I have a String
String str = (a AND b) OR (c AND d)
I tokenise with the help of code below
String delims = "AND|OR|NOT|[!&|()]+"; // Regular expression syntax
String newstr = str.replaceAll(delims, " ");
String[] tokens = newstr.trim().split("[ ]+");
and get String[] below
[a, b, c, d]
To each element of the array I add " =1" so it becomes
[a=1, b=1, c=1, d=1]
NOW I need to replace these values to the initial string making it
(a=1 AND b=1) OR (c=1 AND d=1)
Can someone help or guide me ? The initial String str is arbitrary!
This answer is based on #Michael's idea (BIG +1 for him) of searching words containing only lowercase characters and adding =1 to them :)
String addstr = "=1";
String str = "(a AND b) OR (c AND d) ";
StringBuffer sb = new StringBuffer();
Pattern pattern = Pattern.compile("[a-z]+");
Matcher m = pattern.matcher(str);
while (m.find()) {
m.appendReplacement(sb, m.group() + addstr);
}
m.appendTail(sb);
System.out.println(sb);
output
(a=1 AND b=1) OR (c=1 AND d=1)
Given:
String str = (a AND b) OR (c AND d);
String[] tokened = [a, b, c, d]
String[] edited = [a=1, b=1, c=1, d=1]
Simply:
for (int i=0; i<tokened.length; i++)
str.replaceAll(tokened[i], edited[i]);
Edit:
String addstr = "=1";
String str = "(a AND b) OR (c AND d) ";
String delims = "AND|OR|NOT|[!&|() ]+"; // Regular expression syntax
String[] tokens = str.trim().split( delims );
String[] delimiters = str.trim().split( "[a-z]+"); //remove all lower case (these are the characters you wish to edit)
String newstr = "";
for (int i = 0; i < delimiters.length-1; i++)
newstr += delimiters[i] + tokens[i] + addstr;
newstr += delimiters[delimiters.length-1];
OK now the explanation:
tokens = [a, b, c, d]
delimiters = [ "(" , " AND " , ") OR (" , " AND " , ") " ]
When iterating through delimiters, we take "(" + "a" + "=1".
From there we have "(a=1" += " AND " + "b" + "=1".
And on: "(a=1 AND b=1" += ") OR (" + "c" + "=1".
Again : "(a=1 AND b=1) OR (c=1" += " AND " + "d" + "=1"
Finally (outside the for loop): "(a=1 AND b=1) OR (c=1 AND d=1" += ")"
There we have: "(a=1 AND b=1) OR (c=1 AND d=1)"
How long is str allowed to be? If the answer is "relatively short", you could simply do a "replace all" for every element in the array. This obviously is not the most performance-friendly solution, so if performance is an issue, a different solution would be desireable.
Related
My String:
BByTTheWay .I want to split the string as B By T The Way BByTheWay .That means I want to split string if I get any capital letters and last put the main string as it is. As far I tried in java:
public String breakWord(String fileAsString) throws FileNotFoundException, IOException {
String allWord = "";
String allmethod = "";
String[] splitString = fileAsString.split(" ");
for (int i = 0; i < splitString.length; i++) {
String k = splitString[i].replaceAll("([A-Z])(?![A-Z])", " $1").trim();
allWord = k.concat(" " + splitString[i]);
allWord = Arrays.stream(allWord.split("\\s+")).distinct().collect(Collectors.joining(" "));
allmethod = allmethod + " " + allWord;
// System.out.print(allmethod);
}
return allmethod;
}
It givs me the output: B ByT The Way BByTTheWay . I think stackoverflow community help me to solve this.
You may use this code:
Code 1
String s = "BByTTheWay";
Pattern p = Pattern.compile("\\p{Lu}\\p{Ll}*");
String out = p.matcher(s)
.results()
.map(MatchResult::group)
.collect(Collectors.joining(" "))
+ " " + s;
//=> "B By T The Way BByTTheWay"
RegEx \\p{Lu}\\p{Ll}* matches any unicode upper case letter followed by 0 or more lowercase letters.
CODE DEMO
Or use String.split using same regex and join it back later:
Code 2
String out = Arrays.stream(s.split("(?=\\p{Lu})"))
.collect(Collectors.joining(" ")) + " " + s;
//=> "B By T The Way BByTTheWay"
Use
String s = "BByTTheWay";
Pattern p = Pattern.compile("[A-Z][a-z]*");
Matcher m = p.matcher(s);
String r = "";
while (m.find()) {
r = r + m.group(0) + " ";
}
System.out.println(r + s);
See Java proof.
Results: B By T The Way BByTTheWay
EXPLANATION
--------------------------------------------------------------------------------
[A-Z] any character of: 'A' to 'Z'
--------------------------------------------------------------------------------
[a-z]* any character of: 'a' to 'z' (0 or more
times (matching the most amount possible))
As per requirements, you can write in this way checking if a character is an alphabet or not:
char[] chars = fileAsString.toCharArray();
StringBuilder fragment = new StringBuilder();
for (char ch : chars) {
if (Character.isLetter(ch) && Character.isUpperCase(ch)) { // it works as internationalized check
fragment.append(" ");
}
fragment.append(ch);
}
String.join(" ", fragment).concat(" " + fileAsString).trim(); // B By T The Way BByTTheWay
I have a function which reads stop words from a file and saves it in a HashSet.
HashSet<String> hset = readFile();
This is my string
String words = "the plan crash is invisible";
I am trying to remove all the stop words from the string but it is not working correctly
The output i am getting: plan crash invible
Output i want => plan crash invisible
Code:
HashSet<String> hset = readFile();
String words = "the plan crash is invisible";
String s = words.toLowerCase();
String[] split = s.split(" ");
for(String str: split){
if (hset.contains(str)) {
s = s.replace(str, "");
} else {
}
}
System.out.println("\n" + "\n" + s);
While hset.contains(str) matches full words, s.replace(str, ""); can replace occurrences of the "stop" words which are part of words of the input String. Hence "invisible" becomes "invible".
Since you are iterating over all the words of s anyway, you can construct a String that contains all the words not contained in the Set:
StringBuilder sb = new StringBuilder();
for(String str: split){
if (!hset.contains(str)) {
if (sb.length() > 0) {
sb.append(' ');
}
sb.append(str);
}
}
System.out.println("\n" + "\n" + sb.toString());
No need so check if your string contain the stop word or split your string, you can use replaceAll which use regex, like this :
for (String str : hset) {
s = s.replaceAll("\\s" + str + "|" + str + "\\s", " ");
}
Excample :
HashSet<String> hset = new HashSet<>();
hset.add("is");
hset.add("the");
String words = "the plan crash is invisible";
String s = words.toLowerCase();
for (String str : hset) {
s = s.replaceAll("\\s" + str + "|" + str + "\\s", " ");
}
s = s.replaceAll("\\s+", " ").trim();//comment and idea of #davidxxx
System.out.println(s);
This can gives you :
plan crash invisible
I am trying to split an inputted number such as (123) 456-7890.
String [] split = s.split(delimiters);
I have been searching the web for ways of delimiting the area code inside the set of the parentheses but I haven't found anything that works for my case. I do not know if the array is messing up with it printing either. The array is not required but I did not know what else to do since it is required to use the split method.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class HelloWorld{
public static void main(String[] args){
String phoneNumber = "(123)-456-7890";
String pattern = "\\((\\d+)\\)-(\\d+)-(\\d+)";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(phoneNumber);
if (m.find())
System.out.println(m.group(1) + " " + m.group(2) + " " + m.group(3));
}
}
You can try it here.
If I understand your question, you could use a Pattern like (XXX) XXX-XXXX where X is a digit. You can also use {n} to require n occurences. You group with (). Something like,
String str = "(123) 456-7890";
Pattern p = Pattern.compile("\\((\\d{3})\\) (\\d{3})-(\\d{4})");
Matcher m = p.matcher(str);
if (m.matches()) {
String areaCode = m.group(1);
String first3digits = m.group(2);
String last4digits = m.group(3);
System.out.printf("(%s) %s-%s%n", areaCode, first3digits,
last4digits);
}
Gives your requested output of
(123) 456-7890
or, if you must use split you might first remove the ( and ) with a call to replaceAll and something like
String str = "(123) 456-7890";
String[] arr = str.replaceAll("[()]", "").split("[ -]");
System.out.printf("(%s) %s-%s%n", arr[0], arr[1], arr[2]);
which also gives your requested output of
(123) 456-7890
this works:
String s= "(123) 456-7890";
String[] parts = s.split("[()\\- ]");
System.out.println("(" + parts[1] + ") " + parts[3] + "-" + parts[4]);
If you must use the split method:
String s= "(123) 456-7890"
String[] split = s.split("[()-]");
System.out.println("(" + split[1] + ")" + split[2] + "-" + split[3]);
I wonder why the double quotation marks is not shown in the actual output - just after the equal sign:
String word = "" + c1 + c2 + "ll";
The full code as follows:
public class InstantMethodsIndexOf
{
public void start()
{
String greeting = new String ("Hello World");
int position = greeting.indexOf('r');
char c1 = greeting.charAt(position + 2);
char c2 = greeting.charAt(position - 1);
**String word = "" + c1 + c2 + "ll";**
System.out.println(word);
}
}
When you pass "" to a String you are passing an empty String. You need to escape the quotation with a back slash if you want to print them.
Example:
String word = "\"" + c1 + c2 + "ll\"";
then System.out.println(word) will print:
"Hell"
As you can see I am escaping one double quotation at the beginning and another at the end
(Assuming c1 == 'H' and c2 == 'e')
The quotation mark does not appear because you have none being printed. What you have is an empty string being concatenated with other contents.
If you need the quotation mark, then you shoud do the following:
String word = "\"" + c1 + c2 + "ll";
It's a way to let Java know that it will be a string straight from the beginning, since "" is a String object of an empty string.
In your code, it doesn't really look useful. But following is an example where it would be:
int a=10, b=20;
String word = a + b + "he"; // word = "30he"
String word2 = "" + a + b + "he"; // word2 = "1020he"
I wonder why the double quotation marks is not shown in the actual
output - just after the equal sign:
String word = "" + c1 + c2 + "ll";
You are declaring a String that concatenates:
The empty String ""
c1
c2
The String literal "ll"
To show the quotes and make the code easier to read, try:
String word = '\u0022' + c1 + c2 + "ll"
which uses the unicode character value to print the double quote
I wonder why the double quotation marks is not shown in the actual
output - just after the equal sign:
In java String represented by the use of double quotes means the data between double quotes is considered as String value but if you want to include double quotes you have to use escape character \".
Moreover I suggest you to use StringBuilder and append your characters and String into it and use toString to print.
String str="ABC";//So value of String literal is ABC not "ABC"
String empty="";//is just empty but NOT Null
String quote="\"";//Here String has value " (One Double Quote)
This code
String greeting = "Hello World"; // <-- no need for new String()
int position = greeting.indexOf('r'); // <-- 8
char c1 = greeting.charAt(position + 2); // <-- 'd'
char c2 = greeting.charAt(position - 1); // <-- 'o'
String word = "" + c1 + c2 + "ll"; // <-- "" + 'd' + 'o' + "ll"
The empty String "" is used to coerce the arithmetic to a String, so it could also be written as
StringBuilder sb = new StringBuilder();
sb.append(c1).append(c2).append("ll");
String word = sb.toString();
or
StringBuilder sb = new StringBuilder("ll");
sb.insert(0, c2);
sb.insert(0, c1);
String word = sb.toString();
If you wanted to include double quotes in your word, your could escape them with a \\ or use a character -
char c1 = greeting.charAt(position + 2); // <-- 'd'
char c2 = greeting.charAt(position - 1); // <-- 'o'
String word = "\"" + c1 + c2 + "ll\""; // <-- "\"" + 'd' + 'o' + "ll\""
or
String word = "" + '"' + c1 + c2 + "ll" + '"';
I want to execute a query like
select ID from "xyz_DB"."test" where user in ('a','b')
so the corresponding code is like
String s="(";
for(String user:selUsers){
s+= " ' " + user + " ', ";
}
s+=")";
Select ID from test where userId in s;
The following code is forming the value of s as ('a','b',)
i want to remove the comma after the end of array how to do this ?
Here is one way to do this:
String s = "(";
boolean first = true;
for(String user : selUsers){
if (first) {
first = false;
} else {
s += ", ";
}
s += " ' " + user + " '";
}
s += ")";
But it is more efficient to use a StringBuilder to assemble a String if there is looping involved.
StringBuilder sb = new StringBuilder("(");
boolean first = true;
for(String user : selUsers){
if (first) {
first = false;
} else {
sb.append(", ");
}
sb.append(" ' ").append(user).append(" '");
}
sb.append(")");
String s = sb.toString();
This does the trick.
String s = "";
for(String user : selUsers)
s += ", '" + user + "'";
if (selUsers.size() > 0)
s = s.substring(2);
s = "(" + s + ")";
But, a few pointers:
When concatenating strings like this, you are advised to work with StringBuilder and append.
If this is part of an SQL-query, you probably want to sanitize the user-names. See xkcd: Exploits of a Mom for an explanation.
For fun, a variation of Stephen C's answer:
StringBuilder sb = new StringBuilder("(");
boolean first = true;
for(String user : selUsers){
if (!first || (first = false))
sb.append(", ");
sb.append('\'').append(user).append('\'');
}
sb.append(')');
you could even do the loop it like this :-)
for(String user : selUsers)
sb.append(!first || (first=false) ? ", \'" : "\'").append(user).append('\'');
Use the 'old style' of loop where you have the index, then you add the comma on every username except the last:
String[] selUsers = {"a", "b", "c"};
String s="(";
for(int i = 0; i < selUsers.length; i++){
s+= " ' " + selUsers[i] + " ' ";
if(i < selUsers.length -1){
s +=" , ";
}
}
s+=")";
But as others already mentioned, use StringBuffer when concatenating strings:
String[] selUsers = {"a", "b", "c"};
StringBuffer s = new StringBuffer("(");
for(int i = 0; i < selUsers.length; i++){
s.append(" ' " + selUsers[i] + " ' ");
if(i < selUsers.length -1){
s.append(" , ");
}
}
s.append(")");
Use StringUtils.join from apache commons.
Prior to adding the trailing ')' I'd strip off the last character of the string if it's a comma, or perhaps just replace the trailing comma with a right parenthesis - in pseudo-code, something like
if s.last == ',' then
s = s.left(s.length() - 1);
end if;
s = s + ')';
or
if s.last == ',' then
s.last = ')';
else
s = s + ')';
end if;
Share and enjoy.
i would do s+= " ,'" + user + "'"; (place the comma before the value) and add a counter to the loop where i just do s = "'" + user + "'"; if the counter is 1 (or 0, depending on where you start to count).
(N.B. - I'm not a Java guy, so the syntax may be wrong here - apologies if it is).
If selUsers is an array, why not do:
selUsers.join(',');
This should do what you want.
EDIT:
Looks like I was wrong - I figured Java had this functionality built-in. Looks like the Apache project has something that does what I meant, though. Check out this SO answer: Java: function for arrays like PHP's join()?
I fully support Stephen C's answer - that's the one I wanted to suggest aswell: simply avoid adding the additional comma.
But in your special case, the following should work too:
s = s.replace(", \\)", ")");
It simply replaces the substring ", )" with a single bracket ")".
Java 1.4+
s = s.replaceFirst("\\, \\)$", ")");
Edited: I forgot last space before parethesis
StringBuilder has a perfectly good append(int) overload.
String [] array = {"1","2","3" ...};
StringBuilder builder = new StringBuilder();
builder.append(s + "( ")
for(String i : array)
{
if(builder.length() != 0)
builder.append(",");
builder.append(i);
}
builder.append(" )")
Answer shamelessly copied from here