Non-greedy regular expression search/replace

i want to be able to do non-greedy regular expressions as
dreamweaver defaults to greedy and eats all of the match it can
match.
Is there an add-on to add a checkbox or soemthing in the
Search and Replace Dialog for non-greedyness?
i guess i could throw something in there to match some simple
expressions like...
text to search n' replace:
$Query .= ", dateReceived='".addslashes($dateReceived)."'" .
", dateReleased='".addslashes($dateReleased)."'";
Instead of using this now
Find: '"\.addslashes$\$([^$]*)\)\."'
Replace: ". dateInsert($$1) ."
i want to just use this wihout it getting greedy with the .*
Find: '"\.addslashes$\$(.*)$\."'
Replace: ". dateInsert($$1) ."

In regular expression, if you want the .* to not be greedy
you add a '?' after so yo you have '.*?'
Find: '".addslashes($dateReleased)."'
Hope that help,
Chris
Adobe Dreamweaver Engineering

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Grouping & Back-references with regular expressions on Replace Text window

I really appreciate the inclusion of the Regular Expressions in the search & replace feature. One thing I am missing is back-references in the replacement expression. For instance, in the unix tools vi or sed, I might do something like this:
s/$firstPart$ $secondPart$ $oldThirdPart$/\2 \1 newThirdPart/g
which would allow me to switch the places of firstPart and secondPart, and totally replace thirdPart. If grouping and back-references are already present in the Replace Text window, how does one correctly invoke them?

duplicate of Grouping & Back-references with regular expressions on Replace Text window

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.

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.

Regular Expression Search in Forms Builder

Hi everyone, can anyone help me out?
I'm working on a legacy forms migration project (3 to 10g) where item references often do not include the block name.
eg.
:stock_level := :default_level;
instead of
:stk.stock_level := :ctrl.default_level;
There are also instances of set_item_property('stock_level',....)
In complex multi-block forms this obviously makes the form difficult to understand.
Is there a way in which I can search the triggers and program units of a form for items that don't have a block prefix? The Forms search and replace function has a Regular Expression feature, but I have no experience in this area.
I assume I'm looking for something like this:- ":", followed by a character string that doesn't contain a "." or "="
Thanks

You might have a look at the commercial product formsAPIMaster, it has the ability to do such searches (and also automatic replacements, if needed)

Regular Expression for replacing Leading Spaces

I don't claim to be any expert on Regular Expressions and even after reading CD's introduction to Reg Exp, I still can't figure out this one which I'm sure must be very basic.
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:])*', '.')
2 as result
3 from dual;
RESULT
.      FRED BLOGS    WAS HERE
SQL>Which is replacing the start of line with a "." and not the spaces.
But I want my result to be:-
RESULT
......FRED BLOGS    WAS HERE
SQL>Cheers

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
SQL> SELECT col1, REGEXP_REPLACE(col1, ' ([^ ]+.*)|','*\1')
2 FROM (SELECT ' FRED BLOGS WAS HERE ' col1 FROM dual );
COL1
REGEXP_REPLACE(COL1,'([^]+.*)|','*\1')
FRED BLOGS WAS HERE
*FRED BLOGS    WAS HERE
SQL> SELECT col1, REGEXP_REPLACE(col1, ' ([^ ]+.*)| ','*\1')
2 FROM (SELECT ' FRED BLOGS WAS HERE ' col1 FROM dual );
COL1
REGEXP_REPLACE(COL1,'([^]+.*)|','*\1')
FRED BLOGS WAS HERE
******FRED BLOGS WAS HERE

Regular Expression Search for Case Statement in VBA

Hi,
I'm having trouble trying to use regular expressions in a case statement. I have a CSV spreadsheet of a server's netstat output and am trying to plot everything into Visio. I have been able to do that, however I'm not trying to expand this capability and
resuse the same code for many different servers.
I have the mainServer variable set as a Variant and in my current example it is set as "INTPXY001" (internal proxy server 001). I have tried different regex statements for the potential to have INTPXY001 - INTPXY999, EXTPXY001 - EXTPXY999, and
SVCPXY001 - SVCPXY999 in place of the Case "INTPXY001", but nothing I have tried seems to work.
'========================================
Set mainServer As Variant
Set AppVisio = CreateObject("visio.application")
AppVisio.Visible = True
AppVisio.Documents.AddEx "", visMSDefault, 0
AppVisio.Documents.OpenEx "server_u.vss", visOpenRO + visOpenDocked
mainServer = ActiveSheet.Cells(1, 2) 'sets mainServer to INTPXY001
With AppVisio.ActiveWindow.Page
Select Case mainServer
Case "INTPXY001"
.Drop AppVisio.Documents.Item("SERVER_U.VSS").Masters.ItemU("Proxy server"), 2.25, 9.25
Case Else
.Drop AppVisio.Documents.Item("SERVER_U.VSS").Masters.Item(("Server"), 2.25, 9.25
End Select
End With
'========================================

You cannot declare variables As Variant in VBScript. All variables in VBScript are implicitly variants.
If you are asking about VBA (Visual Basic for Applications), then you're not asking in the correct forum.
-- Bill Stewart [Bill_Stewart]

Regular expressions for replacing text with sms language text

Hi, I'm trying to write a function which converts normal, correctly spelled text into the shorter sms language format but struggling to come up with the regular expressions i need to do so, can anyone help?
1: remove surplus white space at the beginning of a sentence and at the end of a sentence.
e.g. " hello." --> "hello." OR "hello ." --> "hello."
2: remove preceeding and/or proceeding space if there's a word then a number possibly followed by another word
e.g. "come 2 me" --> "come2me" OR "dnt 4get" --> "dnt4get"
3: remove "aeiou" if word starts and ends with "!aeiou"
e.g. "text" --> "txt"

You can make the whitespace on either side optional: text = text.replaceAll("\\s*(\\d)\\s*", "$1");1. Use String's trim() method.
3. This one has to be done in two steps: import java.util.regex.*;
public class Test
public static void main(String... args) throws Exception
    String text = "The quick brown fox jumps over the lazy dog.";
    System.out.println(devowelize(text));
public static String devowelize(String str)
    Pattern p = Pattern.compile(
      "[a-z&&[^aeiou]]++(?:[aeiou]++[a-z&&[^aeiou]]++)+",
      Pattern.CASE_INSENSITIVE);
    Matcher m = p.matcher(str);
    StringBuffer sb = new StringBuffer();
    while (m.find())
      m.appendReplacement(sb, m.group().replaceAll("[aeiou]+", ""));
    m.appendTail(sb);
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}

Feedback / Submit an idea don't Work (Page not found) || Multiline Regular Expression Search

Hello,
I reached for some months an idea with a (multiline edeting, regexp search).
Nice to see this has already been integrated, making me but disappointing is the regex search, in combination with the Multiline Edeting.
But what it brings me to use the regexp search, when I cant edit all hits.
In this feature, I would like to point again.
But unfortunately I cant find the feedback system, and here in the forum I always get the error message, page not found.
Beautiful would be if both would be patched;)
For me, the code is currently the "http://www.sublimetext.com/" editor of my choice if Adobe would therefore produce some its functions, edge code would be a real alternative.
Yet I find Edge Code is not ready for Daily-use.
Greetings
Dirk Persky

Hi Dirk,
Thanks for reporting the broken links -- we'll look into that. As Randy said, you're welcome to just post suggestions & issues here in the forum, though.
Regarding multiline regexp search: that is possible in Edge Code, with some limitations:
To search multiple files, use the Find in Files command, and then use any regexp containing "\n" -- it will match the newlines correctly.
To search a single file, right-click the file and chose "Find In..." (it will behave like Find in Files, but limited to that one file)
Caveat: you can't use the multiline $/^ anchors in your regexp
Caveat: there's no workaround for using the Replace command with multiline regexps
The latest Brackets release 0.41 fixes both the caveats listed above -- though you'll still have to use the "Find In..." workaround (or "Replace In..." for Replace). We eventually plan to further improve this -- you can watch the full multiline search/replace item on our feature backlog for updates.
Hope that helps,
- Peter

Match pattern: change regular expression search via front panel?

Hello,
I have an application that I am developing which is making use of Serial VISA.
I am scanning the output of a serial port which is constantly spitting out a long string of data.
Data is being pulled from the string with several combinations of SCAN FROM STRING functions and MATCH PATTERN Functions.
Question:
How can I use a button or TEXT box on the FRONT PANEL that can change the MATCH PATTERN Functions Regular expression?
for example the string may spit out the following:
Weight\s\s\s\s\s\s\s\s\s\s\s\s-0.00\slb\s\s\s\s\s\s-16\sbits\s+74.40\s\B0F\sCorrected\s\s\s\s\s\s-0.00\slb\s+0.999987\s%\s\r
in this case the serial device is spitting out data in LB.
It could be possible for the device to spit out data in KG ... thus I need to change the REGULAR EXPRESSION.
Thank you for your time

Hi,
just a quick example.
there are other way of doing this but I think the ComboBox is quite an easy one.
hope this helps
When my feet touch the ground each morning the devil thinks "bloody hell... He's up again!"
Attachments:
ComboBox.vi ‏12 KB

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
How can I write a regular expression where I can extract the above values as follows
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

What is regular expression search function

hi
what is use CHARTOROWID ,COMPOSE fuction in oracle

http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions025.htm#SQLRF00615
http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions032.htm#SQLRF00618

Regular Expression replacement not working

I am trying to use a regular expression to replace non-ascii characters on a file, and I'm afraid I've reached the end of my regex knowledge.
Here is the specific code
'Set the Regular Expression paramaters
Set RegEx = CreateObject("VBScript.Regexp")
RegEx.Global = True
RegEx.Pattern = "[^\u0000-\u007F]"
RegEx.IgnoreCase = True
'Replace the UTF-8 characters
ReplacedText = RegEx.Replace(FileText, "\u0020")
If I understand regular expressions correctly the pattern of "[^\u0000-\u007F]" should replace any character that is not an ascii character, and then replace it with a space (which I understand is "\u0020"). What am I doing wrong?

Simply use
ReplacedText = RegEx.Replace(FileText, " ")
Regards, Hans Vogelaar (http://www.eileenslounge.com)

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.

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.

Non-greedy regular expression search/replace

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