
"ed ");
sentence += " " + words[i];
// System.out.print(words[i]);
}
}
return sentence.trim();
}

Your commented out call to System.out.print is printing each element of the array from inside the loop. Your method is returning a String[]. When you try to print an array, you will get the java representation of the array as you are seeing. You either need to change your method to build and return a string with all the array entries concatenated together, or your calling code needs to loop through the returned array and print each entry.