Regular expression to replace "emtpy space" ( ) bitween words with +

Hallo!
When I wish to find in code something like this:
12144541 FirstWord SecondWord
regular expression for that is:
(\d{1,100})[\s-]\D{1,100}[\s-]\D{1,100}
Now, please help me tu find regular expression to replace
"emtpy space" ( ) bitween words with +
12144541 FirstWord SecondWord to become
12144541+FirstWord+SecondWord
Thank you very, very, very much!

A simple-minded solution is to use \s to match all
whitespace; e.g. find \s and replace with +. DW CS3, at least, is
smart enough to not replace end of line characters with the '+'
character if you limit your search & replace to text.

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Regular Expression for replacing Leading Spaces

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I want to replace all the leading spaces in a string with "." chrs. I could do this using the common replace/substr/instr functions, but I reckoned it would be possible in a single regular regexp_replace call.
So far I've got this...
SQL> select regexp_replace('      FRED BLOGS    WAS HERE    ', '^([:space:])*', '.')
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That was very good solution .
Can you explain me the significance of "| " in the code, other things I could trace out.
I try to run the code with the 2 cases
when I give a space after | symbol it prints the * many times
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2 FROM (SELECT ' FRED BLOGS WAS HERE ' col1 FROM dual );
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3. This one has to be done in two steps: import java.util.regex.*;
public class Test
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public static String devowelize(String str)
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}

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Thought of that, but it would remove spaces from crucial areas of my pages. For example, conditional logic within a script, like:
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Regular expression for replace

I have the following sample values in a table column ( first name, middle initial, last name)
Assemblymember Von K. Wright
Assemblyman Kalvin J. Rowling
Assemblywoman Debby J. Sanders
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Wright V
Rowling K
Sander D
Thanks

possibility of some other ways too...
with rt as
(select 'Von K. Wright' str from dual union all
select 'Kalvin J. Rowling' from dual union all
select 'Debby J. Sanders' from dual)
select str,regexp_replace(str, '(.*) (.*) (.*)', '\3 \1') str from rt;
Row#     STR     STR_1
1     Von K. Wright     Wright Von
2     Kalvin J. Rowling     Rowling Kalvin
3     Debby J. Sanders     Sanders Debby
with rt as
(select 'Von K. Wright' str from dual union all
select 'Kalvin J. Rowling' from dual union all
select 'Debby J. Sanders' from dual)
select str,regexp_replace(str, '(.*) (.*) (.*)', '\3 ') || substr(regexp_replace(str, '(.*) (.*) (.*)', '\1'),1,1) str from rt;
Row#     STR     STR_1
1     Von K. Wright     Wright V
2     Kalvin J. Rowling     Rowling K
3     Debby J. Sanders     Sanders D

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Hi,
I'm trying to figure out how to find spaces between word and digit.
I am limited to use only word 'WORD' (either capital or small caps) with different endings like -ing -s -y and more (using \S+).
I wrote something like (?i)(?<=WORD\S+)\s(?=\d+) but this does not seem to work due to some limitation of lookahead I belive?
Any suggestions?
Peter

Peter Stnsz wrote:
… find what: (\<WORD\S+)(\s)(?=\d)
change to: $1~s
Hmmh?
Your Grep do not find your first example: WORD 0,2 (WORD without any ending)
And you don't need the second ()
And please do not use \S
Use this instead:
(\<WORD\l*)\s(?=\d)
(l is the little L)
Have fun

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Hello,
I'm new on Powershell and I'm trying to make the script that:
searches over file and replaces all the spaces which have been found between quotes;
removes all quotes (except these which has not value eg "").
For example:
Source file:
string3=string4 string="string1 string2 string23" string8="" string5="string7 string8"
Destination file:
string3=string4 string=string1_string2_string23 string8="" string5=string7_string8
I have been created script that searches the data correctly
$file="c:\scripts\mk.txt"
$data=Get-Content $file
$1
$regex = [regex]@'
(?x) # ignore pattern whitespace option
(?<test>(["'])(?:(?=(\\?))\2.)*?\1)
$data |% {
if ($_ -match $regex){
new-object psobject -property @{
test = $matches['test']
}#when adding "|select-object test" I'm getting the correct data
I have stopped here on search / replace operation (from space to underscore, and by removing the quotes leaving the empty quotes non touched). Can you help me to finish the script?
Thanks!

Try this. It uses the [Regex] Replace static method, with a script block delegate:
$file="c:\scripts\mk.txt"
$data=Get-Content $file
$regex = '(\S+=[^"\s]+)|(\S+="[^"]+")'
$delegate = { $args[0].value.replace(' ','_') -replace '"(.+)"','$1' }
$data |% { [regex]::Replace($_,$regex,$delegate) }
[string](0..33|%{[char][int](46+("686552495351636652556262185355647068516270555358646562655775 0645570").substring(($_*2),2))})-replace " "

Regular Expression / String matching other than equality check with groovy.

I want to have a string comparison operation using a groovy expression.
Not an equality check, but something like
#{variableName like 'IC%'?'true':'false'}
I saw some operators like
#{bindings.gradeName.inputValue =~ 'IC'?true:false}
But it doesnt seem to be working.
Is it possible with groovy?
Arun

Hi
nything will do.
And not using bean.
This expression has to added in an AMX page (ADF Mobile
It depends on where you want to execute this. Groovy in a validation script would be different to EL in a page as they would be in different stages in the lifecycle.
Now for ADF MObile this may be different, I dont know.'
If its to be part of a validation then you can call the source methods in groovy and do the comparison.
from a validator for instance:
source.getGradeName() returns the grade name before commiting.
String enteredgrade = source.getGradeName();
if (enteredgrade.index(0).equals('I') and enteredgrade(1).equals('C'){
do what you want here. i.e. return true or false etc.
el in a page would look entirly different but you can implement this same functionality in a bean, access it via the accessors and then return true or false.
if you are already using EL and you managed to link your page with JSTL, you will have to define the access variable in the header section inorder to access it and then use "startswith", though I havent tried it this way.

Introduction to regular expressions ... last part.

Continued from Introduction to regular expressions ... continued., here's the third and final part of my introduction to regular expressions. As always, if you find mistakes or have examples that you think could be solved through regular expressions, please post them.
Having fun with regular expressions - Part 3
In some cases, I may have to search for different values in the same column. If the searched values are fixed, I can use the logical OR operator or the IN clause, like in this example (using my brute force data generator from part 2):
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE data IN ('abc', 'xyz', '012');There are of course some workarounds as presented in this asktom thread but for a quick solution, there's of course an alternative approach available. Remember the "|" pipe symbol as OR operator inside regular expressions? Take a look at this:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)$')
;I can even use strings composed of values like 'abc, xyz , 012' by simply using another regular expression to replace "," and spaces with the "|" pipe symbol. After reading part 1 and 2 that shouldn't be too hard, right? Here's my "thinking in regular expression": Replace every "," and 0 or more leading/trailing spaces.
Ready to try your own solution?
Does it look like this?
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(' || REGEXP_REPLACE('abc, xyz , 012', ' *, *', '|') || ')$')
;If I wouldn't use the "^" and "$" metacharacter, this SELECT would search for any occurence inside the data column, which could be useful if I wanted to combine LIKE and IN clause. Take a look at this example where I'm looking for 'abc%', 'xyz%' or '012%' and adding a case insensitive match parameter to it:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)', 'i')
; An equivalent non regular expression solution would have to look like this, not mentioning other options with adding an extra "," and using the INSTR function:
SELECT data
FROM (SELECT data, LOWER(DATA) search
          FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search LIKE 'abc%'
    OR search LIKE 'xyz%'
    OR search LIKE '012%'
SELECT data
FROM (SELECT data, SUBSTR(LOWER(DATA), 1, 3) search
          FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search IN ('abc', 'xyz', '012')
; I'll leave it to your imagination how a complete non regular example with 'abc, xyz , 012' as search condition would look like.
As mentioned in the first part, regular expressions are not very good at formatting, except for some selected examples, such as phone numbers, which in my demonstration, have different formats. Using regular expressions, I can change them to a uniform representation:
WITH t AS (SELECT '123-4567' phone
             FROM dual
            UNION
           SELECT '01 345678'
             FROM dual
            UNION
           SELECT '7 87 8787'
             FROM dual
SELECT t.phone, REGEXP_REPLACE(REGEXP_REPLACE(phone, '[^0-9]'), '(.{3})(.*)', '(\1)-\2')
FROM t
;First, all non digit characters are beeing filtered, afterwards the remaining string is put into a "(xxx)-xxxx" format, but not cutting off any phone numbers that have more than 7 digits. Using such a conversion could also be used to check the validity of entered data, and updating the value with a uniform format afterwards.
Thinking about it, why not use regular expressions to check other values about their formats? How about an IP4 address? I'll do this step by step, using 127.0.0.1 as the final test case.
First I want to make sure, that each of the 4 parts of an IP address remains in the range between 0-255. Regular expressions are good at string matching but they don't allow any numeric comparisons. What valid strings do I have to take into consideration?
Single digit values: 0-9
Double digit values: 00-99
Triple digit values: 000-199, 200-255 (this one will be the trickiest part)
So far, I will have to use the "|" pipe operator to match all of the allowed combinations. I'll use my brute force generator to check if my solution works for a single value:
SELECT data
FROM TABLE(regex_utils.gen_data('0123456789', 3))
WHERE REGEXP_LIKE(data, '^(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$')
; More than 255 records? Leading zeros are allowed, but checking on all the records, there's no value above 255. First step accomplished. The second part is to make sure, that there are 4 such values, delimited by a "." dot. So I have to check for 0-255 plus a dot 3 times and then check for another 0-255 value. Doesn't sound to complicated, does it?
Using first my brute force generator, I'll check if I've missed any possible combination:
SELECT data
FROM TABLE(regex_utils.gen_data('03.', 15))
WHERE REGEXP_LIKE(data,
                   '^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; Looks good to me. Let's check on some sample data:
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
            UNION
           SELECT '256.128.64.32'
             FROM dual
SELECT t.ip
FROM t WHERE REGEXP_LIKE(t.ip,
                   '^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; No surprises here. I can take this example a bit further and try to format valid addresses to a uniform representation, as shown in the phone number example. My goal is to display every ip address in the "xxx.xxx.xxx.xxx" format, using leading zeros for 2 and 1 digit values.
Regular expressions don't have any format models like for example the TO_CHAR function, so how could this be achieved? Thinking in regular expressions, I first have to find a way to make sure, that each single number is at least three digits wide. Using my example, this could look like this:
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
SELECT t.ip, REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2')
FROM t
; Look at this: leading zeros. However, that first value "00127" doesn't look to good, does it? If you thought about using a second regular expression function to remove any excess zeros, you're absolutely right. Just take the past examples and think in regular expressions. Did you come up with something like this?
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
SELECT t.ip, REGEXP_REPLACE(REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2'),
                            '[0-9]*([0-9]{3})(\.?)', '\1\2'
FROM t
; Think about the possibilities: Now you can sort a table with unformatted IP addresses, if that is a requirement in your application or you find other values where you can use that "trick".
Since I'm on checking INET (internet) type of values, let's do some more, for example an e-mail address. I'll keep it simple and will only check on the
"[email protected]", "[email protected]" and "[email protected]" format, where x represents an alphanumeric character. If you want, you can look up the corresponding RFC definition and try to build your own regular expression for that one.
Now back to this one: At least one alphanumeric character followed by an "@" at sign which is followed by at least one alphanumeric character followed by a "." dot and exactly 3 more alphanumeric characters or 2 more characters followed by a "." dot and another 2 characters. This should be an easy one, right? Use some sample e-mail addresses and my brute force generator, you should be able to verify your solution.
Here's mine:
SELECT data
FROM TABLE(regex_utils.gen_data('a1@.', 9))
WHERE REGEXP_LIKE(data, '^[[:alnum:]]+@[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})$', 'i'); Checking on valid domains, in my opinion, should be done in a second function, to keep the checks by itself simple, but that's probably a discussion about readability and taste.
How about checking a valid URL? I can reuse some parts of the e-mail example and only have to decide what type of URLs I want, for example "http://", "https://" and "ftp://", any subdomain and a "/" after the domain. Using the case insensitive match parameter, this shouldn't take too long, and I can use this thread's URL as a test value. But take a minute to figure that one out for yourself.
Does it look like this?
WITH t AS (SELECT 'Introduction to regular expressions ... last part. URL
             FROM dual
            UNION
           SELECT 'http://x/'
             FROM dual
SELECT t.URL
FROM t
WHERE REGEXP_LIKE(t.URL, '^(https*|ftp)://(.+\.)*[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})/', 'i')
Update: Improvements in 10g2
All of you, who are using 10g2 or XE (which includes some of 10g2 features) may want to take a look at several improvements in this version. First of all, there are new, perl influenced meta characters.
Rewriting my example from the first lesson, the WHERE clause would look like this:
WHERE NOT REGEXP_LIKE(t.col1, '^\d+$')Or my example with searching decimal numbers:
'^(\.\d+|\d+(\.\d*)?)$'Saves some space, doesn't it? However, this will only work in 10g2 and future releases.
Some of those meta characters even include non matching lists, for example "\S" is equivalent to "[^ ]", so my example in the second part could be changed to:
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions', '\S')), 0)
FROM dual
;Other meta characters support search patterns in strings with newline characters. Just take a look at the link I've included.
Another interesting meta character is "?" non-greedy. In 10g2, "?" not only means 0 or 1 occurrence, it means also the first occurrence. Let me illustrate with a simple example:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +')
FROM dual
;This is old style, "greedy" search pattern, returning everything until the last space.
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +?')
FROM dual
;In 10g2, you'd get only "Having " because of the non-greedy search operation. Simulating that behavior in 10g1, I'd have to change the pattern to this:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^[^ ]+ +')
FROM dual
;Another new option is the "x" match parameter. It's purpose is to ignore whitespaces in the searched string. This would prove useful in ignoring trailing/leading spaces for example. Checking on unsigned integers with leading/trailing spaces would look like this:
SELECT REGEXP_SUBSTR(' 123 ', '^[0-9]+$', 1, 1, 'x')
FROM dual
;However, I've to be careful. "x" would also allow " 1 2 3 " to qualify as valid string.
I hope you enjoyed reading this introduction and hope you'll have some fun with using regular expressions.
C.
Fixed some typos ...
Message was edited by:
cd
Included 10g2 features
Message was edited by:
cd

Can I write this condition with only one reg expr in Oracle (regexp_substr in my example)?I meant to use only regexp_substr in select clause and without regexp_like in where clause.
but for better understanding what I'd like to get
next example:
a have strings of two blocks separated by space.
in the first block 5 symbols of [01] in the second block 3 symbols of [01].
In the first block it is optional to meet one (!), in the second block it is optional to meet one (>).
The idea is to find such strings with only one reg expr using regexp_substr in the select clause, so if the string does not satisfy requirments should be passed out null in the result set.
with t as (select '10(!)010 10(>)1' num from dual union all
select '1112(!)0 111' from dual union all --incorrect because of '2'
select '(!)10010 011' from dual union all
select '10010(!) 101' from dual union all
select '10010 100(>)' from dual union all
select '13001 110' from dual union all -- incorrect because of '3'
select '100!01 100' from dual union all --incorrect because of ! without (!)
select '100(!)1(!)1 101' from dual union all -- incorrect because of two occurencies of (!)
select '1001(!)10 101' from dual union all --incorrect because of length of block1=6
select '1001(!)10 1011' from dual union all) --incorrect because of length of block2=4
select '10110 1(>)11(>)0' from dual union all)--incorrect because of two occurencies of (>)
select '1001(>)1 11(!)0' from dual)--incorrect because (!) and (>) are met not in their blocks
--end of test data

Want to replace a string containing consecutive repeating words to one using regular expression

Hi Experts,
I need a regular expression to replace all duplicate words in a string with one.
eg: 'Hello Hello World 4-4-5 etc etc' should be changed to 'Hello World 4-4-5 etc'.
I tried many of them but they had one or the other problem. like (\w+\S\W)\1+' replace with ' \1' and ' (\w+\W)\1+' replaced with ' \1' , etc
Thanks in advance
Tarique

Hi,
Translating what frank said to JAVA would be something like this:
        StringBuffer result = new StringBuffer();
        String myString = "This is right right, that is wrong.";
        String[] words = myString.split(" ");
        String lastWord = "";
        for (String str : words){
            if (!str.contains(lastWord))
                result.append(str);
            else
                result.append(str.substring((lastWord.length() >= 0 ? lastWord.length() : 0 ) , str.length()));
            lastWord = str;
            result.append(" ");
        System.out.println(result);
If you didnt have points and commas in your message then would be easier. But the code is not 100% correct and you will need to make it work according to yours requirements.

Regular expression on words with % wildcard

Hi,
I've got some processing working using regular expression where I need to process words e.g.
regexp_replace('word1 word2','(\w+)','myprefix{\1}') - results in - 'myprefixword1 myprefixword2'
However, if I'm presented with this; '%word0 word1% wo%d2 word3', then I need to treat % as special case and leave the word as is, so result here would be; - '%word0 word1% wo%d2 myprefixword3', is this achievable using regexp ?

And for those who don't know, I guess we should explain why we're having to expand single spaces to double spaces...
(I'll use the "¬" character to represent spaces to make it clearer to see)
If we have a string such as
word1¬word2¬word3and we want to identify the words in the string (without using any special regexp word identifier) then we are going to use the spaces to identify the start and end of words. To make life easy, we manually put a space at the start and end of the string so we can say that each word in the string will have a space before and after it regardless of where it is in the string...
¬word1¬word2¬word3¬However, when we specify what we want to search for we are going to say we want a space, followed by a number of characters (not spaces), followed by a space...
¬[^¬]*¬So, ideally, you'd expect it to look through the string and say
¬word1¬word2¬word3¬
\_____/... found word1
¬word1¬word2¬word3¬
      \_____/... found word2
¬word1¬word2¬word3¬
            \_____/... found word3
Unfortunately, there is a problem. Once the first word has been found the pointer for searching the rest of the string is located on the next character after the match i.e.
¬word1¬word2¬word3¬
       ^So it won't be able to pick out word2 and will only get to word3. Let's see it in action...
SQL> ed
Wrote file afiedt.buf
1 with t as (select ' word1 word2 word3 ' as txt from dual)
2 --
3 select regexp_replace(txt, ' [^ ]* ', 'xxxxx') as txt
4* from t
SQL> /
TXT
xxxxxword2xxxxx
SQL>In order to deal with this, if we replace the single spaces with double spaces (not required at the start and end) our string looks like...
¬word1¬¬word2¬¬word3¬So as it searches it finds word1 as a match and then the pointer in the string is located...
¬word1¬¬word2¬¬word3¬
       ^... so the next match for the pattern of space-characters-space is word2 and then the pointer is located...
¬word1¬¬word2¬¬word3¬
              ^... ready to find word 3. Example...
SQL> ed
Wrote file afiedt.buf
1 with t as (select ' word1 word2 word3 ' as txt from dual)
2 --
3 select regexp_replace(txt, ' [^ ]* ', 'xxxxx') as txt
4* from t
SQL> /
TXT
xxxxxxxxxxxxxxx
SQL>Hopefully that's a little clearer. You just have to remember the "pointer" principle and the fact that once a match is found it is located on the character after the match.
;)

Regular expression and pattern matching/replacing

I have a list of key words. It has around 1000 key word now but can grow to 5000 keywords.
My web application displays lot of texts which are stored in the database. My requirement is to scan each text for the occurance of any of the above keywords. If any keyword is present I have to replace that with some custom values, before showing it to the user.
I was thinking of using using regular expression for replacing the keyword in the text using matcher.replaceAll method as follows:
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(inputStr);
String output = matcher.replaceAll(replacementStr);
But My pattern string will have around 5000 keywords with the 'OR' Logical Operator like- keyword1| keyword2 I keyword3 | ..........
Will such a big pattern string adversly affect the performance? What can I do to speed up the performance? (Since my keyword list is not static i would prefer to do the replacement just before showing the text to the user)
Any suggestions are most welcome.

I don't think a pure regex approach would be that slow, but it would be a maintenance nightmare. I think a combined regex/table-lookup approach would be best: use a regex to identify potential keywords, then look them up in the table to confirm. For instance, to find all Java keywords you could use the regex "\\b[a-z]{2,12}+\\b" to filter out anything that can't possibility be a keyword.
What are you going to replace the keywords with? Will it vary depending on which keyword is found? If so, you'll have to use a table--and you won't be able to use the replaceAll method, because it can't handle dynamically generated replacement values. You would have to use the lower-level appendReplacement and appendTail method instead.

Regular expression to replace "emtpy space" ( ) bitween words with +

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