Hi, am trying to split the word rtmp://xx.yyy.in/sample/test?22082208,False#&all this word.The word sample is dynamically added I don't know the count.
I want to split /sample/ how to do this kindly help me?
You want the string.split() method
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/String.html#split%28%29
var array:Array = myString.split("/"); //returns an array of everything in between /
In your case this will return
[0]->?rtmp:/ [1]->xx.yy.in [2]->sample [3]->test?22082208,False#&all
If you're looking for everything aside from the test?22082208,False#&all part and your URL will always be in this format you can use string.lastIndexOf()
var pos:int = string.lastIndexOf("/", 0); //returns the position of the last /
var newString:String = string.substr(0, pos); //creates a new string starting at 0 and ending at the last index of /
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/String.html#substr%28%29
You can do this (and almost everything) with regex:
var input:String = "rtmp://xx.yyy.in/sample/test?22082208,False#&all";
var pattern:RegExp = /^rtmp:\/\/.*\/([^\/]*)\/.*$/;
trace(input.replace(pattern, "$1")); //outputs "sample"
Here is the regex in details:
^ : start of the string
rtmp:\/\/ first string to find "rtmp://"
.* anything
\/ first slash
([^\/]) capture everything but a slash until...
\/ ...second slash
.* anything
$ the end
Then $1 represents the captured group between the parenthesis.
Related
I have a scenario where i want to extract some substring based on following condition.
search for any pattern myvalue=123& , extract myvalue=123
If the "myvalue" present at end of the line without "&", extract myvalue=123
for ex:
The string is abcdmyvalue=123&xyz => the it should return myvalue=123
The string is abcdmyvalue=123 => the it should return myvalue=123
for first scenario it is working for me with following regex - myvalue=(.?(?=[&,""]))
I am looking for how to modify this regex to include my second scenario as well. I am using https://regex101.com/ to test this.
Thanks in Advace!
Some notes about the pattern that you tried
if you want to only match, you can omit the capture group
e* matches 0+ times an e char
the part .*?(?=[&,""]) matches as least chars until it can assert eiter & , or " to the right, so the positive lookahead expects a single char to the right to be present
You could shorten the pattern to a match only, using a negated character class that matches 0+ times any character except a whitespace char or &
myvalue=[^&\s]*
Regex demo
function regex(data) {
var test = data.match(/=(.*)&/);
if (test === null) {
return data.split('=')[1]
} else {
return test[1]
}
}
console.log(regex('abcdmyvalue=123&3e')); //123
console.log(regex('abcdmyvalue=123')); //123
here is your working code if there is no & at end of string it will have null and will go else block there we can simply split the string and get the value, If & is present at the end of string then regex will simply extract the value between = and &
if you want to use existing regex then you can do it like that
var test = data1.match(/=(.*)&|=(.*)/)
const result = test[1] ? test[1] : test[2];
console.log(result);
I have been trying to get a regex expression to return me the following in the following situations.
XX -> XX
XXX -> XXX
XX/XX -> XX
XX/XX/XX -> XX/XX
XXX/XXX/XX -> XXX/XXX
I had the following Regex, however they do no work.
^[^/]+ => https://regex101.com/r/xvCbNB/1
=========
([A-Z])\w+ => https://regex101.com/r/xvCbNB/2
They are close but are not there.
Any Help would be appreciated.
You want to get all text from the start till the last occurrence of a specific character or till the end of string if the character is missing.
Use
^(?:.*(?=\/)|.+)
See the regex demo and the regex graph:
Details
^ - start of string
(?:.*(?=\/)|.+) - a non-capturing group that matches either of the two alternatives, and if the first one matches first the second won't be tried:
.*(?=\/) - any 0+ chars other than line break chars, as many as possible upt to but excluding /
| - or
.+ - any 1+ chars other than line break chars, as many as possible.
It will be easier to use a replace here to match / followed by non-slash characters before end of line:
Search regex:
/[^/]*$
Replacement String:
""
Updated RegEx Demo 1
If you're looking for a regex match then use this regex:
^(.*?)(?:/[^/]*)?$
Updated RegEx Demo 2
Any special reason it has to be a regular expression? How about just splitting the string at the slashes, remove the last item and rejoin:
function removeItemAfterLastSlash(string) {
const list = string.split(/\//);
if (list.length == 1) [
return string;
}
list.pop();
return list.join("/");
}
Or look for the last slash an remove it:
function removeItemAfterLastSlash(string) {
const index = string.lastIndexOf("/");
if (index === -1) {
return string;
}
return string.splice(0, index);
}
I want to search all the words in a line/sentence and detect any word with ize and convert it to ise except for certain words listed.
Find: ^(?!size)(?!resize)(?!Belize)(?!Bizet)(?!Brize)(?!Pfizer)(?!assize)(?!baize)(?!bedizen)(?!citizen)(?!denizen)(?!filesize)(?!maize)(?!prize)(?!netizen)(?!seize)(?!wizen)(?!outsize)(?!oversize)(?!misprize)(?!supersize)(?!undersize)(?!unsized)(?!upsize)([a-zA-Z-\s]+)ize
Replace: $1ise
So far all i get is the first word of the line with ize to work, or the last word with ize to work.
Example Organize to socialize whatever size.
To Organise to socialise whatever size.
Find (?i)(?!size|resize|Belize|so&so|unsized|upsize)(?<!\w)(\w+)ize
Replace $1ise
worked as intended. Capitalisation issues added (?i)
The regex ([a-zA-Z-\s]+)ize has the whitespace marker in it (\s) so it will will match anything beyond the word boundary. You might want to work with \w and/or \b to match only characters from the word where the "ize" is located. Additionally, you don't want the ^ at the beginning since this would match the start of the string.
Possible regex: (?!....your list....)(\w+)ize
Example input: "Organize to socialize whatever size."
Found matches: "Organize" and "socialize", but not "size", see https://regex101.com/r/UIfoa8/1
After that you can use your replacement $1ise to replace the found string with the captured group and "ise".
Make a Whitelist Array
Make the excluded words (whitelist) an array of strings
.split(' ') the text being searched through (searchStr) into an array
then .map() through each word of the array
using .indexOf() to compare a word vs. the whitelist
using .test() to see if it's a x+"ize" word to .replace()
Once the searchArray is complete, .join() it into a string (resultString).
Demo
"organize", "mesmerized", "socialize", and "baptize" was mixed into the search string of some whitelist words
var searchStr = `organize Belize Bizet mesmerized Brize Pfizer assize baize bedizen citizen denizen filesize socialize maize prize netizen seize wizen outsize baptize`;
var whitelist = ["size", "resize", "Belize", "Bizet", "Brize", "Pfizer", "assize", "baize", "bedizen", "citizen", "denizen", "filesize", "maize", "prize", "netizen", "seize", "wizen", "outsize", "oversize", "misprize", "supersize", "undersize", "unsized", "upsize"];
var searchArray = searchStr.split(' ').map(function(word) {
var match;
if (whitelist.indexOf(word) !== -1) {
match = word;
} else if (/([a-z]+?)ize/i.test(word)) {
match = word.replace(/([a-z]+?)ize/i, '$1ise');
} else {
match = word;
}
return match;
});
var resultString = searchArray.join(', ');
console.log(resultString);
Using notepad++, how can I replace the -s noted by the carats? The dashes I want to replace occurs every 7th character in the string.
11.871-2-2.737-2.00334-2
^ ^ ^
123456781234567812345678
It's pretty simple since it's only dashes:
(\S*?)-
Begin capture group.............................. (
Find any number of non-space chars... \S*
Lazily until...............................................?
End capture group...................................)
No capture find hyphen...........................-
Demo 1
var str = `11.871-2-2.737-2.00334-2`;
var sub = `$1`;
var rgx = /(\S*?)-/g;
var res = str.replace(rgx, sub);
console.log(res);
"There is a dash (right above 1) that I would like to preserve. This seems to get rid of all the dashes in the string"
The question clearly shows that there isn't a dash at the "1 position", but since there's a possibility that it's possible considering the pattern (n7). Don't have time to break it down, but I can refer you to a proper definition of the meta char \b.
Demo 2
var str = `-11.871-2-2.737-2.00334-2`;
var sub = `$1$2`;
var rgx = /\b[-]{1}(\S*?)-(\S*?)\b/g;
var res = str.replace(rgx, sub);
console.log(res);
Search for ([0-9\.-]{6,6})-
Replace with: $1MY_SEPARATOR
I have the following string of characters:
73746174652C313A312C310D
|
- extract the value at this position
I would like to extract the value 1 (the 1 at the end of the string) using regex.
So basically a regex that acts as a charAt(index).
I need this solution for a 3rd party application that only supports regular expressions. Note that the application cannot access capture groups and does not support negative lookbehinds.
In C#:
(?<=^.{21})(.)
in JS:
/.(?=.{2}$)/
You could try:
(?<=^.{21}).
It won't work in Javascript, but perhaps it will work in your app.
It means: a single character preceded (?<= ... ) by the beginning of the string ^ plus 21 characters .{21} . So, in the end, it returns the 22th character.
The 22nd character is in capture group 1.
/^.{21}(.)/
But what system are you in that requires this instead of normal string processing?
Depends how you want to match it ( x distance from the beginning or x distance from the end )
/(.).{2}$/ Third from the end (capturing group 1)
/^.{21}(.)/ 22nd character (capturing group 1)
//PHP
$str = '73746174652C313A312C310D';
$char = preg_replace('/(.).{2}$/','$1',$str); //3rd from last
preg_match('/(.).{2}$/',$str,$chars); //3rd from last
$char = $chars[1];
preg_match('/^.{21}(.)/',$str,$chars); //22nd character
$char = $chars[1];
//JS
var str = '73746174652C313A312C310D';
var ch = str.replace(/(.).{2}$/,'$1'); //3rd from last
var ch = str.match(/(.).{2}$/)[1]; //3rd from last
var ch = str.match(/^.{21}(.)/)[1]; //22nd character
If you're having to use the result of the First match: bit of your tool, run it twice:
73746174652C313A312C310D - ^.{21}. = 73746174652C313A312C31
73746174652C313A312C31 - .$ = 1