Develop Reference

Deleting content of every string after first empty space

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));

Difficulty splitting string at delimiter and keeping it

I have a string that is read in pairs, separated by comma. However, I do not always want to split at the comma because there is not always 1 comma in the input. For example, the string,
(http://www.wolframalpha.com/input/?i=103%2F30+%3D+4a-3b,+71%2F60+%3D+a+%2B+b
,http://www.wolframalpha.com/input/?i=x%5E2%2B5x%2B6,file:///tmp/foo/bar/p,d,f.pdf)
Is read in all one line. For this case, I only want to split at the ,h, and no where else in the string. Essentially, after the split, the strings should be:
http://www.wolframalpha.com/input/?i=103%2F30+%3D+4a-3b,+71%2F60+%3D+a+%2B+b
http://www.wolframalpha.com/input/?i=x%5E2%2B5x%2B6
file:///tmp/foo/bar/p,d,f.pdf
Maintaining the order of the comma in the first string. (I will get rid of parenthesis). I have looked at this stack overflow question, and while helpful, does not correctly split this string. This is in Java. Any help is appreciated.

You can use regex to do the split. Please see below code snippet.
String str = "(http://www.wolframalpha.com/input/?i=103%2F30+%3D+4a-3b,+71%2F60+%3D+a+%2B+b,http://www.wolframalpha.com/input/?i=x%5E2%2B5x%2B6)";
String[] strArr = str.split("(,(?=http))");
You will have Array of all the value which would be possible according to your requirement.

Split on 'http' then re-add it.
Psuedo-code
String input = "http://www.wolframalpha.com/input/?i=103%2F30+%3D+4a-3b,+71%2F60+%3D+a+%2B+b
,http://www.wolframalpha.com/input/?i=x%5E2%2B5x%2B6"
List<String> split = input.split('http');
List<String> finalList = new ArrayList<String>();
for(String fixup in split)
{
finalList.put( "http" + fixup );
}
Final should contain the two URLs.

Best way to trim exactly one quote from each side of Java string

I want to be able to trim one quote from each side of a java string. Here are some examples.
"foo" -> foo
"foo\"" -> foo\"
"\"foo\"" -> \"foo\"
I'm currently using StringUtils.trim from common lang but when I end the string with a escaped quote, it trims that too because they are consecutive. I want to be able to trim exactly one quote.

I ended up using org.apache.commons.lang3.StringUtils.substringBetween and it works.

You may also use the substring() method and trim the first and last characters on condition although it's a bit long.
trimedString= s.substring((s.charAt(0)=='"')?1:0 , (s.charAt(s.length()-1)=='"')?s.length()-1:s.length());

I prefer to use this String method
public String[] split(String regex)
basically if you feed in the quotation mark then you will get an array of strings holding all of the chunks between your quotation marks.
String[] parts = originalString.split("\"");
String quoteReduced = parts[0];
for (int i = 1; i < (parts.length() -1); i++){
quoteReduced = quoteReduced.concat( parts[i] +"\"" );
}
quoteReduced = quoteReduced.concat( "\"" +parts[parts.length()-1]);
While it may not be the most straight forward it is the way that I would get around this. The first piece and last piece could be included in the loop but would require an if statement.

Java - Changing multiple words in a string at once?

I'm trying to create a program that can abbreviate certain words in a string given by the user.
This is how I've laid it out so far:
Create a hashmap from a .txt file such as the following:
thanks,thx
your,yr
probably,prob
people,ppl
Take a string from the user
Split the string into words
Check the hashmap to see if that word exists as a key
Use hashmap.get() to return the key value
Replace the word with the key value returned
Return an updated string
It all works perfectly fine until I try to update the string:
public String shortenMessage( String inMessage ) {
String updatedstring = "";
String rawstring = inMessage;
String[] words = rawstring.replaceAll("[^a-zA-Z ]", "").toLowerCase().split("\\s+");
for (String word : words) {
System.out.println(word);
if (map.containsKey(word) == true) {
String x = map.get(word);
updatedstring = rawstring.replace(word, x);
}
}
System.out.println(updatedstring);
return updatedstring;
}
Input:
thanks, your, probably, people
Output:
thanks, your, probably, ppl
Does anyone know how I can update all the words in the string?
Thanks in advance

updatedstring = rawstring.replace(word, x);
This keeps replacing your updatedstring with the rawstring with a the single replacement.
You need to do something like
updatedstring = rawstring;
...
updatedString = updatedString.replace(word, x);
Edit:
That is the solution to the problem you are seeing but there are a few other problems with your code:
Your replacement won't work for things that you needed to lowercased or remove characters from. You create the words array that you iterate from altered version of your rawstring. Then you go back and try to replace the altered versions from your original rawstring where they don't exist. This will not find the words you think you are replacing.
If you are doing global replacements, you could just create a set of words instead of an array since once the word is replaced, it shouldn't come up again.
You might want to be replacing the words one at a time, because your global replacement could cause weird bugs where a word in the replacement map is a sub word of another replacement word. Instead of using String.replace, make an array/list of words, iterate the words and replace the element in the list if needed and join them. In java 8:
String.join(" ", elements);

split method leaving space in array

{
ArrayList<String> node_array = new ArrayList<String>();
String allValues[] = node.split("[(,)]");
for(String value : allValues){
node_array.add(value);
}
node is a string, for example: (3,4,5,6,3)
for some reason when I verify the content of the arraylist the split seems to leave a trail of space as elements, specifically where ( and ) is supposed to be. What am I doing wrong?

You're asking split() to split at parentheses and commas. In your string, there is a blank substring right before the first separator, the opening parenthesis. split() is keeping that blank substring and returning it at the zeroth element of the resulting array.
There are plenty of examples in the documentation that illustrate how the function works.
To work around this, you can either ignore the empty strings, or flip the regex on its head and match the numbers instead of splitting at the punctuation characters.

You have defined a separator to be the one of the characters that's the first character in your String, so an empty string "" will show up in your ArrayList, because that what occurs before the first separator. However, for your application you can easily fix it like this:
ArrayList<String> node_array = new ArrayList<String>();
String allValues[] = node.split("[(,)]");
for(String value : allValues){
if(!value.equals("")) node_array.add(value);
}
return node_array;

node.replace("(","").replace(")","").split(",");
or
node.substring(1,node.length()-1).split(",");