Find/Replace Using Regular Expressions

Can someone help me with this...I am using Regular expressions to
FIND:
http.*lid=([^&"]*)[^"]*
REPLACE:
$set(\1,ID_id,code)$
So that in the following it will change this:
a href="http://www.test.com/shc/s/home_10153_12605?lid=Search" rilt="Search"
To this:
a href="$set(Search,ID_id,code)$" rilt="Search
Those expressions work in Notepad++ but when i use dreamweaver it just replaces the http... with "$set(\1,ID_id,code)$" and doesnt reference the "search"
Any help?
Thanks

Let me begin by saying I'm a complete idiot with DW's Reg Ex. I use Search Specific Tag whenever possible. See screenshot below.
Try this on your Current Document to see if it works. Then make a back-up copy of site before attempting it on Entire Local Site as you cannot "Undo" this process.
Good luck,
Nancy O.

Similar Messages

Find text using regular expression and add highlight annotation

Hi Friends
Is it possible to find text using regular expression and add highlight annotation using plugin

A plugin can use the PDWordFinder to get a list of the words on a page, and their location. That's all that the API offers for searching. Of course, you can use a regular expression library to work with that word list.

Finding URLs using regular expression.

I have an requirement where user will type some text containing URLs like "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747. Thank you". This text has to be modified as below before saving it to the database.
"Please visit this site <a href='http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747'>http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747</a>. Thank you"
I am using regular expression (http|https)://.+?\\s which marks the end of the url with a white space character.This pattern doesn't work if the URL is located at the end of the string since there will be no space at the end.
For example if the string is "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747" the regex will fail.
My acutal problem is to find the URL irrespective its position within the string.
Pattern urlPattern = Pattern.compile("(http|https)://.+?\\s", Pattern.CASE_INSENSITIVE);
Matcher matcher = urlPattern.matcher(plainText);
Map stringIndexMap = new HashMap();
//Searching the input string for urlPattern...
while(matcher.find()) {
String urlString = matcher.group();
//Storing the urls in a hashmap with their indices as keys....
stringIndexMap.put(new Integer(matcher.start()), urlString.trim());
Set keySet = stringIndexMap.keySet();
Iterator it = keySet.iterator();
//Iterating over the hashmap containing urls...
while(it.hasNext()) {
String urlString = (String) stringIndexMap.get(it.next());
* Replacing the url string in the input text with <a href="#" onclick="window.open('<urlString>')"
* using String index
clickableURLString.replace(clickableURLString.indexOf(urlString),
clickableURLString.indexOf(urlString) + urlString.length(),
"<a href=\"#\" onclick=\"window.open('" + urlString
+ "')\">" + urlString + "</a>");
return clickableURLString.toString();

The end of the input is '$' as a regex.
import java.util.regex.*;
public class Prasanna{
public static void main(String[] args){
 String text
= "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747";
// String regex = "(http|https)://.+?(?:\\s|$)"; // this works
 String regex = "(http|https)://[^ ]+"; // this also works
 Pattern pat = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
 Matcher mat = pat.matcher(text);
 while (mat.find()){
 System.out.println(mat.group());
}

GUI Based tool for search and replace using regular expression

Hi,
I have developed this small tool which can be used for search and replace in multiple files using reqular expressions.
Features:
1. Full regular expression
2. GUI based with Highlighted results
3. Preview for replace available
4. Pure Java based.
5. Its like unix sed and grep
Please visit below site for download/more information :
http://sourceforge.net/projects/regexsearchrepl/
Thanks,
Hitesh Viseria

I agree with you, it cannot compete grep/sed/awk combination. I couldnt find anything even close to grep/sed for windows which will also have a preview and most important, free. That is what made me write this tool. I am trying to
improve its performance and more features like history etc.
Any suggestions on additional features are most welcome.

Find/replace and regular expression problem

Hello, i'm using find and replace with a regular expression
for the first time. I have it checkmarked and it's finding my text
but it's missing (not highlighting) the ')' at the end of the line.
Here's my code:
[($[0-9]+US)]
it's supposed to find everything inside the square brackets -
but it misses the closing parenthesis after . I need
to find this string and replace with nothing to remove the string
from any/all pages. Is there a reason why it's missing the closing
parenthesis? I was actually able to add a few more parenthesis
(e.g. "))))") before OR after the closing square bracket and it
still found the original text minus the closing bracket and the
extra parenthesis didn't prevent the text from being found.
Any help is appreciated!
James...

WyattEA wrote:
> Hello, i'm using find and replace with a regular express
for the first time. I
> have it checkmarked and it's finding my text but it's
missing (not
> highlighting) the ')' at the end of the line. Here's my
code:
>
> [($[0-9]+US)]
That's not how square brackets work
Try:
$\$\d+US$
A left parens, followed by the dollar sign, followed by at
least one
digit, followed by US,
followed by a right parens.
Mick
>
> it's supposed to find everything inside the square
brackets - but it misses
> the closing parenthesis after . I need to
find this string and replace
> with nothing to remove the string from any/all pages. Is
there a reason why
> it's missing the closing parenthesis? I was actually
able to add a few more
> parenthesis (e.g. "))))") before OR after the closing
square bracket and it
> still found the original text minus the closing bracket
and the extra
> parenthesis didn't prevent the text from being found.
>
> Any help is appreciated!
>
> James...
>

String replace using regular expressions

I'm not very good at regular expressions, but I would like my script to replace
<a href="somepage.html">
by
<a href="event:somepage">
How do I do this? Thanks in advance!

Replacing a string that matches a certain pattern with another string is one of the more common RegEx tasks. There is documentation on using them here:
http://livedocs.adobe.com/flex/3/html/help.html?content=12_Using_Regular_Expressions_01.ht ml
hth,
matt horn
flex docs

Using Regular Expressions to replace Quotes in Strings

I am writing a program that generates Java files and there are Strings that are used that contain Quotes. I want to use regular expressions to replace " with \" when it is written to the file. The code I was trying to use was:
String temp = "\"Hello\" i am a \"variable\"";
temp = temp.replaceAll("\"","\\\\\"");
however, this does not work and when i print out the code to the file the resulting code appears as:
String someVar = ""Hello" i am a "variable"";
and not as:
String someVar = "\"Hello\" i am a \"variable\"";
I am assumming my regular expression is wrong. If it is, could someone explain to me how to fix it so that it will work?
Thanks in advance.

Thanks, appearently I'm just doing something weird that I just need to look at a little bit harder.

Using Regular Expressions in Numbers 09?

Is there any way to use regular expressions in Numbers 09 in Find & Replace?

kilowattradio wrote:
Is there any way to use regular expressions in Numbers 09 in Find & Replace?
NO !
_Go to "Provide Numbers Feedback" in the "Numbers" menu_, describe what you wish.
Then, cross your fingers, and wait _at least_ for iWork'10
Yvan KOENIG (VALLAURIS, France) vendredi 25 septembre 2009 14:49:49

One for the Tekkies: How to get this output using REGULAR EXPRESSIONS?

How to get the below output using REGULAR EXPRESSIONS??
SQL> ed
Wrote file afiedt.buf
1* CREATE TABLE cus___addresses (full_address VARCHAR2(200 BYTE))
SQL> /
Table created.
SQL> PROMPT Address Format is: House #/Housename, street, City, Zip Code, COUNTRY
House #/Housename, street, City, Zip Code, COUNTRY
SQL> INSERT INTO cus___addresses VALUES('1, 3rd street, Lansing, MI 49001, USA');
1 row created.
SQL> INSERT INTO cus___addresses VALUES('3B, fifth street, Clinton, OK 74103, USA');
1 row created.
SQL> INSERT INTO cus___addresses VALUES('Rose Villa, Stanton Grove, Murray, TN 37183, USA');
1 row created.
SQL> SELECT * FROM cus___addresses;
FULL_ADDRESS
1, 3rd street, Lansing, MI 49001, USA
3B, fifth street, Clinton, OK 74103, USA
Rose Villa, Stanton Grove, Murray, TN 37183, USA
SQL> The REG EXP query shouLd output the ZIP codes: i.e. 49001, 74103, 37183 in 3 rows.Edited by: user12240205 on Jun 18, 2012 3:19 AM

Hi,
user12240205 wrote:
... Frank, ʃʃp's method, I understand. But your method, although correct, I find it difficult to understand.
Could you explain how you did this?? What does '.*(\d{5})\D*' and '\1' mean???
Your method is better because it uses only ONE reg expression function. ʃʃp's uses 2.In Oracle 10.2 (I believe) and higher, '\d' is equivalent to '[[:digit:]]', and '\D' is equivalent to '[^[:digit:]]'. I find '\d' and '\D' easier to type, but there's nothing wrong with using '[[:digit:]]' and '[^[:digit:]]'.
'.*' means "0 or more of any character".
'\D*' means "0 or more non-digits".
The whole expression, '.*(\d{5})\D*' means:
a. 0 or more characters (any characters)
b. 5 digits
c. 0 or more non-digits.
'\1' is a Backreference . It means the sub-string that matched the pattern after the 1st '(', up to (but not including) its matching ')'. In this case, that means the sub-string that matched '\d{5}', or b. using the explanation immediately above.
So the entire REGEXP_REPLACE call means "When you see a sub-string consisting of a., follwed immediately by b., followed immedately by c., replace that sub-string with b. alone."

Help in query using regular expression

HI,
I need a help to get the below output using regular expression query. Please help me.
SELECT REGEXP_SUBSTR ('PWRPKG(P/W+P/L+CC)', '[^+]+', 1, lvl) val, lvl
FROM DUAL,(SELECT LEVEL lvl FROM DUAL
CONNECT BY LEVEL <=(SELECT MAX ( LENGTH ('PWRPKG(P/W+P/L+CC)') - LENGTH (REPLACE ('PWRPKG(P/W+P/L+CC)','+',NULL))+ 1) FROM DUAL));
I need the output as
correct result:
==============
val lvl
P/W 1
P/L 2
CC 3
But i tried the above it is not coming the above result. Please help me where i did a mistake.
Thanks in advance

Frank gave you a solution in your other thread. You could simplify it if you are on 11g:
SQL> select * from table_x
2 /
TXT
TECHPKG(INTELLI CC+FRT SONAR)
PWRPKG(P/W+P/L+CC)
select txt,
 regexp_substr(
 txt,
 '(.*\()*([^+)]+)',
 1,
 column_value,
 null,
 2
 ) element,
 column_value element_number
from table_x,
 table(
 cast(
 multiset(
 select level
 from dual
 connect by level <= regexp_count(txt,'\+') + 1
 as sys.OdciNumberList
order by rowid,
 column_value
TXT ELEMENT ELEMENT_NUMBER
TECHPKG(INTELLI CC+FRT SONAR) INTELLI CC 1
TECHPKG(INTELLI CC+FRT SONAR) FRT SONAR 2
PWRPKG(P/W+P/L+CC) P/W 1
PWRPKG(P/W+P/L+CC) P/L 2
PWRPKG(P/W+P/L+CC) CC 3
SQL> SY.

Rplacing space with &nbsb; in html using regular expressions

Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What number does this represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

sorry i missed some information in above,The modified information was in red color
Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADIN G="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What&nbsb;number&nbsb;does&nbsb;this&nbsb;represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

Changeparticular characters in a string by using regular expressions ...

Hello Everyone,
I am trying to write a function by using oracles regular expression function REGEXP_REPLACE but I could not succed till now.
My problem as follows, I have a text in a column for example let say 'sdfsdf Sdfdfs Sdfd' I want replace all s and S characters with X and make the text look like 'XdfXdf XdfdfX Xdfd'.
Is it possible by using regular expressions in oracle ?
Can you give me some clues ?
Thank you

SSU wrote:
Hello Everyone,
I am trying to write a function by using oracles regular expression function REGEXP_REPLACE but I could not succed till now.
My problem as follows, I have a text in a column for example let say 'sdfsdf Sdfdfs Sdfd' I want replace all s and S characters with X and make the text look like 'XdfXdf XdfdfX Xdfd'.
Is it possible by using regular expressions in oracle ?
Can you give me some clues ?
Thank you
SQL> SELECT
2 regexp_replace('sdfsdf Sdfdfs Sdfd','s|S','X') from dual;
REGEXP_REPLACE('SD
XdfXdf XdfdfX XdfdRegards,
Achyut

Using regular expressions for validating time fields

Similar to my problem with converting a big chunk of validation into smaller chunks of functions I am trying to use Regular Expressions to handle the validation of many, many time fields in a flexible working time sheet.
I have a set of FormCalc scripts to calculate the various values for days, hours and the gain/loss of hours over a four week period. For these scripts to work the time format must be in HH:MM.
Accessibility guidelines nix any use of message box pop ups so I wanted to get around this by having a hidden/visible field with warning text but can't get it to work.
So far I have:
var r = new RegExp(); // Create a new Regular Expression Object
r.compile ("^[00-99]:\\] + [00-59]");
var result = r.test(this.rawValue);
if (result == true){
true;
form1.flow.page.parent.part2.part2body.errorMessage.presence = "visible";
else (result == false){
false;
form1.flow.page.parent.part2.part2body.errorMessage.presence = "hidden";
Any help would be appreciated!

Date and time fields are tricky because you have to consider the formattedValue versus the rawValue. If I am going to use regular expressions to do validation I find it easier to make them text fields and ignore the time patterns (formattedValue). Something like this works (as far as my very brief testing goes) for 24 hour time where time format is HH:MM.
// form1.page1.subform1.time_::exit - (JavaScript, client)
var error = false;
form1.page1.subform1.errorMsg.rawValue = "";
if (!(this.isNull)) {
var time_ = this.rawValue;
if (time_.length != 5) {
    error = true;
else {
    var regExp = /^([01]?[0-9]|2[0-3]):[0-5][0-9]$/;
    if (!(regExp.test(time_))) {
      error = true;
if (error == true) {
form1.page1.subform1.errorMsg.rawValue = "The time must be in the format HH:MM where HH is 00-23 and MM is 00-59.";
form1.page1.subform1.errorMsg.presence = "visible";
Steve

Using regular expressions to get a customized output

Hi,
I have a string/varchar variable with the data ',a,b,c,' in it.
I want the display as follows:
a
b
c
I would like to get the similar output using regular expressions.
How do I get this output using REGEXP_REPLACE or REGEXP_SUBSTR?
Please do the needful.
Thanks & Regards,
Rakshit

I remember that, however if we look closer, that one has a little flaw: The 2nd row should be null, because ",," indicates an empy field. The MODEL clause solution works just fine in this case:
with t as (select 'aaaa,,bbbb,cccc,dddd,eeee,ffff' col1 from dual)
-- end of sample data
SELECT col_new
FROM t
MODEL
 PARTITION BY (ROWNUM rn)
 DIMENSION BY (0 dim)
 MEASURES(col1, col1 col_new)
 RULES ITERATE(99) UNTIL (ITERATION_NUMBER = LENGTH(REGEXP_REPLACE(col1[0], '[^,]')))
 (col_new[ITERATION_NUMBER] = REPLACE(REGEXP_SUBSTR(col1[0], '(^|,)[^,]*', 1, ITERATION_NUMBER+1), ','))
COL_NEW
aaaa
bbbb
cccc
dddd
eeee
ffff
7 Zeilen ausgewählt.Update: I had this nagging feeling that I missed something, and there it was. If you want to see what the problem with my solution is, change the example to
with t as (select ',aaaa,,bbbb,cccc,dddd,eeee,ffff' col1 from dual)So I went back and tried to fix BlueShadows approach. Here it is:
with t as (select 'aaaa,,bbbb,cccc,dddd,eeee,ffff' txt from dual)
-- end of sample data
SELECT REPLACE(REGEXP_SUBSTR(',' || txt, ',[^,]*', 1, level), ',') col_new
FROM t
CONNECT BY level <= length(regexp_replace(txt,'[^,]*'))+1
;C.

How to fetch substring using regular expression

Hi,
I am new to using regular expression and would like to know some basic details of how to use them in Java.
I have a String example= "http://www.google.com/foobar.html#*q*=database&aq=f&aqi=g10&fp=c9fe100d9e542c1e" and would like to get the value of "q" parameter (in bold) using regular expression in java.
For the same example, when we tried using javascript:
match = example.match("/^http:\/\/(?:(?!mail\.)[^\.]+?\.)?google\.[^\?#]+(?:.*[\?#&](?:as_q|q)=([^&]+))?/i}");
document.write('
' + match);
We are getting the output as: http://www.google.com/foobar.html#q=database,*database* where the bold text is the value of "q" parameter.
In Java we are trying to get the value of the q parameter separately or atleast resembles the output given by JavaScript. Please help me resolving this issue.
Regards
Praveen

BalusC wrote:
Regex is a cumbersome solution for fixed patterns like URL's. String#substring() in combination with String#indexOf would most likely already suffice.I usually agree, although, in this case, finding the exact parameter might be difficult without a small regex, perhaps:
"\\wq=\\s*"in conjunction with Pattern/Matcher, used similarly to an indexOf() to find the start of the parameter value.
Winston

Find/Replace Using Regular Expressions

Similar Messages

Maybe you are looking for