Regular Expression to Remove Part of URLs

I need to accomplish the following:
Search for all URLs that contain "
http://www.mycompany.com/dept/catalogs/"
and strip off anything past the .../catalogs/"
Basically, I have 3000+ links that begin with "
http://www.mycompany.com/dept/catalogs/"
and I need to strip whatever they may have after the .../catalogs/
piece of the URL.
Anyone know of an automatic way to accomplish this?
Thanks in advance!

In what programming environment are you doing this?
In DW's Find/Replace?
If so, then try this carefully on a test page, as I have not
tested it:
Find:
http:\/\/www\.mycompany\.com\/dept\/catalogs\/[^\"]*\"
Replace with:
http://www.mycompany.com/dept/catalogs/"
If you need this in some other environment, let us know.
E. Michael Brandt
www.divahtml.com
www.divahtml.com/products/scripts_dreamweaver_extensions.php
Standards-compliant scripts and Dreamweaver Extensions
www.valleywebdesigns.com/vwd_Vdw.asp
JustSo PictureWindow
JustSo PhotoAlbum, et alia

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Hi there,
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Strange regular expression difference when using an URL

Hello all,
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Use your original pattern but URL decode the URL first using class URLDecoder.
P.S. I have not checked your regex BUT it looks over complicated.

Regular Expression to remove space in HTML Tag

Hello All,
My HTML string is like below.
select '<CityName>RICHMOND</CityName>
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<StateCd/>
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Hi,
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Dear All,
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MichaelS wrote:
I am trying to achieve this using a SINGLE regexp_replace.Here we go:
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select 'THIS Number 124356 Is to Change.This Number 5 Also' str from dual
select str, regexp_replace(str, '(\d{0,})\d{1}', '\1$') str2
from t
STR                                                     STR2
ABC124556def568gh236klJ258                              ABC12455$def56$gh23$klJ25$
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Nice..
Learning for me ..
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Hii
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But there's no difference between two group of lines which you posted. Could you please format your input and desired output with please?
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> Hi,
> How to use regular expressions to remove email addresses
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> Thanks in advance.
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Hello,
I’m new to JAVA and Regular Expressions; I’m trying to write a regular expression that will find any records that contain a non digit byte in a multiple byte field.
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\D{1,}
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Below is my sample data. I would like the regular expression to find all of the records that are not all numeric. However when I use the regular expression \D{1,} it is only finding the 2 records that all bytes are non digits. (i.e. “ “ and “A “)
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“A0111229”
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Any assistance someone can provide would be greatly appreciated.

You don't show the code you are using but I surmise you are using String.matches() which requires that the whole target must match the regular expression not just part of it. Instead you should create a Pattern and then a Matcher and use the Matcher.find() method. Check the Javadoc for Pattern and Matcher and look at the Java regex tutorial - http://docs.oracle.com/javase/tutorial/essential/regex/ .
P.S. You can re-use the Pattern object - you don't have to create it every time you need one.
P.P.S. Java regular expressions work with characters not bytes and characters are not not not bytes.

Regular Expressions

I am trying to use a regular expression to filter out certain urls from getting aliased in my httpd.conf file.
I currently have
ScriptAliasMatch /someDir/ "C:/myCGIBin/"
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Ex. http://www.mySite.com/someDir/someApp.exe maps internally to c:/myCgiBin/someApp.exe
The catch is that there is 1 subdirectory which should not point to the cgi-bin
/someDir/specialDir/*
So any requests to this special directory should fail the ScriptAliasMatch, while all others pass.
ex.
/someDir/somefile.exe PASS
/someDir/someOtherDir/* PASS
/someDir/specialDir/* FAILI am not sure how to build a regular expression to achieve this.

public static void test3() {
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    // so the regex does not match XXXXX right after /dirA/
    String str = "/dirA/(?!dirB/).*";
    Pattern pat = Pattern.compile(str);
    String [] testCases = { "/dirA/file.exe",
                            "/dirA/goodDir/*",
                            "/dirA/dirB/*",
                            "/dirA/dirB.txt"}; // this last one is valid!
    Matcher m = pat.matcher("");
    for (int i = 0, maxi = testCases.length; i < maxi; i++)
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System.out.println(testCases[i] + " passes: " + m.find());

Regular expression - find repeating +++ signs

I'm trying to use regular expressions to remove duplicate +++ signs in a string. When I test my pattern using the expresso test (www.ultrapico.com) it parses the string correctly, in Java 1.5 it doesn't work. .. mp.matches() is always false. Any suggestions would be appreciated.
finalLongstring = "TTL1,clip1+TTL2+++clip3,TTL4,clip4,TTL5,clip5+TTL6+clip6+TTL7+clip7,TTL8,clip8,TTL9,clip9,TTL10,clip10,TTL11,clip11,TTL12,clip12,TTL13,clip13+TTL14+clip14,TTL15,clip15,TTL16,clip16,TTL17,clip17,TTL18,clip18,TTL19,clip19,TTL20,clip20,TTL21,clip21,TTL22,clip22,TTL23,clip23,TTL24,clip24,TTL25,clip25,TTL26,clip26,TTL27,clip27,TTL28,clip28,TTL29,clip29";
Pattern multiplePunctuation=null;
          multiplePunctuation=Pattern.compile("[,+]{2,6}");
          //                                     | |
          //                                     | 2 or more times
          //                                     a comma or plus character
          Matcher mp=multiplePunctuation.matcher(finalLongstring);
          if(mp.matches()){
               finalLongstring=mp.replaceAll("+");
/code]

Answere in your other thread.
http://forum.java.sun.com/thread.jspa?threadID=5143654

Regular Expression with comma and encapsulated charaters

Would appreciate some help. Looking for a regular expression to remove comma's from encapsulated text as follows
For example
- Input
1,"This is a string, need to remove the comma",Another text string,10
- Required output
1,"This is a string; need to remove the comma",Another text string,10
Have tried to use the REGEXP_REPLACE but could not grasp the pattern matching.
Thanks John

John Heaton wrote:
Thanks for the solution,this works great for a single field encapsulated by " and containing ,. I am parsing several different file definitions so it would need to cascade through the string for a undetermined number of times and replace all occurrences. Then try (performance-wise) MODEL solution:
{code}
with t as (
select '1,"This is a string, need to remove the comma",Another text string,10' txt from dual union all
select '1,"remove this comma,",Another text string,10,"remove this comma,",xxx,"remove this comma,",11' txt from dual
select txt_original,
txt
from t
model
partition by(row_number() over(order by 1) p)
dimension by(1 rn)
measures(txt txt_original,txt,0 quote)
rules
iterate(
1e9
until(
iteration_number + 1 = length(txt[1])
quote[1] = case substr(txt[1],iteration_number + 1,1)
when '"' then quote[1] + 1
else quote[1]
end,
txt[1] = case substr(txt[1],iteration_number + 1,1)
when ',' then case mod(quote[1],2)
when 1 then substr(txt[1],1,iteration_number) || ';' || substr(txt[1],iteration_number + 2)
else txt[1]
end
else txt[1]
end
TXT_ORIGINAL TXT
1,"This is a string, need to remove the comma",Another text string,10 1,"This is a string; need to remove the comma",Another text string,10
1,"remove this comma,",Another text string,10,"remove this comma,",xxx,"remove this comma,",11 1,"remove this comma;",Another text string,10,"remove this comma;",xxx,"remove this comma;",11
SQL>
SY.

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

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