Develop Reference

How to add prefix to each word in a given string

I want to add prefix to every word in a given string. My code is :-
StringBuilder strColsToReturns = new StringBuilder();
String strPrefix = "abc.";
strColsToReturns.append(String.format(" %sId, %sname, %stype,", strPrefix, strPrefix, strPrefix));
This is fine, for small string, but I have a very large static string like this. So, this method of adding string prefix looks like a tedious method. Is there any other sophisticated way to achieve this.

You can use replaceAll(regex, replacement). As regex we may want to select place which has
space or start of string before it
and non-space after it.
(we can use look-around mechanisms here)
So something like this should work:
String replaced = originalString.replaceAll("(?<=\\s|^)(?=\\S)", "abc.");
But if replacement can contain $ or \ characters you will need to escape them with \, or with Matcher.quoteReplacement(replacement) method.
String originalString = "foo bar baz";
String replaced = originalString.replaceAll("(?<=\\s|^)(?=\\S)", Matcher.quoteReplacement("abc."));
System.out.println(replaced);
Output: abc.foo abc.bar abc.baz

Regex Java when we have specific text upto a pattern

As i haven't much worked on regex, can someone help me out in getting the answer for below thing:
(1)I want to remove a text say Element
(2)It may of may not followed by delimiter say pipe(||)
I tried below thing, but it is not working in the way i want:
String str = "String:abc||Element:abc||Value:abc"; // Sample text 1
String str1 = "String:abc||Element:abc"; // Sample text 2
System.out.println(str.replaceFirst("Element.*\\||", ""));
System.out.println(str1.replaceFirst("Element.*\\||", ""));
Required output in above cases:
String:abc||Value:abc //for the first case
String:abc //for the second case

Assuming that you can decide to give another value to the original pattern which is Element in this case, you can use Pattern.quote to escape it as below:
String str = "String:abc||Element:abc||Value:abc"; // Sample text 1
String str1 = "String:abc||Element:abc"; // Sample text 2
String originalPattern = "Element";
String pattern = String.format("\\|{2}%s[^\\|]+", Pattern.quote(originalPattern));
System.out.println(str.replaceFirst(pattern, ""));
System.out.println(str1.replaceFirst(pattern, ""));
Your patter is then generic and its value is String.format("\\|{2}%s[^\\|]+", Pattern.quote(originalPattern))
Output:
String:abc||Value:abc
String:abc

You put the escape wrong. It should be:
Element(.*?\|\||.*$)
Put the escape on each pipe, and use ? for non greedy Regex so you only replace just enough string, not everything.

String text = "String:abc||Element:abc||Value:abc";
text = text.replaceAll("\\belement\\b", "");
you might need to use replace all this will replace all element from your string here i am using '\b' word boundary in java regular expression in between the words

Remove all characters before special character in android

I have to remove ## before $$ from string ##$$abxcyhshbhs##xcznbx##. I am using:
string.split("\\#");
The problem is that it also removes # after $$.

Use replace() instead.
String text = "##$$abxcyhshbhs##xcznbx##";
text = text.replace("##$$", "$$");

You can use substring method like below
string.substring(2);

If you really want to use String.split() you can do what you want by limiting the number of results by doing:
String str = "##$$abxcyhshbhs##xcznbx##";
str = str.split("##", 2)[1];
I don't know your exact issue but as has already been said, replace() or substring() is probably a better option.

If you have unknown number of # symbols before $$ and they appear not just at the beginning of the string, you can use the following replaceAll with a regex:
String re = "#+\\${2}";
String str = "##$$abxcyh###$$shbhs##xcznbx##";
System.out.println(str.replaceAll(re, "\\$\\$")); // Note escaped $ !!!
// => $$abxcyh$$shbhs##xcznbx##
// or
re = "#+(\\${2})"; // using capturing and back-references
System.out.println(str.replaceAll(re, "$1"));
See IDEONE demo.
Do not forget to assign the variable a new value when using in your code:
str = str.replaceAll("#+(\\${2})", "$1")

If your purpose is to remove ## from first occurrence of ##$$ in the string, then following code snippet will be helpful:
if(yourString.startsWith("##$$")){
yourString.replaceFirst("##$$","$$");
}
OR considering there is only single $$ in your string, following would be helpful:
String requiredString="";
String[] splitArr = yourString.split("\\$");
if ( splitArr.length > 1 ) {
requiredString = "$$" + splitArr[splitArr.length-1];
}
I have written a code snippet here. You can make changes and execute on your own.

To literally remove the first two characters, use the following:
String s = "##$$abxcyhshbhs##xcznbx##";
s.substring(2, s.length());
This doesn't do any pattern matching to look for the $$.

Extracting numbers into a string array

I have a string which is of the form
String str = "124333 is the otp of candidate number 9912111242.
Please refer txn id 12323335465645 while referring blah blah.";
I need 124333, 9912111242 and 12323335465645 in a string array. I have tried this with
while (Character.isDigit(sms.charAt(i)))
I feel that running the above said method on every character is inefficient. Is there a way I can get a string array of all the numbers?

Use a regex (see Pattern and matcher):
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(<your string here>);
while (m.find()) {
//m.group() contains the digits you want
}
you can easily build ArrayList that contains each matched group you find.
Or, as other suggested, you can split on non-digits characters (\D):
"blabla 123 blabla 345".split("\\D+")
Note that \ has to be escaped in Java, hence the need of \\.

You can use String.split():
String[] nbs = str.split("[^0-9]+");
This will split the String on any group of non-numbers digits.

And this works perfectly for your input.
String str = "124333 is the otp of candidate number 9912111242. Please refer txn id 12323335465645 while referring blah blah.";
System.out.println(Arrays.toString(str.split("\\D+")));
Output:
[124333, 9912111242, 12323335465645]
\\D+ Matches one or more non-digit characters. Splitting the input according to one or more non-digit characters will give you the desired output.

Java 8 style:
long[] numbers = Pattern.compile("\\D+")
.splitAsStream(str)
.mapToLong(Long::parseLong)
.toArray();
Ah if you only need a String array, then you can just use String.split as the other answers suggests.

Alternatively, you can try this:
String str = "124333 is the otp of candidate number 9912111242. Please refer txn id 12323335465645 while referring blah blah.";
str = str.replaceAll("\\D+", ",");
System.out.println(Arrays.asList(str.split(",")));
\\D+ matches one or more non digits
Output
[124333, 9912111242, 12323335465645]

First thing comes into my mind is filter and split, then i realized that it can be done via
String[] result =str.split("\\D+");
\D matches any non-digit character, + says that one or more of these are needed, and leading \ escapes the other \ since \D would be parsed as 'escape character D' which is invalid

Split the string

abcd+xyz
i want to split the string and get left and right components with respect to "+"
that is i need to get abcd and xyz seperatly.
I tried the below code.
String org = "abcd+xyz";
String splits[] = org.split("+");
But i am getting null value for splits[0] and splits[1]...
Please help..

The string you send as an argument to split() is interpreted as a regex (documentation for split(String regex)). You should add an escape character before the + sign:
String splits[] = org.split("\\+");
You might also find the Summary of regular-expression constructs worth reading :)

"+" is wild character for regular expression.
So just do
String splits[] = org.split("\\+");
This will work

the expression "+" means one or many in java regular expression.
split takes Regex as a argument hence the comparion given by you fails
So use
String org = "abcd+xyz";
String splits[] = org.split(""\+");
regards!!

Try:
String splits[] = org.split("\\+");