I have one string which i need to divide into two parts using regex
String string = "2pbhk";
This string i need to divide into 2p and bhk
More over second part should always be bhk or rk, as strings can be one of 1bhk, 5pbhk etc
I have tried
String pattern = ([^-])([\\D]*);
You can use the following regex "(?=bhk|rk)" with split.
str.split("(?=bhk|rk)");
This will split it if there is one of bhk or rk.
This should do the trick:
(.*)(bhk|rk)
First capture holds the "number" part, and the second bhk OR rk.
Regards
String string = "2pbhk";
String first_part, second_part = null;
if(string.contains("bhk")){
first_part = string.substring(0, string.indexOf("bhk"));
second_part = "bhk";
}
else if(string.contains("rk")){
first_part = string.substring(0, string.indexOf("rk"));
second_part = "rk";
}
Try the above once, not using regex but should work.
In case you are looking to split strings that end with rk or bhk but not necessarily at the end of the string (i.e. at the word boundaries), you need to use a regex with \\b:
String[] arr = "5ddddddpbhk".split("(?=(?:rk|bhk)\\b)");
System.out.println(Arrays.toString(arr));
If you want to allow splitting inside a longer string, remove the \\b.
If you only split individual words, use $ instead of \\b (i.e. end of string):
(?=(?:rk|bhk)$)
Here is my IDEONE demo
Related
I have this original string and I want to insert new string between two dots of original string. I did it this way, but having errors.
String originalString ="asdASfasdlpe.hereNeedToPutNewString.asdasfdfepw";
String stringForReplace = "NewString";
String new = originalString.replace(originalString.substring(originalString.indexOf(".") + 1), stringForReplace);
it gives me: "asdASfasdlpe.NewString"
Result should be: "asdASfasdlpe.NewString.asdasfdfepw"
I would do it like so.
from the question it looks like you want to replace the first occurrence so use replaceFirst
(?<=\\.) - look behind assertion - so start with following character
(?=\\.) - look ahead assertion - so end prior to that
.*? - reluctant quantifier to limit to just characters between two periods. Use * in case you have two adjacent periods since the string could be empty.
String s = "first.oldstring.third.fourth.fifth";
String n = "second";
s = s.replaceFirst("(?<=\\.).*?(?=\\.)",n);
System.out.println(s);
prints
first.second.third.fourth.fifth
String originalString ="asdASfasdlpe.hereNeedToPutNewString.asdasfdfepw";
String stringForReplace = "NewString";
String a[]=originalString.split("[.]");
String newString="";
if(a.length==3) {
newString=originalString.replace(a[1], stringForReplace);
}
System.out.println(newString);
Or with ternary operator:
newString=(a.length== 3 ? originalString.replace(a[1], stringForReplace):null);
System.out.println(newString);
One shorter solution is to use regex with a lookahead and lookbehind
String replaced = originalString.replaceAll("(?<=\\.).+(?=\\.)", stringForReplace);
The problem with your code is due to using this particular piece of code:
originalString.substring(originalString.indexOf(".") + 1)
The reason is that indexof() function will only give the index on which "." was found on, and substring will only know where to start taking the substring from, but it wouldn't know where to end it.
Try this:
String originalString ="asdASfasdlpe.hereNeedToPutNewString.asdasfdfepw";
String stringForReplace = "NewString";
String newString = originalString.replace(originalString.split("[.]", 3)[1], stringForReplace);
System.out.println(newString);
The split function in this piece of code will break the whole string by "."
and you will have the string you want to replace available to you.
originalString.split("[.]", 3)[1]
You could try the following:
public static void main(String[] args) {
String originalString ="asdASfasdlpe.hereNeedToPutNewString.asdasfdfepw";
String stringForReplace = "NewString";
String newStr = originalString.replaceAll("(?<=\\.).*(?=\\.)", stringForReplace);
//using lookahead and lookbehind regex
String newStr2 = originalString.replaceAll("\\..*\\.", "."+stringForReplace+".");
System.out.println(newStr);
System.out.println(newStr2);
}
One option uses lookahead and lookbehinds, you could opt to not use that if it is not supported.
Output:
asdASfasdlpe.NewString.asdasfdfepw
asdASfasdlpe.NewString.asdasfdfepw
Here You go:
String new = originalString.replace(originalString.substring(originalString.indexOf(".") + 1), stringForReplace)+originalString.substring(originalString.indexOf(".",originalString.indexOf(".")+1),originalString.length());
What I have done is adding the resultant string to the new String.indexOf function takes another argument too, which is the position the search will start
I need to split a string based on a pattern and again i need to merge it back on a portion of string.
for ex: Below is the actual and expected strings.
String actualstr="abc.def.ghi.jkl.mno";
String expectedstr="abc.mno";
When i use below, i can store in a Array and iterate over to get it back. Is there anyway it can be done simple and efficient than below.
String[] splited = actualstr.split("[\\.\\.\\.\\.\\.\\s]+");
Though i can acess the string based on index, is there any other way to do this easily. Please advise.
You do not understand how regexes work.
Here is your regex without the escapes: [\.\.\.\.\.\s]+
You have a character class ([]). Which means there is no reason to have more than one . in it. You also don't need to escape .s in a char class.
Here is an equivalent regex to your regex: [.\s]+. As a Java String that's: "[.\\s]+".
You can do .split("regex") on your string to get an array. It's very simple to get a solution from that point.
I would use a replaceAll in this case
String actualstr="abc.def.ghi.jkl.mno";
String str = actualstr.replaceAll("\\..*\\.", ".");
This will replace everything with the first and last . with a .
You could also use split
String[] parts = actualString.split("\\.");
string str = parts[0]+"."+parts[parts.length-1]; // first and last word
public static String merge(String string, String delimiter, int... partnumbers)
{
String[] parts = string.split(delimiter);
String result = "";
for ( int x = 0 ; x < partnumbers.length ; x ++ )
{
result += result.length() > 0 ? delimiter.replaceAll("\\\\","") : "";
result += parts[partnumbers[x]];
}
return result;
}
and then use it like:
merge("abc.def.ghi.jkl.mno", "\\.", 0, 4);
I would do it this way
Pattern pattern = Pattern.compile("(\\w*\\.).*\\.(\\w*)");
Matcher matcher = pattern.matcher("abc.def.ghi.jkl.mno");
if (matcher.matches()) {
System.out.println(matcher.group(1) + matcher.group(2));
}
If you can cache the result of
Pattern.compile("(\\w*\\.).*\\.(\\w*)")
and reuse "pattern" all over again this code will be very efficient as pattern compilation is the most expensive. java.lang.String.split() method that other answers suggest uses same Pattern.compile() internally if the pattern length is greater then 1. Meaning that it will do this expensive operation of Pattern compilation on each invocation of the method. See java.util.regex - importance of Pattern.compile()?. So it is much better to have the Pattern compiled and cached and reused.
matcher.group(1) refers to the first group of () which is "(\w*\.)"
matcher.group(2) refers to the second one which is "(\w*)"
even though we don't use it here but just to note that group(0) is the match for the whole regex.
I have string like this String s="ram123",d="ram varma656887"
I want string like ram and ram varma so how to seperate string from combined string
I am trying using regex but it is not working
PersonName.setText(cursor.getString(cursor.getColumnIndex(cursor
.getColumnName(1))).replaceAll("[^0-9]+"));
The correct RegEx for selecting all numbers would be just [0-9], you can skip the +, since you use replaceAll.
However, your usage of replaceAll is wrong, it's defined as follows: replaceAll(String regex, String replacement). The correct code in your example would be: replaceAll("[0-9]", "").
You can use the following regex: \d for representing numbers. In the regex that you use, you have a ^ which will check for any characters other than the charset 0-9
String s="ram123";
System.out.println(s);
/* You don't need the + because you are using the replaceAll method */
s = s.replaceAll("\\d", ""); // or you can also use [0-9]
System.out.println(s);
To remove the numbers, following code will do the trick.
stringname.replaceAll("[0-9]","");
Please do as follows
String name = "ram varma656887";
name = name.replaceAll("[0-9]","");
System.out.println(name);//ram varma
alternatively you can do as
String name = "ram varma656887";
name = name.replaceAll("\\d","");
System.out.println(name);//ram varma
also something like given will work for you
String given = "ram varma656887";
String[] arr = given.split("\\d");
String data = new String();
for(String x : arr){
data = data+x;
}
System.out.println(data);//ram varma
i think you missed the second argument of replace all. You need to put a empty string as argument 2 instead of actually leaving it empty.
try
replaceAll(<your regexp>,"")
you can use Java - String replaceAll() Method.
This method replaces each substring of this string that matches the given regular expression with the given replacement.
Here is the syntax of this method:
public String replaceAll(String regex, String replacement)
Here is the detail of parameters:
regex -- the regular expression to which this string is to be matched.
replacement -- the string which would replace found expression.
Return Value:
This method returns the resulting String.
for your question use this
String s = "ram123", d = "ram varma656887";
System.out.println("s" + s.replaceAll("[0-9]", ""));
System.out.println("d" + d.replaceAll("[0-9]", ""));
Hi please help me out in getting regular expression for the
following requirement
I have string type as
String vStr = "Every 1 nature(s) - Universe: (Air,Earth,Water sea,Fire)";
String sStr = "Every 1 form(s) - Earth: (Air,Fire) ";
from these strings after using regex I need to get values as "Air,Earth,Water sea,Fire" and "Air,Fire"
that means after
String vStrRegex ="Air,Earth,Water sea,Fire";
String sStrRegex ="Air,Fire";
All the strings that are input will be seperated by ":" and values needed are inside brackets always
Thanks
The regular expression would be something like this:
: \((.*?)\)
Spelt out:
Pattern p = Pattern.compile(": \\((.*?)\\)");
Matcher m = p.matcher(vStr);
// ...
String result = m.group(1);
This will capture the content of the parentheses as the first capture group.
Try the following:
\((.*)\)\s*$
The ending $ is important, otherwise you'll accidentally match the "(s)".
If you have each string separately, try this expression: \(([^\(]*)\)\s*$
This would get you the content of the last pair of brackets, as group 1.
If the strings are concatenated by : try to split them first.
Ask yourself if you really need a regex. Does the text you need always appear within the last two parentheses? If so, you can keep it simple and use substring instead:
String vStr = "Every 1 nature(s) - Universe: (Air,Earth,Water sea,Fire)";
int lastOpeningParens = vStr.lastIndexOf('(');
int lastClosingParens = vStr.lastIndexOf(')');
String text = vStr.substring(lastOpeningParens + 1, lastClosingParens);
This is much more readable than a regex.
I assume that there are only whitespace characters between : and the opening bracket (:
Pattern regex = Pattern.compile(":\\s+\\((.+)\\)");
You'll find your results in capturing group 1.
Try this regex:
.*\((.*)\)
$1 will contain the required string
I need to split a string base on delimiter - and .. Below are my desired output.
AA.BB-CC-DD.zip ->
AA
BB
CC
DD
zip
but my following code does not work.
private void getId(String pdfName){
String[]tokens = pdfName.split("-\\.");
}
I think you need to include the regex OR operator:
String[]tokens = pdfName.split("-|\\.");
What you have will match:
[DASH followed by DOT together] -.
not
[DASH or DOT any of them] - or .
Try this regex "[-.]+". The + after treats consecutive delimiter chars as one. Remove plus if you do not want this.
You can use the regex "\W".This matches any non-word character.The required line would be:
String[] tokens=pdfName.split("\\W");
The string you give split is the string form of a regular expression, so:
private void getId(String pdfName){
String[]tokens = pdfName.split("[\\-.]");
}
That means to split on any character in the [] (we have to escape - with a backslash because it's special inside []; and of course we have to escape the backslash because this is a string). (Conversely, . is normally special but isn't special inside [].)
Using Guava you could do this:
Iterable<String> tokens = Splitter.on(CharMatcher.anyOf("-.")).split(pdfName);
For two char sequence as delimeters "AND" and "OR" this should be worked. Don't forget to trim while using.
String text ="ISTANBUL AND NEW YORK AND PARIS OR TOKYO AND MOSCOW";
String[] cities = text.split("AND|OR");
Result : cities = {"ISTANBUL ", " NEW YORK ", " PARIS ", " TOKYO ", " MOSCOW"}
pdfName.split("[.-]+");
[.-] -> any one of the . or - can be used as delimiter
+ sign signifies that if the aforementioned delimiters occur consecutively we should treat it as one.
I'd use Apache Commons:
import org.apache.commons.lang3.StringUtils;
private void getId(String pdfName){
String[] tokens = StringUtils.split(pdfName, "-.");
}
It'll split on any of the specified separators, as opposed to StringUtils.splitByWholeSeparator(str, separator) which uses the complete string as a separator
String[] token=s.split("[.-]");
It's better to use something like this:
s.split("[\\s\\-\\.\\'\\?\\,\\_\\#]+");
Have added a few other characters as sample. This is the safest way to use, because the way . and ' is treated.
Try this code:
var string = 'AA.BB-CC-DD.zip';
array = string.split(/[,.]/);
You may also specified regular expression as argument in split() method ..see below example....
private void getId(String pdfName){
String[]tokens = pdfName.split("-|\\.");
}
s.trim().split("[\\W]+")
should work.
you can try this way as split accepts varargs so we can pass multiple parameters as delimeters
String[]tokens = pdfName.split("-",".");
you can pass as many parameters that you want.
If you know the sting will always be in the same format, first split the string based on . and store the string at the first index in a variable. Then split the string in the second index based on - and store indexes 0, 1 and 2. Finally, split index 2 of the previous array based on . and you should have obtained all of the relevant fields.
Refer to the following snippet:
String[] tmp = pdfName.split(".");
String val1 = tmp[0];
tmp = tmp[1].split("-");
String val2 = tmp[0];
...