I want to split the following string "Good^Evening" i used split option it is not split the value. please help me.
This is what I've been trying:
String Val = "Good^Evening";
String[] valArray = Val.Split("^");
I'm assuming you did something like:
String[] parts = str.split("^");
That doesn't work because the argument to split is actually a regular expression, where ^ has a special meaning. Try this instead:
String[] parts = str.split("\\^");
The \\ is really equivalent to a single \ (the first \ is required as a Java escape sequence in string literals). It is then a special character in regular expressions which means "use the next character literally, don't interpret its special meaning".
The regex you should use is "\^" which you write as "\\^" as a Java String literal; i.e.
String[] parts = "Good^Evening".split("\\^");
The regex needs a '\' escape because the caret character ('^') is a meta-character in the regex language. The 2nd '\' escape is needed because '\' is an escape in a String literal.
try this
String str = "Good^Evening";
String newStr = str.replaceAll("[^]+", "");
Related
I am trying to split the string with combination of {^
How to use combination of delimiter for splitting the string.
The sample data is :
String str = "0002{^000000000000001157{^000006206210015461{^PR{^ID{^62499{^";
The delimiter passed to String.split() is a regex. As { and ^ are characters with special meaning within a regex, you need to escape them if you want to use them as literals:
String[] tokens = str.split("\\{\\^");
split method in java takes an regex as an input.
so if you want to split the string using '{' and '^' then you need to do the following:
String str = "0002{^000000000000001157{^000006206210015461{^PR{^ID{^62499{^";
String[] splitted = str.split("\\{\\^"); //note \\ before { and ^
You have to escape { and ^ in your split Statement, because both are Special character in regex:
s.split("\\{\\^");
I want to remove all Unicode Characters and Escape Characters like (\n, \t) etc. In short I want just alphanumeric string.
For example :
\u2029My Actual String\u2029
\nMy Actual String\n
I want to fetch just 'My Actual String'. Is there any way to do so, either by using a built in string method or a Regular Expression ?
Try
String stg = "\u2029My Actual String\u2029 \nMy Actual String";
Pattern pat = Pattern.compile("(?!(\\\\(u|U)\\w{4}|\\s))(\\w)+");
Matcher mat = pat.matcher(stg);
String out = "";
while(mat.find()){
out+=mat.group()+" ";
}
System.out.println(out);
The regex matches all things except unicode and escape characters. The regex pictorially represented as:
Output:
My Actual String My Actual String
Try this:
anyString = anyString.replaceAll("\\\\u\\d{4}|\\\\.", "");
to remove escaped characters. If you also want to remove all other special characters use this one:
anyString = anyString.replaceAll("\\\\u\\d{4}|\\\\.|[^a-zA-Z0-9\\s]", "");
(I guess you want to keep the whitespaces, if not remove \\s from the one above)
what is wrong in the following code?
String selectedCountriesStr = countries.replaceAll("[", "").replaceAll("]", "").trim();
String[] selectedCountriesArr = selectedCountriesStr.split(",");
Input String [10000,20000,304050,766666]
Getting error java.util.regex.PatternSyntaxException: Unclosed character class near index 0
You have to escape square brackets because replaceAll() interprets the first argument as a regular expression:
replaceAll("\\[", "")
^^
because, as the error message tells you, the are used for character classes in a regex. Double backslashes are necessary, because "\[" would be an invalid escape sequence. Since the backslash is escaped, the regex engine only receives one backslash.
Also, you can use
replace("[", "")
it will also replace all occurrences of the given CharSequence as is.
You can read more about it in JavaDoc.
Brackets are regex metacharacters, you need to prefix them with a backslash:
.replaceAll("\\[", "").replaceAll("\\]", "")
Also, since this is a simple string substitution, you'd better use .replace():
.replace("[", "").replace("]", "")
String str = "hi,hello,abc,example,problems";
String[] splits = str.split(",");
System.out.println("splits.size: " + splits.length);
for(String asset: splits){
System.out.println(asset);
}
Split function will easily split your string like this
public String getPriceString() {
String priceString = "45.0";
String[] priceStringArray = priceString.split(".");
return priceStringArray.length + "";
}
Why does this give me a 0, zero? Shouldn't this be 2?
The argument to split() is a regular expression, and dot has a special meaning in regular expressions (it matches any character).
Try priceString.split("[.]");
You need to escape . like that
String[] priceStringArray = priceString.split("\\.");
split takes regular expression as a parameter and . means any character.
http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html#sum
escape . with backslash like \\.. . is a regex metacharacter for anything. you will have to escape it with \\. in order to make it treat as a normal character
String priceString = "45.0";
String[] priceStringArray = priceString.split("\\.");
String.split takes a regular expression pattern. You're passing in . which means you want to split on any character.
You could use "\\." as the pattern to split on - but personally I'd use Guava instead:
private static final Splitter DOT_SPLITTER = Splitter.on('.');
...
(If you're not already using Guava, you'll find loads of goodies in there.)
You need to escape . as \\. because . has special meaning in regex.
String priceString = "45.0";
String[] priceStringArray = priceString.split("\\.");
return priceStringArray.length + "";
Use String[] priceStringArray = priceString.split("\\.");
You will have to use escape sequence.
I have a String called persons.name
I want to replace the DOT . with /*/ i.e my output will be persons/*/name
I tried this code:
String a="\\*\\";
str=xpath.replaceAll("\\.", a);
I am getting StringIndexOutOfBoundsException.
How do I replace the dot?
You need two backslashes before the dot, one to escape the slash so it gets through, and the other to escape the dot so it becomes literal. Forward slashes and asterisk are treated literal.
str=xpath.replaceAll("\\.", "/*/"); //replaces a literal . with /*/
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replaceAll(java.lang.String,%20java.lang.String)
If you want to replace a simple string and you don't need the abilities of regular expressions, you can just use replace, not replaceAll.
replace replaces each matching substring but does not interpret its argument as a regular expression.
str = xpath.replace(".", "/*/");
Use Apache Commons Lang:
String a= "\\*\\";
str = StringUtils.replace(xpath, ".", a);
or with standalone JDK:
String a = "\\*\\"; // or: String a = "/*/";
String replacement = Matcher.quoteReplacement(a);
String searchString = Pattern.quote(".");
String str = xpath.replaceAll(searchString, replacement);
return sentence.replaceAll("\s",".");