Develop Reference

Java extract only first letters/characters from String

Hello guys I want to extract only first letters from this String:
String str = "使 徒 行 傳 16:31 ERV-ZH";
I only want to get these characters:
使 徒 行 傳
and not include
ERV-ZH
Only the letters or characters before the numbers plus the colon.
Note that Chinese letters can also be English and other letters.
this is what I've tried:
str.split(" ")[0];
But I'm only getting the first letter. Do you have an idea how to achieve my requirement? Any help will be appreciated. Thanks.
NOTE:
Also, strings are dynamic so I only presented sample characters.

This should give you the desired output
String str = "使 徒 行 傳 16:31 ERV-ZH";
String[] test = str.split("\\d\\d:\\d\\d");
for (String s : test) {
System.out.println(s);
}
The first element will be the part before the time and so on
Edit: if you are in need to be more dynamic for times like 6:31 or 16:6 then you could use this regex "\\d{1,2}:\\d{1,2}"

You can use the following regex ^([\\D\\s]+), this is what you need:
String str = "使 徒 行 傳 16:31 ERV-ZH";
String pattern = "^([\\D\\s]+)";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(str);
if (m.find( )) {
System.out.println("Found value: " + m.group(0) );
} else {
System.out.println("NO MATCH");
}
}
This is a live DEMO here.
In the following regex ^([\\D\\s]+):
^ will match only in the begginnig.
\\D will avoid matching any number.
Note that this will be the case for any string.

If you don't always have a date pattern that can be used as a delimiter in the middle, and are looking for a more generic solution, you could go with this: str.replaceAll("[^\\p{L}\\s]+.*", "")

Regex extract last numbers from String

I have some strings which are indexed and are dynamic.
For example:
name01,
name02,
name[n]
now I need to separate name from index.
I've come up with this regex which works OK to extract index.
([0-9]+(?!.*[0-9]))
But, there are some exceptions of these names. Some of them may have a number appended which is not the index.(These strings are limited and I know them, meaning I can add them as "exceptions" in the regex)
For example,
panLast4[01]
Here the last '4' is not part of the index, so I need to distinguish.
So I tried:
[^panLast4]([0-9]+(?!.*[0-9]))
Which works for panLast4[123] but not panLast4[43]
Note: the "[" and "]" is for explanation purposes only, it's not present in the strings
What is wrong?
Thanks

You can use the split method with this pattern:
(?<!^panLast(?=4)|^nm(?=14)|^nm1(?=4))(?=[0-9]+$)
The idea is to find the position where there are digits until the end of the string (?=[0-9]+$). But the match will succeed if the negative lookbehind allows it (to exclude particular names (panLast4 and nm14 here) that end with digits). When one of these particular names is found, the regex engine must go to the next position to obtain a match.
Example:
String s ="panLast412345";
String[] res = s.split("(?<!^panLast(?=4)|^nm(?=14)|^nm1(?=4))(?=[0-9]+$)", 2);
if ( res.length==2 ) {
System.out.println("name: " + res[0]);
System.out.println("ID: " + res[1]);
}
An other method with matches() that simply uses a lazy quantifier as last alternative:
Pattern p = Pattern.compile("(panLast4|nm14|.*?)([0-9]+)");
String s = "panLast42356";
Matcher m = p.matcher(s);
if ( m.matches() && m.group(1).length()>0 ) {
System.out.println("name: "+ m.group(1));
System.out.println("ID: "+ m.group(2));
}

match ;ABC12;10;250.3 using regex java

String regex = "^;[A-Z0-9]{5};[\\d]{1,};[\\d]{1,}.[\\d]{1,}";
String str = ";ABC12;10;250.3";
System.out.println(str.matches(regex));
The above regex works fine.
Consider the following strings
str1=";ABC12;10;250.3"
str2=;ABB62;5;2.3
str3=;ABF02;8;25120.3
str4=;AKC12;11;2504.303
Now i have the string as String strToMatch= str1,str2,str3,str4
How do i convert my regex expression above inorder to match the above string.
Note : There can be n number of comma separated values in the above string. And i also need to take care that the string strToMatch doesnot end with comma.

You can capture the regex with round brackets and repeat one or more times:
String regex = "^(;[A-Z0-9]{5};\\d+;\\d+\\.\\d+){1,}";

Try this pattern instead: (;[A-Z0-9]{5};[\\d]{1,};[\\d]{1,}\\.[\\d]{1,},?)+
This has two differences to your pattern: first I use \\. to denote that this has to be a . because a single dot means "any character" in regex.
Then I used the grouping brackets (...) and the + at the end to say: "Look for this once or more". As the , is optional at the end, I added a ?
If you want to get single matches to process using a Matcher later on, a simple modification should do the trick: (;[A-Z0-9]{5};[\\d]{1,};[\\d]{1,}\\.[\\d]{1,}),?
The + is gone and the ,? is outside the grouping brackets, because those are now capturing brackets (as well).
Example:
final Pattern pattern = Pattern.compile("(;[A-Z0-9]{5};[\\d]{1,};[\\d]{1,}\\.[\\d]{1,}),?");
final Matcher matcher = pattern.matcher(";ABC12;10;250.3,;ABB62;5;2.3,;ABF02;8;25120.3,;AKC12;11;2504.303");
while (matcher.find()) {
System.out.println("Whole match: " + matcher.group());
for (int i = 1; i <= matcher.groupCount(); ++i) {
System.out.println("Group #" + i + ": " + matcher.group(i));
}
}

I have found below way of solving the problem.
String strToMatch = ";ABC12;10;250.3,;ABB62;5;2.3,;ABF02;8;25120.3,;AKC12;11;2504.303";
if(strToMatch.endsWith(",") || strToMatch.startsWith(","))
return false;
else{
String[] str = strToMatch.split(",");
int count = 0;
for (String s : str){
String regex = ";[A-Z0-9]{5};\\d+;\\d+\\.\\d+";
if(s.matches(regex))
return false;
}
return true;
}
Any simpler way than this?

How do I find multiple substrings from one string using regex in Java?

I want to find every instance of a number, followed by a comma (no space), followed by any number of characters in a string. I was able to get a regex to find all the instances of what I was looking for, but I want to print them individually rather than all together. I'm new to regex in general, so maybe my pattern is wrong?
This is my code:
String test = "1 2,A 3,B 4,23";
Pattern p = Pattern.compile("\\d+,.+");
Matcher m = p.matcher(test);
while(m.find()) {
System.out.println("found: " + m.group());
}
This is what it prints:
found: 2,A 3,B 4,23
This is what I want it to print:
found: 2,A
found: 3,B
found: 4,23
Thanks in advance!

try this regex
Pattern p = Pattern.compile("\\d+,.+?(?= |$)");

You could take an easier route and split by space, then ignore anything without a comma:
String values = test.split(' ');
for (String value : values) {
if (value.contains(",") {
System.out.println("found: " + value);
}
}

What you apparently left out of your requirements statement is where "any number of characters" is supposed to end. As it stands, it ends at the end of the string; from your sample output, it seems you want it to end at the first space.
Try this pattern: "\\d+,[^\\s]*"

Java Split not working as expected

I am trying to use a simple split to break up the following string: 00-00000
My expression is: ^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])
And my usage is:
String s = "00-00000";
String pattern = "^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])";
String[] parts = s.split(pattern);
If I play around with the Pattern and Matcher classes I can see that my pattern does match and the matcher tells me my groupCount is 7 which is correct. But when I try and split them I have no luck.

String.split does not use capturing groups as its result. It finds whatever matches and uses that as the delimiter. So the resulting String[] are substrings in between what the regex matches. As it is the regex matches the whole string, and with the whole string as a delimiter there is nothing else left so it returns an empty array.
If you want to use regex capturing groups you will have to use Matcher.group(), String.split() will not do.

for your example, you could simply do this:
String s = "00-00000";
String pattern = "-";
String[] parts = s.split(pattern);

I can not be sure, but I think what you are trying to do is to get each matched group into an array.
Matcher matcher = Pattern.compile(pattern).matcher();
if (matcher.matches()) {
String s[] = new String[matcher.groupCount()) {
for (int i=0;i<matches.groupCount();i++) {
s[i] = matcher.group(i);
}
}
}

From the documentation:
String[] split(String regex) -- Returns: the array of strings computed by splitting this string around matches of the given regular expression
Essentially the regular expression is used to define delimiters in the input string. You can use capturing groups and backreferences in your pattern (e.g. for lookarounds), but ultimately what matters is what and where the pattern matches, because that defines what goes into the returned array.
If you want to split your original string into 7 parts using regular expression, then you can do something like this:
String s = "12-3456";
String[] parts = s.split("(?!^)");
System.out.println(parts.length); // prints "7"
for (String part : parts) {
System.out.println("Part [" + part + "]");
} // prints "[1] [2] [-] [3] [4] [5] [6] "
This splits on zero-length matching assertion (?!^), which is anywhere except before the first character in the string. This prevents the empty string to be the first element in the array, and trailing empty string is already discarded because we use the default limit parameter to split.
Using regular expression to get individual character of a string like this is an overkill, though. If you have only a few characters, then the most concise option is to use foreach on the toCharArray():
for (char ch : "12-3456".toCharArray()) {
System.out.print("[" + ch + "] ");
}
This is not the most efficient option if you have a longer string.
Splitting on -
This may also be what you're looking for:
String s = "12-3456";
String[] parts = s.split("-");
System.out.println(parts.length); // prints "2"
for (String part : parts) {
System.out.print("[" + part + "] ");
} // prints "[12] [3456] "