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]", ""));
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 have following string:
mydata ="\nmyName=ram\nmySalaryL=$2,256.00\n";
How to get my name and salary values using regex?
Try using Regex Capture Groups. They are placed in parenthesis and give you access to a matched substring.
https://docs.oracle.com/javase/tutorial/essential/regex/groups.html
A reasonable regex might be:
java.util.regex.Pattern regex = Pattern.compile("\nmyName=(\w+)\nmySalayL=$((\d{1,3},)*(\d{1,3})\.\d{2}))\n");
java.util.regex.Matcher match = regex.matcher(inputString);
if(match.matches()) {
String myName = match.group(1);
String mySalary = match.group(2);
}
Please pay attention that capture group 0 is the entire string and matches() needs to be called before accessing the capture groups, because it does the actual regex-matching.
No need to use regular expressions at all! It is enough to split by \n character and then by = to get those information:
String mydata ="\nmyName=ram\nmySalaryL=$2,256.00\n";
String[] arr = mydata.split("\n");
System.out.println(arr[1].split("=")[1]);
System.out.println(arr[2].split("=")[1]);
So, I'm trying to parse a String input in Java that contains (opening) square brackets. I have str.replace("\\[", ""), but this does absolutely nothing. I've tried replaceAll also, with more than one different regex, but the output is always unchanged. Part of me wonders if this is possibly caused by the fact that all my back-slash characters appear as yen symbols (ever since I added Japanese to my languages), but it's been that way for over a year and hasn't caused me any issues like this before.
Any idea what I might be doing wrong here?
Strings are immutable in Java. Make sure you re-assign the return value to the same String variable:
str = str.replaceAll("\\[", "");
For the normal replace method, you don't need to escape the bracket:
str = str.replace("[", "");
public String replaceAll(String regex, String replacement)
As shown in the code above, replaceAll method expects first argument as regular expression and hence you need to escape characters like "(", ")" etc (with "\") if these exists in your replacement text which is to be replaced out of the string. For example :
String oldString = "This is (stringTobeReplaced) with brackets.";
String newString = oldString.replaceAll("\\(stringTobeReplaced\\)", "");
System.out.println(newString); // will output "This is with brackets."
Another way of doing this is to use Pattern.quote("str") :
String newString = oldString.replaceAll(Pattern.quote("(stringTobeReplaced)"), "");
This will consider the string as literal to be replaced.
As always, the problem is not that "xxx doesn't work", it is that you don't know how to use it.
First things first:
a String is immutable; if you read the javadoc of .replace() and .replaceAll(), you will see that both specify that a new String instance is returned;
replace() accepts a string literal as its first argument, not a regex literal.
Which means that you probably meant to do:
str = str.replace("[", "");
If you only ever do:
str.replace("[", "");
then the new instance will be created but you ignore it...
In addition, and this is a common trap with String (the other being that .matches() is misnamed), in spite of their respective names, .replace() does replace all occurrences of its first argument with its second argument; the only difference is that .replaceAll() accepts a regex as a first argument, and a "regex aware" expression as its second argument; for more details, see the javadoc of Matcher's .replaceAll().
For it to work it has to be inside a method.
for example:
public class AnyClass {
String str = "gtrg4\r\n" + "grtgy\r\n" + "grtht\r\n" + "htrjt\r\n" + "jtyjr\r\n" + "kytht";
public String getStringModified() {
str.replaceAll("\r\n", "");
return str;
}
}
How to replace a sub-string in java of form " :[number]: "
example:
string="Hello:6:World"
After replacement,
HelloWorld
ss="hello:909:world";
do as below:
String value = ss.replaceAll("[:]*[0-9]*[:]*","");
You can use a regex to define your desired pattern
String pattern = "(:\d+:)";
string EXAMPLE_TEST = ':12:'
System.out.println(EXAMPLE_TEST.replaceAll(pattern, "text to replace with"));
should work depending on what exactly you want to replace...
Do like this
String s = ":6:";
s = s.replaceAll(":", "");
Edit 1: After the question was changed, one should use
:\d+:
and within Java
:\\d+:
This is the answer for replacing :: as well.
This is the regexp you should use:
:\d*:
Debuggex Demo
And here is a running JavaCode snipped:
String str = "Hello :4: World";
String s = str.replaceAll(":\\d*:","");
System.out.println(s);
One problem with replaceAll is often, that the corrected String is returned. The string object from which replaceAll was called is not modified.
What's the difference between java.lang.String 's replace() and replaceAll() methods,
other than later uses regex? For simple substitutions like, replace . with / ,
is there any difference?
In java.lang.String, the replace method either takes a pair of char's or a pair of CharSequence's (of which String is a subclass, so it'll happily take a pair of String's). The replace method will replace all occurrences of a char or CharSequence. On the other hand, the first String arguments of replaceFirst and replaceAll are regular expressions (regex). Using the wrong function can lead to subtle bugs.
Q: What's the difference between the java.lang.String methods replace() and replaceAll(), other than that the latter uses regex.
A: Just the regex. They both replace all :)
http://docs.oracle.com/javase/8/docs/api/java/lang/String.html
PS:
There's also a replaceFirst() (which takes a regex)
Both replace() and replaceAll() replace all occurrences in the String.
Examples
I always find examples helpful to understand the differences.
replace()
Use replace() if you just want to replace some char with another char or some String with another String (actually CharSequence).
Example 1
Replace all occurrences of the character x with o.
String myString = "__x___x___x_x____xx_";
char oldChar = 'x';
char newChar = 'o';
String newString = myString.replace(oldChar, newChar);
// __o___o___o_o____oo_
Example 2
Replace all occurrences of the string fish with sheep.
String myString = "one fish, two fish, three fish";
String target = "fish";
String replacement = "sheep";
String newString = myString.replace(target, replacement);
// one sheep, two sheep, three sheep
replaceAll()
Use replaceAll() if you want to use a regular expression pattern.
Example 3
Replace any number with an x.
String myString = "__1_6____3__6_345____0";
String regex = "\\d";
String replacement = "x";
String newString = myString.replaceAll(regex, replacement);
// __x_x____x__x_xxx____x
Example 4
Remove all whitespace.
String myString = " Horse Cow\n\n \r Camel \t\t Sheep \n Goat ";
String regex = "\\s";
String replacement = "";
String newString = myString.replaceAll(regex, replacement);
// HorseCowCamelSheepGoat
See also
Documentation
replace(char oldChar, char newChar)
replace(CharSequence target, CharSequence replacement)
replaceAll(String regex, String replacement)
replaceFirst(String regex, String replacement)
Regular Expressions
Tutorial
List of patterns
The replace() method is overloaded to accept both a primitive char and a CharSequence as arguments.
Now as far as the performance is concerned, the replace() method is a bit faster than replaceAll() because the latter first compiles the regex pattern and then matches before finally replacing whereas the former simply matches for the provided argument and replaces.
Since we know the regex pattern matching is a bit more complex and consequently slower, then preferring replace() over replaceAll() is suggested whenever possible.
For example, for simple substitutions like you mentioned, it is better to use:
replace('.', '\\');
instead of:
replaceAll("\\.", "\\\\");
Note: the above conversion method arguments are system-dependent.
Both replace() and replaceAll() accepts two arguments and replaces all occurrences of the first substring(first argument) in a string with the second substring (second argument).
replace() accepts a pair of char or charsequence and replaceAll() accepts a pair of regex.
It is not true that replace() works faster than replaceAll() since both uses the same code in its implementation
Pattern.compile(regex).matcher(this).replaceAll(replacement);
Now the question is when to use replace and when to use replaceAll().
When you want to replace a substring with another substring regardless of its place of occurrence in the string use replace(). But if you have some particular preference or condition like replace only those substrings at the beginning or end of a string use replaceAll(). Here are some examples to prove my point:
String str = new String("==qwerty==").replaceAll("^==", "?"); \\str: "?qwerty=="
String str = new String("==qwerty==").replaceAll("==$", "?"); \\str: "==qwerty?"
String str = new String("===qwerty==").replaceAll("(=)+", "?"); \\str: "?qwerty?"
To throw more light with an example into how both are going to work for below code:
public static void main(String[] args)
{
String s = "My\\s aaab\\s is\\s aaab\\s name";
String s1 = s.replace("\\s", "c");
System.out.println(s1);
String s2 = s.replaceAll("\\s", "c");
System.out.println(s2);
}
Output:
Myc aaabc isc aaabc name
My\scaaab\scis\scaaab\scname
Explanation
s.replace replaces "\\s" sequence of characters with c. Hence, the output in first line.
s.replaceAll considers \\s as a regex rather(equivalent to space) and replaces spaces with c. \\s in String s is escaped with first \ encountered and becomes \s.
Intellij Idea is smart enough to notify you of the usage as well. If you take a closer look at below image, you will see the difference in interpretation by Intellij idea for replace and replaceAll usage.
String replace(char oldChar, char newChar)
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
String replaceAll(String regex, String replacement
Replaces each substring of this string that matches the given regular expression with the given replacement.
Old thread I know but I am sort of new to Java and discover one of it's strange things. I have used String.replaceAll() but get unpredictable results.
Something like this mess up the string:
sUrl = sUrl.replaceAll( "./", "//").replaceAll( "//", "/");
So I designed this function to get around the weird problem:
//String.replaceAll does not work OK, that's why this function is here
public String strReplace( String s1, String s2, String s )
{
if((( s == null ) || (s.length() == 0 )) || (( s1 == null ) || (s1.length() == 0 )))
{ return s; }
while( (s != null) && (s.indexOf( s1 ) >= 0) )
{ s = s.replace( s1, s2 ); }
return s;
}
Which make you able to do:
sUrl=this.strReplace("./", "//", sUrl );
sUrl=this.strReplace( "//", "/", sUrl );
As alluded to in wickeD's answer, with replaceAll the replacement string is handled differently between replace and replaceAll. I expected a[3] and a[4] to have the same value, but they are different.
public static void main(String[] args) {
String[] a = new String[5];
a[0] = "\\";
a[1] = "X";
a[2] = a[0] + a[1];
a[3] = a[1].replaceAll("X", a[0] + "X");
a[4] = a[1].replace("X", a[0] + "X");
for (String s : a) {
System.out.println(s + "\t" + s.length());
}
}
The output of this is:
\ 1
X 1
\X 2
X 1
\X 2
This is different from perl where the replacement does not require the extra level of escaping:
#!/bin/perl
$esc = "\\";
$s = "X";
$s =~ s/X/${esc}X/;
print "$s " . length($s) . "\n";
which prints
\X 2
This can be quite a nuisance, as when trying to use the value returned by java.sql.DatabaseMetaData.getSearchStringEscape() with replaceAll().
From Java 9 there is some optimizations in replace method.
In Java 8 it uses a regex.
public String replace(CharSequence target, CharSequence replacement) {
return Pattern.compile(target.toString(), Pattern.LITERAL).matcher(
this).replaceAll(Matcher.quoteReplacement(replacement.toString()));
}
From Java 9 and on.
And Stringlatin implementation.
Which perform way better.
https://medium.com/madhash/composite-pattern-in-a-nutshell-ad1bf78479cc?source=post_internal_links---------2------------------
replace() method doesn't uses regex pattern whereas replaceAll() method uses regex pattern. So replace() performs faster than replaceAll().
To add to the already selected "Best Answer" (and others that are just as good like Suragch's), String.replace() is constrained by replacing characters that are sequential (thus taking CharSequence). However, String.replaceAll() is not constrained by replacing sequential characters only. You could replace non-sequential characters as well as long as your regular expression is constructed in such a way.
Also (most importantly and painfully obvious), replace() can only replace literal values; whereas replaceAll can replace 'like' sequences (not necessarily identical).
replace works on char data type but replaceAll works on String datatype and both replace the all occurrences of first argument with second argument.