Develop Reference

How can i remove a particular comma at end

I have an output like
1054273,
1148244,
1174481,
1175759,
1180656,
1181151,
I need to remove the comma at the end.
I tried the code :
str = str.replaceAll(", $", "");
But the output shows like all the commas were removed.
Can anyone help me to solve this??
Thanks,
SK

String text = "1054273, 1148244, 1174481, 1175759, 1180656, 1181151,".trim();
if(text.endsWith(","))
text = text.substring(0, text.lastIndexOf(","))
The original text is trimmed to ensure that you don't have trailing space. Assuming you have a valid-length string.

If you want to remove a character, why are you replacing it?
Say that you want to remove the last character, because, in fact, the comma is at the end (and I guess it will always be):
str = str.substring(0, str.length() - 1);

you can use substring like this
String str = "1054273, 1148244, 1174481, 1175759, 1180656, 1181151,";
String output = str.substring(0, str.length()-1);
System.out.println(output);
output is :
1054273, 1148244, 1174481, 1175759, 1180656, 1181151
or if you want to create a custom output you'd better use split and create an array of string like this :
String[] outputArrays = str.split(",");
Arrays.asList(outputArrays).stream().forEach(System.out::println);
output is (and you free to change it):
1054273
1148244
1174481
1175759
1180656
1181151

In your case you can simply remove the last char.
With substring :
str = str.substring(0, str.length() - 1);
With regex :
str = str.replaceAll(".$", "");

why split() produces extra , after sets limit -1

I want to split Area Code and preceding number from Telephone number without brackets so i did this.
String pattern = "[\\(?=\\)]";
String b = "(079)25894029".trim();
String c[] = b.split(pattern,-1);
for (int a = 0; a < c.length; a++)
System.out.println("c[" + a + "]::->" + c[a] + "\nLength::->"+ c[a].length());
Output:
c[0]::-> Length::->0
c[1]::->079 Length::->3
c[2]::->25894029 Length::->8
Expected Output:
c[0]::->079 Length::->3
c[1]::->25894029 Length::->8
So my question is why split() produces and extra blank at the start, e.g
[, 079, 25894029]. Is this its behavior, or I did something go wrong here?
How can I get my expected outcome?

First you have unnecessary escaping inside your character class. Your regex is same as:
String pattern = "[(?=)]";
Now, you are getting an empty result because ( is the very first character in the string and split at 0th position will indeed cause an empty string.
To avoid that result use this code:
String str = "(079)25894029";
toks = (Character.isDigit(str.charAt(0))? str:str.substring(1)).split( "[(?=)]" );
for (String tok: toks)
System.out.printf("<<%s>>%n", tok);
Output:
<<079>>
<<25894029>>

From the Java8 Oracle docs:
When there is a positive-width match at the beginning of this string
then an empty leading substring is included at the beginning of the
resulting array. A zero-width match at the beginning however never
produces such empty leading substring.
You can check that the first character is an empty string, if yes then trim that empty string character.

Your regex has problems, as does your approach - you can't solve it using your approach with any regex. The magic one-liner you seek is:
String[] c = b.replaceAll("^\\D+|\\D+$", "").split("\\D+");
This removes all leading/trailing non-digits, then splits on non-digits. This will handle many different formats and separators (try a few yourself).
See live demo of this:
String b = "(079)25894029".trim();
String[] c = b.replaceAll("^\\D+|\\D+$", "").split("\\D+");
System.out.println(Arrays.toString(c));
Producing this:
[079, 25894029]

How can I split a string except when the delimiter is protected by quotes or brackets?

I asked How to split a string with conditions. Now I know how to ignore the delimiter if it is between two characters.
How can I check multiple groups of two characters instead of one?
I found Regex for splitting a string using space when not surrounded by single or double quotes, but I don't understand where to change '' to []. Also, it works with two groups only.
Is there a regex that will split using , but ignore the delimiter if it is between "" or [] or {}?
For instance:
// Input
"text1":"text2","text3":"text,4","text,5":["text6","text,7"],"text8":"text9","text10":{"text11":"text,12","text13":"text14","text,15":["text,16","text17"],"text,18":"text19"}
// Output
"text1":"text2"
"text3":"text,4"
"text,5":["text6","text,7"]
"text8":"text9"
"text10":{"text11":"text,12","text13":"text14","text,15":["text,16","text17"],"text,18":"text19"}

You can use:
text = "\"text1\":\"text2\",\"text3\":\"text,4\",\"text,5\":[\"text6\",\"text,7\"],\"text8\":\"text9\",\"text10\":{\"text11\":\"text,12\",\"text13\":\"text14\",\"text,15\":[\"text,16\",\"text17\"],\"text,18\":\"text19\"}";
String[] toks = text.split("(?=(?:(?:[^\"]*\"){2})*[^\"]*$)(?![^{]*})(?![^\\[]*\\]),+");
for (String tok: toks)
System.out.printf("%s%n", tok);
- RegEx Demo
OUTPUT:
"text1":"text2"
"text3":"text,4"
"text,5":["text6","text,7"]
"text8":"text9"
"text10":{"text11":"text,12","text13":"text14","text,15":["text,16","text17"],"text,18":"text19"}

java string split based on new line

I have following string
String str="aaaaaaaaa\n\n\nbbbbbbbbbbb\n \n";
I want to break it on \n so at the end i should two string aaaaaaaa and bbbbbbbb. I dont want last one as it only contain white space. so if i split it based on new line character using str.split() final array should have two entry only.
I tried below:
String str="aaaaaaaaa\n\n\nbbbbbbbbbbb\n \n".replaceAll("\\s+", " ");
String[] split = str.split("\n+");
it ignore all \n and give single string aaaaaaaaaa bbbbbbbb.

Delete the call to replaceAll(), which is removing the newlines too. Just this will do:
String[] split = str.split("\n\\s*");
This will not split on just spaces - the split must start at a newline (followed by optional further whitespace).
Here's some test code using your sample input with edge case enhancement:
String str = "aaaaaaaaa\nbbbbbb bbbbb\n \n";
String[] split = str.split("\n\\s*");
System.out.println(Arrays.toString(split));
Output:
[aaaaaaaaa, bbbbbb bbbbb]

This should do the trick:
String str="aaaaaaaaa\n\n\nbbbbbbbbbbb\n \n";
String[] lines = str.split("\\s*\n\\s*");
It will also remove all trailing and leading whitespace from all lines.

The \ns are removed by your first statement: \s matches \n

Remove all spaces and punctuation (anything not a letter) from a string?

In Java, how can I take a string as a parameter, and then remove all punctuation and spaces and then convert the rest of the letters to uppercase?
Example 1:
Input: How's your day going?
Output: HOWSYOURDAYGOING
Example 2:
Input: What's your name again?
Output: WHATSYOURNAMEAGAIN

This should do the trick
String mystr= "How's your day going?";
mystr = mystr.replaceAll("[^A-Za-z]+", "").toUpperCase();
System.out.println(mystr);
Output:
HOWSYOURDAYGOING
The regex [^A-Za-z]+ means one or more characters that do not match anything in the range A-Za-z, and we replace them with the empty string.

String yourString = "How's your day going";
yourString=yourString.replaceAll("\\s+",""); //remove white space
yourString=yourString.replaceAll("[^a-zA-Z ]", ""); //removes all punctuation
yourString=yourString.toUpperCase(); //convert to Upper case

I did it with
inputText = inputText.replaceAll("\\s|[^a-zA-Z0-9]","");
inputText.toUpper(); //and later uppercase the complete string
Though #italhourne 's answer is correct but you can just reduce it in single step by just removing the spaces as well as keeping all the characters from a-zA-Z and 0-9, in a single statement by adding "or".
Just a help for those who need it!!

public static String repl1(String n){
n = n.replaceAll("\\p{Punct}|\\s","");
return n;
}

Well, I did it the long way, take a look if you want. I used the ACII code values (this is my main method, transform it to a function on your own).
String str="How's your day going?";
char c=0;
for(int i=0;i<str.length();i++){
c=str.charAt(i);
if(c<65||(c>90&&c<97)||(c>122)){
str=str.replace(str.substring(i,i+1) , "");
}
}
str=str.toUpperCase();
System.out.println(str);