I need to replace parts of a string by looking up the System properties.
For example, consider the string It was {var1} beauty killed {var2}
I need to parse the string, and replace all the words contained within the parenthesis by looking up their value in System properties. If System.getProperty() returns null, then simply replace with empty character. This is pretty straightforward when I know the variables well ahead. But the string that I need to parse is not defined ahead. I wouldn't know how many number of variables are in the string and what the variable names are. Assuming a simple, well formatted string (no nested parenthesis, open - close matches), what is the simplest or the most elegant way to parse through the string and replace all the character sequences that are enclosed in the parenthesis?
Only solution I could come up with is to traverse the string from the first character, note down the positions of the start and end positions of the parenthesis, replace the string between them, and then continue until reaching the end of the string. Is there simpler way to do this?
You can use the parentheses to break the initial string into substrings, and then replace every other substring.
String[] substituteValues = {"the", "str", "other", "another"};
int substituteValuesIndex = 0;
String test = "Here is {var1} string called {var2}";
// split the string up into substrings
test = test.replaceAll("\\}", "\\{");
String[] splitString = test.split("\\{");
// now sub in your values
for (int k=1; k < splitString.length; k = k+2) {
splitString[k] = substituteValues[substituteValuesIndex];
substituteValuesIndex++;
}
String result = "";
for (String s : splitString) {
result = result + s;
}
Related
How can I delete everything after first empty space in a string which user selects? I was reading this how to remove some words from a string in java. Can this help me in my case?
You can use replaceAll with a regex \s.* which match every thing after space:
String str = "Hello java word!";
str = str.replaceAll("\\s.*", "");
output
Hello
regex demo
Like #Coffeehouse Coder mention in comment, This solution will replace every thing if the input start with space, so if you want to avoid this case, you can trim your input using string.trim() so it can remove the spaces in start and in end.
Assuming that there is no space in the beginning of the string.
Follow these steps-
Split the string at space. It will create an array.
Get the first element of that array.
Hope this helps.
str = "Example string"
String[] _arr = str.split("\\s");
String word = _arr[0];
You need to consider multiple white spaces and space in the beginning before considering the above code.
I am not native to JAVA Programming but have an idea that it has split function for string.
And the reference you cited in the question is bit complex, while you can achieve the desired thing very easily.
P.S. In future if you make a mind to get two words or three, splitting method is better (assuming you have already dealt with multiple white-spaces) else substring is better.
A simple way to do it can be:
System.out.println("Hello world!".split(" ")[0]);
// Taking 'str' as your string
// To remove the first space(s) of the string,
str = str.trim();
int index = str.indexOf(" ");
String word = str.substring(0, index);
This is just one method of many.
str = str.replaceAll("\\s+", " "); // This replaces one or more spaces with one space
String[] words = str.split("\\s");
String first = words[0];
The simplest solution in my opinion would be to just locate the index which the user wants it to be cut off at and then call the substring() method from 0 to the index they wanted. Set that = to a new string and you have the string they want.
If you want to replace the string then just set the original string = to the result of the substring() method.
Link to substring() method: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)
There are already 5 perfectly good answers, so let me add a sixth one. Variety is the spice of life!
private static final Pattern FIRST_WORD = Pattern.compile("\\S+");
public static String firstWord(CharSequence text) {
Matcher m = FIRST_WORD.matcher(text);
return m.find() ? m.group() : "";
}
Advantages over the .split(...)[0]-type answers:
It directly does exactly what is being asked, i.e. "Find the first sequence of non-space characters." So the self-documentation is more explicit.
It is more efficient when called on multiple strings (e.g. for batch processing a large list of strings) because the regular expression is compiled only once.
It is more space-efficient because it avoids unnecessarily creating a whole array with references to each word when we only need the first.
It works without having to trim the string.
(I know this is probably too late to be of any use to the OP but I'm leaving it here as an alternative solution for future readers.)
This would be more efficient
String str = "Hello world!";
int spaceInd = str.indexOf(' ');
if(spaceInd != -1) {
str = str.substring(0, spaceInd);
}
System.out.println(String.format("[%s]", str));
I was working on some string formatting, and I was curious if I was doing it the most efficient way.
Assume I have a String Array:
String ArrayOne[] = {"/test/" , "/this/is/test" , "/that/is/" "/random/words" }
I want the result Array to be
String resultArray[] = {"test", "this_is_test" , "that_is" , "random_words" }
It's quite messy and brute-force-like.
for(char c : ArrayOne[i].toCharArray()) {
if(c == '/'){
occurances[i]++;
}
}
First I count the number of "/" in each String like above and then using these counts, I find the indexOf("/") for each string and add "_" accordingly.
As you can see though, it gets very messy.
Is there a more efficient way to do this besides the brute-force way I'm doing?
Thanks!
You could use replaceAll and replace, as follows:
String resultArray[] = new String[ArrayOne.length];
for (int i = 0; i < ArrayOne.length; ++i) {
resultArray[i] = ArrayOne[i].replaceAll("^/|/$", "").replace('/', '_');
}
The replaceAll method searches the string for a match to the regex given in the first argument, and replaces each match with the text in the second argument.
Here, we use it first to remove leading and trailing slashes. We search for slashes at the start of the string (^/) or the end of the string (/$), and replace them with nothing.
Then, we replace all remaining slashes with underscores using replace.
I have a method that converts all the first letters of the words in a sentence into uppercase.
public static String toTitleCase(String s)
{
String result = "";
String[] words = s.split(" ");
for (int i = 0; i < words.length; i++)
{
result += words[i].replace(words[i].charAt(0)+"", Character.toUpperCase(words[i].charAt(0))+"") + " ";
}
return result;
}
The problem is that the method converts each other letter in a word that is the same letter as the first to uppercase. For example, the string title comes out as TiTle
For the input this is a title this becomes the output This Is A TiTle
I've tried lots of things. A nested loop that checks every letter in each word, and if there is a recurrence, the second is ignored. I used counters, booleans, etc. Nothing works and I keep getting the same result.
What can I do? I only want the first letter in upper case.
Instead of using the replace() method, try replaceFirst().
result += words[i].replaceFirst(words[i].charAt(0)+"", Character.toUpperCase(words[i].charAt(0))+"") + " ";
Will output:
This Is A Title
The problem is that you are using replace method which replaces all occurrences of described character. To solve this problem you can either
use replaceFirst instead
take first letter,
create its uppercase version
concatenate it with rest of string which can be created with a little help of substring method.
since you are using replace(String, String) which uses regex you can add ^ before character you want to replace like replace("^a","A"). ^ means start of input so it will only replace a that is placed after start of input.
I would probably use second approach.
Also currently in each loop your code creates new StringBuilder with data stored in result, append new word to it, and reassigns result of output from toString().
This is infective approach. Instead you should create StringBuilder before loop that will represent your result and append new words created inside loop to it and after loop ends you can get its String version with toString() method.
Doing some Regex-Magic can simplify your task:
public static void main(String[] args) {
final String test = "this is a Test";
final StringBuffer buffer = new StringBuffer(test);
final Pattern patter = Pattern.compile("\\b(\\p{javaLowerCase})");
final Matcher matcher = patter.matcher(buffer);
while (matcher.find()) {
buffer.replace(matcher.start(), matcher.end(), matcher.group().toUpperCase());
}
System.out.println(buffer);
}
The expression \\b(\\p{javaLowerCase}) matches "The beginning of a word followed by a lower-case letter", while matcher.group() is equal to whats inside the () in the part that matches. Example: Applying on "test" matches on "t", so start is 0, end is 1 and group is "t". This can easily run through even a huge amount of text and replace all those letters that need replacement.
In addition: it is always a good idea to use a StringBuffer (or similar) for String manipulation, because each String in Java is unique. That is if you do something like result += stringPart you actually create a new String (equal to result + stringPart) each time this is called. So if you do this with like 10 parts, you will in the end have at least 10 different Strings in memory, while you only need one, which is the final one.
StringBuffer instead uses something like char[] to ensure that if you change only a single character no extra memory needs to be allocated.
Note that a patter only need to be compiled once, so you can keep that as a class variable somewhere.
I am trying to break apart a very simple collection of strings that come in the forms of
0|0
10|15
30|55
etc etc. Essentially numbers that are seperated by pipes.
When I use java's string split function with .split("|"). I get somewhat unpredictable results. white space in the first slot, sometimes the number itself isn't where I thought it should be.
Can anybody please help and give me advice on how I can use a reg exp to keep ONLY the integers?
I was asked to give the code trying to do the actual split. So allow me to do that in hopes to clarify further my problem :)
String temp = "0|0";
String splitString = temp.split("|");
results
\n
0
|
0
I am trying to get
0
0
only. Forever grateful for any help ahead of time :)
I still suggest to use split(), it skips null tokens by default. you want to get rid of non numeric characters in the string and only keep pipes and numbers, then you can easily use split() to get what you want. or you can pass multiple delimiters to split (in form of regex) and this should work:
String[] splited = yourString.split("[\\|\\s]+");
and the regex:
import java.util.regex.*;
Pattern pattern = Pattern.compile("\\d+(?=([\\|\\s\\r\\n]))");
Matcher matcher = pattern.matcher(yourString);
while (matcher.find()) {
System.out.println(matcher.group());
}
The pipe symbol is special in a regexp (it marks alternatives), you need to escape it. Depending on the java version you are using this could well explain your unpredictable results.
class t {
public static void main(String[]_)
{
String temp = "0|0";
String[] splitString = temp.split("\\|");
for (int i=0; i<splitString.length; i++)
System.out.println("splitString["+i+"] is " + splitString[i]);
}
}
outputs
splitString[0] is 0
splitString[1] is 0
Note that one backslash is the regexp escape character, but because a backslash is also the escape character in java source you need two of them to push the backslash into the regexp.
You can do replace white space for pipes and split it.
String test = "0|0 10|15 30|55";
test = test.replace(" ", "|");
String[] result = test.split("|");
Hope this helps for you..
You can use StringTokenizer.
String test = "0|0";
StringTokenizer st = new StringTokenizer(test);
int firstNumber = Integer.parseInt(st.nextToken()); //will parse out the first number
int secondNumber = Integer.parseInt(st.nextToken()); //will parse out the second number
Of course you can always nest this inside of a while loop if you have multiple strings.
Also, you need to import java.util.* for this to work.
The pipe ('|') is a special character in regular expressions. It needs to be "escaped" with a '\' character if you want to use it as a regular character, unfortunately '\' is a special character in Java so you need to do a kind of double escape maneuver e.g.
String temp = "0|0";
String[] splitStrings = temp.split("\\|");
The Guava library has a nice class Splitter which is a much more convenient alternative to String.split(). The advantages are that you can choose to split the string on specific characters (like '|'), or on specific strings, or with regexps, and you can choose what to do with the resulting parts (trim them, throw ayway empty parts etc.).
For example you can call
Iterable<String> parts = Spliter.on('|').trimResults().omitEmptyStrings().split("0|0")
This should work for you:
([0-9]+)
Considering a scenario where in we have read a line from csv or xls file in the form of string and need to separate the columns in array of string depending on delimiters.
Below is the code snippet to achieve this problem..
{ ...
....
String line = new BufferedReader(new FileReader("your file"));
String[] splittedString = StringSplitToArray(stringLine,"\"");
...
....
}
public static String[] StringSplitToArray(String stringToSplit, String delimiter)
{
StringBuffer token = new StringBuffer();
Vector tokens = new Vector();
char[] chars = stringToSplit.toCharArray();
for (int i=0; i 0) {
tokens.addElement(token.toString());
token.setLength(0);
i++;
}
} else {
token.append(chars[i]);
}
}
if (token.length() > 0) {
tokens.addElement(token.toString());
}
// convert the vector into an array
String[] preparedArray = new String[tokens.size()];
for (int i=0; i < preparedArray.length; i++) {
preparedArray[i] = (String)tokens.elementAt(i);
}
return preparedArray;
}
Above code snippet contains method call to StringSplitToArray where in the method converts the stringline into string array splitting the line depending on the delimiter specified or passed to the method. Delimiter can be comma separator(,) or double code(").
For more on this, follow this link : http://scrapillars.blogspot.in
In Java, I have a String:
Jamaica
I would like to remove the first character of the string and then return amaica
How would I do this?
const str = "Jamaica".substring(1)
console.log(str)
Use the substring() function with an argument of 1 to get the substring from position 1 (after the first character) to the end of the string (leaving the second argument out defaults to the full length of the string).
public String removeFirstChar(String s){
return s.substring(1);
}
In Java, remove leading character only if it is a certain character
Use the Java ternary operator to quickly check if your character is there before removing it. This strips the leading character only if it exists, if passed a blank string, return blankstring.
String header = "";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
header = "foobar";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
header = "#moobar";
header = header.startsWith("#") ? header.substring(1) : header;
System.out.println(header);
Prints:
blankstring
foobar
moobar
Java, remove all the instances of a character anywhere in a string:
String a = "Cool";
a = a.replace("o","");
//variable 'a' contains the string "Cl"
Java, remove the first instance of a character anywhere in a string:
String b = "Cool";
b = b.replaceFirst("o","");
//variable 'b' contains the string "Col"
Use substring() and give the number of characters that you want to trim from front.
String value = "Jamaica";
value = value.substring(1);
Answer: "amaica"
You can use the substring method of the String class that takes only the beginning index and returns the substring that begins with the character at the specified index and extending to the end of the string.
String str = "Jamaica";
str = str.substring(1);
substring() method returns a new String that contains a subsequence of characters currently contained in this sequence.
The substring begins at the specified start and extends to the character at index end - 1.
It has two forms. The first is
String substring(int FirstIndex)
Here, FirstIndex specifies the index at which the substring will
begin. This form returns a copy of the substring that begins at
FirstIndex and runs to the end of the invoking string.
String substring(int FirstIndex, int endIndex)
Here, FirstIndex specifies the beginning index, and endIndex specifies
the stopping point. The string returned contains all the characters
from the beginning index, up to, but not including, the ending index.
Example
String str = "Amiyo";
// prints substring from index 3
System.out.println("substring is = " + str.substring(3)); // Output 'yo'
you can do like this:
String str = "Jamaica";
str = str.substring(1, title.length());
return str;
or in general:
public String removeFirstChar(String str){
return str.substring(1, title.length());
}
public String removeFirst(String input)
{
return input.substring(1);
}
The key thing to understand in Java is that Strings are immutable -- you can't change them. So it makes no sense to speak of 'removing a character from a string'. Instead, you make a NEW string with just the characters you want. The other posts in this question give you a variety of ways of doing that, but its important to understand that these don't change the original string in any way. Any references you have to the old string will continue to refer to the old string (unless you change them to refer to a different string) and will not be affected by the newly created string.
This has a number of implications for performance. Each time you are 'modifying' a string, you are actually creating a new string with all the overhead implied (memory allocation and garbage collection). So if you want to make a series of modifications to a string and care only about the final result (the intermediate strings will be dead as soon as you 'modify' them), it may make more sense to use a StringBuilder or StringBuffer instead.
I came across a situation where I had to remove not only the first character (if it was a #, but the first set of characters.
String myString = ###Hello World could be the starting point, but I would only want to keep the Hello World. this could be done as following.
while (myString.charAt(0) == '#') { // Remove all the # chars in front of the real string
myString = myString.substring(1, myString.length());
}
For OP's case, replace while with if and it works aswell.
You can simply use substring().
String myString = "Jamaica"
String myStringWithoutJ = myString.substring(1)
The index in the method indicates from where we are getting the result string, in this case we are getting it after the first position because we dont want that "J" in "Jamaica".
Another solution, you can solve your problem using replaceAll with some regex ^.{1} (regex demo) for example :
String str = "Jamaica";
int nbr = 1;
str = str.replaceAll("^.{" + nbr + "}", "");//Output = amaica
My version of removing leading chars, one or multiple. For example, String str1 = "01234", when removing leading '0', result will be "1234". For a String str2 = "000123" result will be again "123". And for String str3 = "000" result will be empty string: "". Such functionality is often useful when converting numeric strings into numbers.The advantage of this solution compared with regex (replaceAll(...)) is that this one is much faster. This is important when processing large number of Strings.
public static String removeLeadingChar(String str, char ch) {
int idx = 0;
while ((idx < str.length()) && (str.charAt(idx) == ch))
idx++;
return str.substring(idx);
}
##KOTLIN
#Its working fine.
tv.doOnTextChanged { text: CharSequence?, start, count, after ->
val length = text.toString().length
if (length==1 && text!!.startsWith(" ")) {
tv?.setText("")
}
}