I keep getting an error about dangling meta character when I use '+', '*', '(', and ')'.
I've already tried escaping those characters in the regex but I still get the error. This is what I have:
"[-\\+\\*/%\\(\\)]"
Update:
test:
String input = "+";
String vals = new WNScanner(input).getNextToken(); //**********
System.out.println("token: " + vals);
System.out.println(vals.matches("[-+*/%()]"));
from another class:
...
String expression = input;
...
public String getNextToken() {
String[] token = {""};
if (expression.length() == 0)
return "";
token = expression.split("\\s");
recentToken = token[0];
expression = expression.replaceFirst(token[0], ""); //*************
expression = expression.trim();
return token[0];
}
* there are exceptions on these lines.
OK, I don't know what you want to achieve there... Especially at this line:
expression = expression.replaceFirst(token[0], "");
If your input string is "+", then your whole regex is +. And that is not legal.
You need to quote the input string in order to use it in any regex-related operation, and that includes String's .replaceFirst() and .replaceAll() (but not .replace()...).
Therefore, do:
final String re = Pattern.quote(token[0]);
expression = expression.replaceFirst(re, "");
Related
String text = "/'Team1 = 6', while /'Team2 = 4', and /'Team3 = 2'";
String[] body = text.split("/|,");
String b1 = body[1];
String b2 = body[2];
String b3 = body[3];
Desired results:
b1 = 'Team1 = 6'
b2 = 'Team2 = 4'
b3 = 'Team3 = 2'
Use regex. Something like this:
String text = "/'Team1 = 6', while /'Team2 = 4', and /'Team3 = 2'";
Matcher m = Pattern.compile("(\\w+\\s=\\s\\d+)").matcher(text);
// \w+ matches the team name (eg: Team1). \s=\s matches " = " and \d+ matches the score.
while (m.find()){
System.out.print(m.group(1)+"\n");
}
This prints:
Team1 = 6
Team2 = 4
Team3 = 2
There's a few ways you can do this, but in your case I'd use regex.
I don't know Java but think something like this regex pattern should work:
Pattern compile("\/'(.*?)'")
A random regex tester site with this pattern is here: https://regex101.com/r/MCRfMm/1
I'm going to say "friends don't let friends use regex" and recommend parsing this out. The built-in class StreamTokenizer will handle the job.
private static void testTok( String in ) throws Exception {
System.out.println( "Input: " + in );
StreamTokenizer tok = new StreamTokenizer( new StringReader( in ) );
tok.resetSyntax();
tok.wordChars( 'a', 'z' );
tok.wordChars( 'A', 'Z' );
tok.wordChars( '0', '9' );
tok.whitespaceChars( 0, ' ' );
String prevToken = null;
for( int type; (type = tok.nextToken()) != StreamTokenizer.TT_EOF; ) {
// System.out.println( tokString( type ) + ": nval=" + tok.nval + ", sval=" + tok.sval );
if( type == '=' ) {
tok.nextToken();
System.out.println( prevToken + "=" + tok.sval );
}
prevToken = tok.sval;
}
}
Output:
Input: /'Team1 = 6', while /'Team2 = 4', and /'Team3 = 2'
Team1=6
Team2=4
Team3=2
BUILD SUCCESSFUL (total time: 0 seconds)
One advantage of this technique is that the individual tokens like "Team1", "=" and "6" are all parsed separately, whereas the regex presented so far is already complex to read and would have to be made even more complex to isolate each of those tokens separately.
You can split on "a slash, optionally preceded by a comma followed by zero or more non-slash characters":
String[] body = text.split("(?:,[^/]*)?/");
public class MyClass {
public static void main(String args[]) {
String text = "/'Team1 = 6', while /'Team2 = 4', and /'Team3 = 2'";
char []textArr = text.toCharArray();
char st = '/';
char ed = ',';
boolean lookForEnd = false;
int st_idx =0;
for(int i =0; i < textArr.length; i++){
if(textArr[i] == st){
st_idx = i+1;
lookForEnd = true;
}
else if(lookForEnd && textArr[i] == ed){
System.out.println(text.substring(st_idx,i));
lookForEnd = false;
}
}
// we still didn't find ',' therefore print everything from lastFoundIdx of '/'
if(lookForEnd){
System.out.println(text.substring(st_idx));
}
}
}
/*
'Team1 = 6'
'Team2 = 4'
'Team3 = 2'
*/
You could use split and a regex using an alternation matching either the start of the string followed by a forward slash or matching a comma, match not a comma one or more times and then a forward slash followed by a positive lookahead to assert that what follows the alternation is a '
(?:^/|,[^,]+/)(?=')
Explanation
(?: Start non capturing group
^/ Assert the start of the string followed by forward slash
| Or
,[^,]+/ Match a comma followed by match not a comma one or more times using a negated character class and then match a forward slash
(?=') Positive lookahead to assert what follows is '
) Close non capturing group
Regex demo - Java demo
Getting a match instead of split
If you want to to match a pattern like 'Team1 = 6', you could use:
'[^=]+=[^']+'
Regex demo - Java demo
I want to remove a part of string from one character, that is:
Source string:
manchester united (with nice players)
Target string:
manchester united
There are multiple ways to do it. If you have the string which you want to replace you can use the replace or replaceAll methods of the String class. If you are looking to replace a substring you can get the substring using the substring API.
For example
String str = "manchester united (with nice players)";
System.out.println(str.replace("(with nice players)", ""));
int index = str.indexOf("(");
System.out.println(str.substring(0, index));
To replace content within "()" you can use:
int startIndex = str.indexOf("(");
int endIndex = str.indexOf(")");
String replacement = "I AM JUST A REPLACEMENT";
String toBeReplaced = str.substring(startIndex + 1, endIndex);
System.out.println(str.replace(toBeReplaced, replacement));
String Replace
String s = "manchester united (with nice players)";
s = s.replace(" (with nice players)", "");
Edit:
By Index
s = s.substring(0, s.indexOf("(") - 1);
Use String.Replace():
http://www.daniweb.com/software-development/java/threads/73139
Example:
String original = "manchester united (with nice players)";
String newString = original.replace(" (with nice players)","");
originalString.replaceFirst("[(].*?[)]", "");
https://ideone.com/jsZhSC
replaceFirst() can be replaced by replaceAll()
Using StringBuilder, you can replace the following way.
StringBuilder str = new StringBuilder("manchester united (with nice players)");
int startIdx = str.indexOf("(");
int endIdx = str.indexOf(")");
str.replace(++startIdx, endIdx, "");
You should use the substring() method of String object.
Here is an example code:
Assumption: I am assuming here that you want to retrieve the string till the first parenthesis
String strTest = "manchester united(with nice players)";
/*Get the substring from the original string, with starting index 0, and ending index as position of th first parenthesis - 1 */
String strSub = strTest.subString(0,strTest.getIndex("(")-1);
I would at first split the original string into an array of String with a token " (" and the String at position 0 of the output array is what you would like to have.
String[] output = originalString.split(" (");
String result = output[0];
Using StringUtils from commons lang
A null source string will return null. An empty ("") source string will return the empty string. A null remove string will return the source string. An empty ("") remove string will return the source string.
String str = StringUtils.remove("Test remove", "remove");
System.out.println(str);
//result will be "Test"
If you just need to remove everything after the "(", try this. Does nothing if no parentheses.
StringUtils.substringBefore(str, "(");
If there may be content after the end parentheses, try this.
String toRemove = StringUtils.substringBetween(str, "(", ")");
String result = StringUtils.remove(str, "(" + toRemove + ")");
To remove end spaces, use str.trim()
Apache StringUtils functions are null-, empty-, and no match- safe
Kotlin Solution
If you are removing a specific string from the end, use removeSuffix (Documentation)
var text = "one(two"
text = text.removeSuffix("(two") // "one"
If the suffix does not exist in the string, it just returns the original
var text = "one(three"
text = text.removeSuffix("(two") // "one(three"
If you want to remove after a character, use
// Each results in "one"
text = text.replaceAfter("(", "").dropLast(1) // You should check char is present before `dropLast`
// or
text = text.removeRange(text.indexOf("("), text.length)
// or
text = text.replaceRange(text.indexOf("("), text.length, "")
You can also check out removePrefix, removeRange, removeSurrounding, and replaceAfterLast which are similar
The Full List is here: (Documentation)
// Java program to remove a substring from a string
public class RemoveSubString {
public static void main(String[] args) {
String master = "1,2,3,4,5";
String to_remove="3,";
String new_string = master.replace(to_remove, "");
// the above line replaces the t_remove string with blank string in master
System.out.println(master);
System.out.println(new_string);
}
}
You could use replace to fix your string. The following will return everything before a "(" and also strip all leading and trailing whitespace. If the string starts with a "(" it will just leave it as is.
str = "manchester united (with nice players)"
matched = str.match(/.*(?=\()/)
str.replace(matched[0].strip) if matched
I need to remove some specific "special" characters and replace them with empty string if they show up.
I am currently having a problem with the regex, probably with the Java escaping. I can't put them all together, it just doesn't work, I tried a lot! T_T
Currently I am doing it one by one which is kinda silly, but for now at least it works, like that :
public static String filterSpecialCharacters(String string) {
string = string.replaceAll("-", "");
string = string.replaceAll("\\[", "");
string = string.replaceAll("\\]", "");
string = string.replaceAll("\\^", "");
string = string.replaceAll("/", "");
string = string.replaceAll(",", "");
string = string.replaceAll("'", "");
string = string.replaceAll("\\*", "");
string = string.replaceAll(":", "");
string = string.replaceAll("\\.", "");
string = string.replaceAll("!", "");
string = string.replaceAll(">", "");
string = string.replaceAll("<", "");
string = string.replaceAll("~", "");
string = string.replaceAll("#", "");
string = string.replaceAll("#", "");
string = string.replaceAll("$", "");
string = string.replaceAll("%", "");
string = string.replaceAll("\\+", "");
string = string.replaceAll("=", "");
string = string.replaceAll("\\?", "");
string = string.replaceAll("|", "");
string = string.replaceAll("\"", "");
string = string.replaceAll("\\\\", "");
string = string.replaceAll("\\)", "");
string = string.replaceAll("\\(", "");
return string;
}
Those are all the character I need to remove:
- [ ] ^ / , ' * : . ! > < ~ # # $ % + = ? | " \ ) (
I am clearly missing something, I can't figure out how to put it all in one line. Help?
Your code does not work in fact because .replaceAll("$", "") replaces an end of string with empty string. To replace a literal $, you need to escape it. Same issue is with the pipe symbol removal.
All you need to do is to put the characters you need to replace into a character class and apply the + quantifier for better performance, like this:
string = string.replaceAll("[-\\[\\]^/,'*:.!><~##$%+=?|\"\\\\()]+", "");
Note that inside a character class, most "special regex metacharacters" lose their special status, you only have to escape [, ], \, a hyphen (if it is not at the start/end of the character class), and a ^ (if it is the first symbol in the "positive" character class).
DEMO:
String s = "-[]^/,'*:.!><~##$%+=?|\"\\()TEXT";
s = s.replaceAll("[-\\[\\]^/,'*:.!><~##$%+=?|\"\\\\()]+", "");
System.out.println(s); // => TEXT
Use these codes
String REGEX = "YOUR_REGEX";
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(yourString);
yourString = m.replaceAll("");
UPDATE :
Your REGEX looks something like
String REGEX = "-|\\[|\\]|\\^|\\/|,|'|\\*|\\:|\\.|!|>|<|\\~|#|#|\\$|%|\\+|=\\?|\\||\\\\|\\\\\\\\|\\)|\\(";
SAPMLE :
String yourString = "#My (name) -is #someth\ing"";
//Use Above codes
Log.d("yourString",yourString);
OUTPUT
I want to add Two java JSON String manually , so for this i need to remove "}" and replace it with comma "," of first JSON String and remove the first "{" of the second JSON String .
This is my program
import java.util.Map;
import org.codehaus.jackson.type.TypeReference;
public class Hi {
private static JsonHelper jsonHelper = JsonHelper.getInstance();
public static void main(String[] args) throws Exception {
Map<String, Tracker> allCusts = null;
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String B = "{\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String totalString = A + B;
if (null != totalString) {
allCusts = (Map<String, Tracker>) jsonHelper.toObject(
totalString, new TypeReference<Map<String, Tracker>>() {
});
}
System.out.println(allCusts);
}
}
When adding two Strings A + B
I want to remove the last character of "}" in A and replace it with "," and remove the FIrst character of "{" in B .
SO this should it look like .
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1},";
String B = "\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
I have tried
String Astr = A.replace(A.substring(A.length()-1), ",");
String Bstr = B.replaceFirst("{", "");
String totalString = Astr + Bstr ;
With this i was getting
Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
please suggest .
{ is a control character for Regular Expressions, and since replaceFirst takes a string representation of a Regular Expression as its first argument, you need to escape the { so it's not treated as a control character:
String Bstr = B.replaceFirst("\\{", "");
I would say that using the replace methods is really overkill here since you're just trying to chop a character off of either end of a string. This should work just as well:
String totalString = A.substring(0, A.length()-1) + "," + B.substring(1);
Of course, regex doesn't look like a very good tool for this. But the following seem to work:
String str = "{..{...}..}}";
str = str.replaceFirst("\\{", "");
str = str.replaceFirst("}$", ",");
System.out.println(str);
Output:
..{...}..},
Some issues in your first two statements. Add 0 as start index in substring method and leave with that. Put \\ as escape char in matching pattern and ut a , in second statement as replacement value.
String Astr = A.substring(0, A.length()-1);//truncate the ending `}`
String Bstr = B.replaceFirst("\\{", ",");//replaces first '{` with a ','
String totalString = Astr + Bstr ;
Please note: There are better ways, but I am just trying to correct your statements.
I try to split a String into tokens.
The token delimiters are not single characters, some delimiters are included into others (example, & and &&), and I need to have the delimiters returned as token.
StringTokenizer is not able to deal with multiple characters delimiters. I presume it's possible with String.split, but fail to guess the magical regular expression that will suits my needs.
Any idea ?
Example:
Token delimiters: "&", "&&", "=", "=>", " "
String to tokenize: a & b&&c=>d
Expected result: an string array containing "a", " ", "&", " ", "b", "&&", "c", "=>", "d"
--- Edit ---
Thanks to all for your help, Dasblinkenlight gives me the solution. Here is the "ready to use" code I wrote with his help:
private static String[] wonderfulTokenizer(String string, String[] delimiters) {
// First, create a regular expression that matches the union of the delimiters
// Be aware that, in case of delimiters containing others (example && and &),
// the longer may be before the shorter (&& should be before &) or the regexpr
// parser will recognize && as two &.
Arrays.sort(delimiters, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
return -o1.compareTo(o2);
}
});
// Build a string that will contain the regular expression
StringBuilder regexpr = new StringBuilder();
regexpr.append('(');
for (String delim : delimiters) { // For each delimiter
if (regexpr.length() != 1) regexpr.append('|'); // Add union separator if needed
for (int i = 0; i < delim.length(); i++) {
// Add an escape character if the character is a regexp reserved char
regexpr.append('\\');
regexpr.append(delim.charAt(i));
}
}
regexpr.append(')'); // Close the union
Pattern p = Pattern.compile(regexpr.toString());
// Now, search for the tokens
List<String> res = new ArrayList<String>();
Matcher m = p.matcher(string);
int pos = 0;
while (m.find()) { // While there's a delimiter in the string
if (pos != m.start()) {
// If there's something between the current and the previous delimiter
// Add it to the tokens list
res.add(string.substring(pos, m.start()));
}
res.add(m.group()); // add the delimiter
pos = m.end(); // Remember end of delimiter
}
if (pos != string.length()) {
// If it remains some characters in the string after last delimiter
// Add this to the token list
res.add(string.substring(pos));
}
// Return the result
return res.toArray(new String[res.size()]);
}
It could be optimize if you have many strings to tokenize by creating the Pattern only one time.
You can use the Pattern and a simple loop to achieve the results that you are looking for:
List<String> res = new ArrayList<String>();
Pattern p = Pattern.compile("([&]{1,2}|=>?| +)");
String s = "s=a&=>b";
Matcher m = p.matcher(s);
int pos = 0;
while (m.find()) {
if (pos != m.start()) {
res.add(s.substring(pos, m.start()));
}
res.add(m.group());
pos = m.end();
}
if (pos != s.length()) {
res.add(s.substring(pos));
}
for (String t : res) {
System.out.println("'"+t+"'");
}
This produces the result below:
's'
'='
'a'
'&'
'=>'
'b'
Split won't do it for you as it removed the delimeter. You probably need to tokenize the string on your own (i.e. a for-loop) or use a framework like
http://www.antlr.org/
Try this:
String test = "a & b&&c=>d=A";
String regEx = "(&[&]?|=[>]?)";
String[] res = test.split(regEx);
for(String s : res){
System.out.println("Token: "+s);
}
I added the '=A' at the end to show that that is also parsed.
As mentioned in another answer, if you need the atypical behaviour of keeping the delimiters in the result, you will probably need to create you parser yourself....but in that case you really have to think about what a "delimiter" is in your code.