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Why the space appears as sub string in this split instruction?

I have string with spaces and some non-informative characters and substrings required to be excluded and just to keep some important sections. I used the split as below:
String myString[]={"01: Hi you look tired today? Can I help you?"};
myString=myString[0].split("[\\s+]");// Split based on any white spaces
for(int ii=0;ii<myString.length;ii++)
System.out.println(myString[ii]);
The result is :
01:
Hi
you
look
tired
today?
Can
I
help
you?
The spaces appeared after the split as sub strings when the regex is “[\s+]” but disappeared when the regex is "\s+". I am confused and not able to find answer in the related stack overflow pages. The link regex-Pattern made me more confused.
Please help, I am new with java.
19/1/2015:Edit
After your valuable advice, I reached to point in my program where a conditional statements is required to be decomposed and processed. The case I have is:
String s1="01:IF rd.h && dq.L && o.LL && v.L THEN la.VHB , av.VHR with 0.4610;";
String [] s2=s1.split(("[\\s\\&\\,]+"));
for(int ii=0;ii<s2.length;ii++)System.out.println(s2[ii]);
The result is fine till now as:
01:IF
rd.h
dq.L
o.LL
v.L
THEN
la.VHB
av.VHR
with
0.4610;
My next step is to add string "with" to the regex and get rid of this word while doing the split.
I tried it this way:
String s1="01:IF rd.h && dq.L && o.LL && v.L THEN la.VHB , av.VHR with 0.4610;";
String [] s2=s1.split(("[\\s\\&\\, with]+"));
for(int ii=0;ii<s2.length;ii++)System.out.println(s2[ii]);
The result not perfect, because I got unwonted extra split at every "h" letter as:
01:IF
rd.
dq.L
o.LL
v.L
THEN
la.VHB
av.VHR
0.4610;
Any advice on how to specify string with mixed white spaces and separation marks?
Many thanks.

inside square brackets, [\s+] will represent the whitespace character class with the plus sign added. it is only one character so a sequence of spaces will split many empty strings as Todd noted, and will also use + as separator.
you should use \s+ (without brackets) as the separator. that means one or more whitespace characters.
myString=myString[0].split("\\s+");

Your biggest problem is not understanding enough about regular expressions to write them properly. One key point you don't comprehend is that [...] is a character class, which is a list of characters any one of which can match. For example:
[abc] matches either a, b or c (it does not match "abc")
[\\s+] matches any whitespace or "+" character
[with] matches a single character that is either w, i, t or h
[.$&^?] matches those literal characters - most characters lose their special regex meaning when in a character class
To split on any number of whitespace, comma and ampersand and consume "with" (if it appears), do this:
String [] s2 = s1.split("[\\s,&]+(with[\\s,&]+)?");
You can try it easily here Online Regex and get useful comments.

Java split by alphabeta char creates an empty value in array

I want to split my string on every occurrence of an alpha-beta character.
for example:
"s1l1e13" to an array of: ["s1","l1","e13"]
when trying to use this simple split by regex i get some weird results:
testStr = "s1l1e13"
Arrays.toString(testStr.split("(?=[a-z])"))
gives me the array of:
["","s1","l1","e13"]
how can i create the split without the empty array element?
I tried a couple more things:
testStr = "s1"
Arrays.toString(testStr.split("(?=[a-z])"))
does return the currect array: ["s1"]
but when trying to use substring
testStr = "s1l1e13"
Arrays.toString(testStr.substring(1).split("(?=[a-z])")
i get in return ["1","l1","e13"]
what am i missing?

Your Lookahead marks each position before any character of a to z; marking the following positions:
s1 l1 e13
^ ^ ^
So by spliting using just the Lookahead, it returns ["", "s1", "l1", "e13"]
You can use a Negative Lookbehind here. This looks behind to see if there is not the beginning of the string.
String s = "s1l1e13";
String[] parts = s.split("(?<!\\A)(?=[a-z])");
System.out.println(Arrays.toString(parts)); //=> [s1, l1, e13]

Your problem is that (?=[a-z]) means "place before [a-z]" and in your text
s1l1e13
you have 3 such places. I will mark them with |
|s1|l1|e13
so split (unfortunately correctly) produces "" "s1" "l1" "e13" and doesn't automatically remove for you first empty elements.
To solve this problem you have at least two options:
make sure that there is something before your place you need to split on (it is not at start of your string). You can use for instance (?<=\\d)(?=[a-z]) if you want to split after digit but before character
(PREFFERED SOLUTION) start using Java 8 which automatically removes empty strings at start of result array if regex used on split is zero-length (look-arounds are zero length).

The first match finds "" to be okay because its looking ahead for any alpha character, which is called zero-width lookahead, so it doesn't need to actually match anything. So "s" at the beginning is alphanumeric, and it matches that at a probable spot.
If you want the regex to match something always, use ".+(?=[a-z])"

The problem is that the initial "s" counts as an alphabetic character. So, the regex is trying to split at s.
The issue is that there is nothing before the s, so the regex machine instead decides to show that there is nothing by adding the null element. It'll do the same thing at the end if you ended with "s" (or any other letter).
If this is the only string you're splitting, or if every array you had starts with a letter but does not end with one, just truncate the array to omit the first element. Otherwise, you'll probably need to loop through each array as you make it so that you can drop empty elements.

So it seems your matches has the pattern x###, where x is a letter, and # is a number.
I'd make the following Regex:
([a-z][0-9]+)

Regular Expression to select first five CSVs from a string

I have a CSV string like apple404, orange pie, wind\,cool, sun\\mooon, earth, in Java. To be precise each value of the csv string could be any thing provided commas and backslash are escaped using a back slash.
I need a regular expression to find the first five values. After some goggling I came up with the following. But it wont allow escaped commas within the values.
Pattern pattern = Pattern.compile("([^,]+,){0,5}");
Matcher matcher = pattern.matcher("apple404, orange pie, wind\\,cool, sun\\\\mooon, earth,");
if (matcher.find()) {
System.out.println(matcher.group());
} else {
System.out.println("No match found.");
}
Does anybody know how to make it work for escaped commas within values?

Following negative look-behind based regex will work:
Pattern pattern = Pattern.compile("(?:.*?(?<!(?:(?<!\\\\)\\\\)),){0,5}");
However for full fledged CSV parsing better use a dedicated CSV parser like JavaCSV.

You can use String.split() here. By specifying the limit as 6 the first five elements (index 0 to 4) would always be the first five column values from your CSV string. If in case any extra column values are present they would all overflow to index 5.
The regex (?<!\\\\), makes sure the CSV string is only split at a , comma not preceded with a \.
String[] cols = "apple404, orange pie, wind\\,cool, sun\\\\mooon, earth, " +
"mars, venus, pluto".split("(?<!\\\\),", 6);
System.out.println(cols.length); // 6
System.out.println(Arrays.toString(cols));
// [apple404, orange pie, wind\,cool, sun\\mooon, earth, mars, venus, pluto]
System.out.println(cols[4]); // 5th = earth
System.out.println(cols[5]); // 6th discarded = mars, venus, pluto

This regular expression works well. It also properly recognizes not only backslash-escaped commas, but also backslash-escaped backslashes. Also, the matches it produces do not contain the commas.
/(?:\\\\|\\,|[^,])*/g
(I am using standard regular expression notation with the understanding that you would replace the delimiters with quote marks and double all backslashes when representing this regular expression within a Java string literal.)
example input
"apple404, orange pie, wind\,cool, sun\\,mooon, earth"
produces this output
"apple404"
" orange pie"
" wind\,cool"
" sun\\"
"mooon"
Note that the double backslash after "sun" is escaped and therefore does not escape the following comma.
The way this regular expression works is by atomizing the input into longest sequences first, beginning with double backslashes (treating them as one possible multi-byte character value alternative), followed by escaped commas (a second possible multi-byte character alternative), followed by any non-comma value. Any number of these atoms are matched, followed by a literal comma.
In order to obtain the first N fields, one may simply splice the array of matches from the previous answer or surround the main expression in additional parentheses, include an optional comma in order to match the contents between fields, anchor it to the beginning of the string to prevent the engine from returning further groups of N fields, and quantify it (with N = 5 here):
/^((?:\\\\|\\,|[^,])*,?){0,5}/g
Once again, I am using standard regular expression notation, but here I will also do the trivial exercise of quoting this as a Java string:
"^((?:\\\\\\\\|\\\\,|[^,])*,?){0,5}"
This is the only solution on this page so far which actually answers both parts of the precise requirements specified by the OP, "...commas and backslash are escaped using a back slash." For the input fi\,eld1\\,field2\\,field3\\,field4\\,field5\\,field6\\,, it properly matches only the first five fields fi\,eld1\\,field2\\,field3\\,field4\\,field5\\,.
Note: my first answer made the same assumption that is implicitly part of the OP's original code and example data, which required a comma to be following every field. The problem was that if input is exactly 5 fields or less, and the last field not followed by a comma (equivalently, by an empty field), then final field would not be matched. I did not like this, and so I updated both of my answers so that they do not require following commas.
The shortcoming with this answer is that it follows the OP's assumption that values between commas contain "anything" plus escaped commas or escaped backslashes (i.e., no distinction between strings in double quotes, etc., but only recognition of escaped commas and backslashes). My answer fulfills the criteria of that imaginary scenario. But in the real world, someone would expect to be able to use double quotes around a CSV field in order to include commas within a field without using backslashes.
So I echo the words of #anubhava and suggest that a "real" CSV parser should always be used when handling CSV data. Doing otherwise is just being a script kiddie and not in any way truly "handling" CSV data.

Parsing CSV with a RegEx in java - escape double quote within cell

I am looking for a java regex which will escape the doublequote within an excel cell.
I have followed this example but need another change in the regular expression to make it work for escaping doublequote within one of the cells.
Parsing CSV input with a RegEx in java
private final Pattern pattern = Pattern.compile("\"([^\"]*)\"|(?<=,|^)([^,]*)(?=,|$)");
Example Data:
"A,B","2" size","text1,text2, text3"
The regex from above fails at 2".
I want the output to be as below .Doesn't matter if the outer double quotes are there or not.
"A,B"
"2" size"
"text1,text2, text3"

while I agree, that using regex for parsing a CVS is not really the best way, a slightly better pattern is:
Pattern pattern = Pattern.compile("^\"([^\"]*)\",|,\"([^\"]*)\",|,\"([^\"]*)\"$|(?<=,|^)([^,]*)(?=,|$)");
This will terminate a cell value only after quote and comma, or start it after a command and a quote.

well as F.J commented, the input data is ambiguous. But for your example input, you could try
string.split("\",\"") method to get a String[].
after this, you got an array with 3 elements:
[
"A,B,
2" size,
text1,text2, text3"
]
remove the first character (which is double quote) of the first element of the array
remove the last character (which is double quote) of the last element of the array

Java String split regex not working as expected

The following Java code will print "0". I would expect that this would print "4". According to the Java API String.split "Splits this string around matches of the given regular expression". And from the linked regular expression documentation:
Predefined character classes
. Any character (may or may not match line terminators)
Therefore I would expect "Test" to be split on each character. I am clearly misunderstanding something.
System.out.println("Test".split(".").length); //0

You're right: it is split on each character. However, the "split" character is not returned by this function, hence the resulting 0.
The important part in the javadoc: "Trailing empty strings are therefore not included in the resulting array. "
I think you want "Test".split(".", -1).length(), but this will return 5 and not 4 (there are 5 ''spaces'': one before T, one between T and e, others for e-s, s-t, and the last one after the final t.)

You have to use two backslashes and it should work fine:
Here is example:
String[] parts = string.split("\\.",-1);

Everything is ok. "Test" is split on each character, and so between them there is no character. If you want iterate your string over each character you can use charAt and length methods.