i have problems with my shared hosting account. the apache server i'm using scrambles utf8 so i can't use Hebrew/Arabic in the url such as www.mydomain.com/אבא.php
So i want to know how can i make it that if someone asks for the page:
www.mydomain.com/%D7%90%D7%91%D7%90.php
he would get to the page: www.mydomain.com/D790D791D790.php (without the percentages)
but his browser url will show the first page he asked for (with no redirection).
I guess using mod_rewrite in .htaccess but have no clue how to approach this.
Please help you guys, this is a 911 for me.
I've been thinking about this for a little while now, and unless you know how many sets there are (6 in your example) I don't think there will be a terribly elegant way to do this. One solution may be to use a rather vague rewrite RewriteRule ^(.*?%.*?)\.php$ foo.php?bar=$1 then process the data in PHP where you have a few more options and quite a bit more flexibility.
Details of the Regex:
^.*?%.*?\.php$
Options: case insensitive
Assert position at the beginning of the string «^»
Match any single character that is not a line break character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “%” literally «%»
Match any single character that is not a line break character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “.” literally «\.»
Match the characters “php” literally «php»
Assert position at the end of the string (or before the line break at the end of the string, if any) «$»
Created with RegexBuddy
Related
I am trying to recover a password I have not used in a long time.
I know the words used in the passphrase, but I do not remember exactly the character substitutions,
and upper/lower case I have used. I only remember some, and know the possibilities for others.
The passphrase I am trying to recover is 15 characters long.
I have installed John the Ripper (jumbo version 1.9), and I tried to create some rules for character
substitutions I know I have used hoping to quickly generate a wordlist with all possible passphrases
based on my rules.
Let's say my passphrase is password with some character substitutions. If I use this set of rules:
sa#
ss$
so0
soO
I get those results:
p#ssword
pa$$word
passw0rd
passwOrd
When I say I am looking for all possible combinations, I am looking for something lookig more like the following (this list is not exhaustive)
p#ssword
p#$sword
p#$$word
pa$sword
pa$$word
p#ssw0rd
p#$sw0rd
p#$$w0rd
pa$sw0rd
pa$$w0rd
p#sswOrd
p#$swOrd
p#$$wOrd
pa$swOrd
pa$$wOrd
Gathering all rules in one line does not help me achieve my goal, and making one rule (line) with substitution by character position is basically generating my list by hand.
I am now wondering how can I achieve my goal, or, if JtR is the right tool for the job.
I have found a solution that fits my use case. the oNx syntax allows to replace the character at Nth position (zero based) with x.
In addition to that, using brackets allow to apply more than one substitution to the same character. So oN[xy] will yield two passwords with the character at Nth position replaced with x, then y.
For my password example above, the rule needed to achieve my goal would be:
o1[a#] o2[sS$] o3[sS$] o5[oO0]
I hope it helps someone with some old database to unlock )
All.
I am used to programming VBA in Excel, but am new to the structures in Word.
I am working through a library of text files to update them. Many of them are either OCR documents, or were manually entered.
Each has a recurring pattern, the most common of which is unnecessary carriage returns.
For example, I am looking at several text files where there is a double return after each line. A search and replace of all double carriage returns removes all paragraph distinctions.
However, each line is approximately 30 characters long, and if I manually perform the following logic, it gives me a functional document.
If there is a double carriage return after 30+ characters, I replace them with a space.
If there were less than 30 characters prior to the double return, I replace them with a single return.
Can anyone help me with some rudimentary code that would help me get started on that? I could then modify it for each "pattern" of text documents I have.
e.g.
In this case, there are more than
thirty characters per line. And I
will keep going to illustrate this
example.
This would be a new paragraph, and
would be separated by another of
the single returns.
I want code that would return:
In this case, there are more than thirty character returns. And I will keep going to illustrate this example.
This would be a new paragraph, and would be separated by another of the single returns.
Let me know if anyone can throw something out that I can play with!
You can do this without code (which RegEx requires), simply using Word's own wildcard Find/Replace tools, where:
Find = ([!^13]{30,})[^13]{1,}
Replace = \1^32
and, to clean up the residual multi-paragraph breaks:
Find = [^13]{2,}
Replace = ^p
You could, of course, record the above as a macro...
Here is a RegEx that might work for you:
(\n\n)(?<!\.(\n\n))
The substitution is just a plain space, you can try it out (and modify / tweak it) here: https://regex101.com/r/zG9GPw/4
This 'pattern' tells the RegEx engine to look for the newline character \n which occurs x2 like this \n\n (worth noting this is from your question and might be different in your files, e.g. could be \r\n) and it assumes that a valid line break will be proceeded by a full stop: \..
In RegEx the full stop symbol is a single character wild card so it needs to be escaped with the '\' (n and r are normal characters, escaping them tells the RegEx engine they represent newline and return characters).
So... the expression is looking for a group of x2 newline characters but then uses a negative look-behind to exclude any matches where the previous character was a full stop.
Anyway, it's all explained on the site:
Here is how you could do a RegEx find and replace using NotePad++ (I'm not sure if it comes with RegEx or if a plugin is needed, either way it is easy). But you can set a location, filters (to target specific file types), and other options (such as search in sub-directories).
Other than that, as #MacroPod pointed out you could also do this with MS Word, document by document, not using any code :)
This is a bit of a puzzler for me. I have a string that looks like:
fanspd<fanspd>3</fanspd>
doorinprocess<doorinprocess>0</doorinprocess>
timeremaining<timeremaining>0</timeremaining>
macaddr<macaddr>60:CB:FB:99:99:C1</macaddr>
ipaddr<ipaddr>10.0.0.6</ipaddr>
model<model>4.4eWHF</model>
softver: <softver>2.14.2</softver>
interlock1: <interlock1>0</interlock1>
interlock2: <interlock2>0</interlock2>
cfm: <cfm>2200</cfm>
power: <power>120</power>
inside: <house_temp>-99</house_temp>
<DNS1>10.0.0.1</DNS1>
attic: <attic_temp>76</attic_temp>
OA: <oa_temp>-99</oa_temp>
server response: <server_response>Ó£àêEE²ç©þ]kõ «jsÐ</server_response>
DIP Switches: <DIPS>11100</DIPS>
Remote Switch: <switch2>1111</switch2>
Setpoint:<Setpoint>0</Setpoint>
The string includes the "/n" so I have split it into corrisponding lines that look like
fanspd<fanspd>0</fanspd>
All I really want is the char(s) in the middle of the line. In the above example it would be 0.
I can match everything with regular expressions but by doing the following:
(.*)(<[a-z]+>)(.*)(</[a-z]+>)
But what I'd like is something more that would exclude or strip away or remove all the junk and grab the middle chars.
(!(.*)(!<[a-z]+>))(.*)(!(</[a-z]+>))
I've tried this and it does not work. I've also thought of doing another [NSstring componentsSeparatedByString:#"(with either < or or >"] but that would leave be with more parsing yet to do and I think there should be a way to get just the chars inbetween the tags with either regular expressions or string compare or some such way to parse out the
Any suggestions or help would be greatly appreciated.
Thanks
Two things.
Your regular expression does not escape the forward slash.
Your regular expression seems overly complicated for what you are trying to do.
If all you want is that lone middle character with regular expressions,
Try this:
<[a-z]+>(.*)<\/[a-z]+>
Here's a great tool to play around with:
http://rubular.com
Heck you could probably even get away with:
<[a-z]+>(.*)<\/
EDIT:
I figured out your problem partially, some of the tags part way down contain characters other than a through z. So here you go:
<.+>(.*)<\/.+>
I know there are a lot of resources with regex for it. But I could not find the one I want.
My problem is:
I want to remove one line comments (//) from obj-c sources, but I don't want to break the code in it. For instance, with this regex: #"//.*" I can remove all comments, but it also corrupts string literal:
#"bsdv//sdfsdf"
I played with non-capturing parentheses (?:(\"*\")*+), but without success.
Also I found this expression for Python:
r'(\".*?\"|\'.*?\')|(/\*.*?\*/|//[^\r\n]*$)'
It should cover my case, but I've not figure out how to make it work with obj-c.
Please, help me to build proper regex.
UPDATE: Yeah, that's a tough one, I know there're a lot of caveats, other than the one I described. I would appreciate if someone post regex that only fix my issue. Anyway, I gonna post my solution, without regex soon, I hope it will be helpful for anyone who struggling with such problem too.
Try this regex:
(?:^|.*;(?!.*")|#(?:define|endif|ifn?def|import|undef|...).*)\s*(//[^\r\n]+$)
Demo
http://regex101.com/r/jT4xC8
Description
Discussion
Besides all the warnings expressed in the comments, I assume that a single line can appear in two distinct cases:
Case 1: Alone on its line preceded or not by blank chars
Case 2: Not Alone on its line preceded or not by blank chars, and other chars.
In the first case, we match the beginning of the line (^ with /m flag). Then we search zero or more blank chars (\s*) and finally the single line comment: //[$\r\n]+$.
In the second case, if there are other chars on the line, they form statements. Any statement is ended by a semicolon ;. So we search the last statement and its corresponding semicolon .*;(?!.*"). Then we search the single line comment. Those other chars can be also preprocessor statements. In this case, they are introduced by a sharp #.
One important keypoint is that I assume the code passed to the regex is a code that compiles.
There is more
Don't forget also to add some other pre-processor directives that may apply in your case. Check this SO answer: https://stackoverflow.com/a/18014883/363573
I have a website on wordpress and I have written a plugin which required SEO friendly URL now i am stuck at the following
http://domain.com/catalogue-category/1/ is working fine but when i replace the /1/ with the name of the category like http://domain.com/catalogue-category/Seating/ it does not work at all and gives me 404 error.
Its also working at /catalogue-category/?cat=Seating
My apache rewrite rule is
RewriteRule ^catalogue-category/^([^/]+)/$ /catalogue-category/?cat=$1 [L]
I am not that good in mod rewrite as that in PHP, so please bear my ignorance and treat me as a rookie.
Looking forward to hear from the gurus :)
Try:
RewriteRule ^catalogue-category/([^/]+)/$ /catalogue-category/?cat=$1 [L]
You were using the caret ^ twice in your regex pattern (^catalogue-category/^([^/]+)/$), the caret asserts that the match should be from the start of the string.
Pattern explanations
Previous pattern:
Assert position at the beginning of the string «^»
Match the characters “catalogue-category/” literally
«catalogue-category/»Assert position at the beginning of the string «^» Match the regular expression below and
capture its match into backreference number 1 «([^/]+)» Match
any character that is NOT a “/” «[^/]+» Between one and
unlimited times, as many times as possible, giving back as needed
(greedy) «+» Match the character “/”
literally «/» Assert position at the end of the string (or
before the line break at the end of the string, if any) «$»
Suggested one
Assert position at the beginning of the string «^»
Match the characters “catalogue-category/” literally
«catalogue-category/» Match the regular expression below and
capture its match into backreference number 1 «([^/]+)» Match
any character that is NOT a “/” «[^/]+» Between one and
unlimited times, as many times as possible, giving back as needed
(greedy) «+» Match the character “/”
literally «/» Assert position at the end of the string (or
before the line break at the end of the string, if any) «$»