I have the following string
234234=AWORDHERE('sdf.'aa')
where I need to extract AWORDHERE.
Sometimes there can be space in between.
234234= AWORDHERE('sdf.'aa')
Can I do this with a regular expression?
Or should I do it manually by finding indexes?
The datasets are huge, so it's important to do it as fast as possible.
Try this regex:
\d+=\s?(\w+)\(
Check Demo
in Javascript it would like that:
var myString = "234234=AWORDHERE('sdf.'aa')";// or 234234= AWORDHERE('sdf.'aa')
var myRegexp = /\d+=\s?(\w+)\(/g;
var match = myRegexp.exec(myString);
console.log(match[1]); // AWORDHERE
You could do this at least three ways. You need to benchmark to see what's fastest.
Substring w/ indexes
function extract(from) {
var ixEq = from.indexOf("=");
var ixParen = from.indexOf("(");
return from.substring(ixEq + 1, ixParen);
}
.
Splits
function extract(from) {
var spEq = from.split("=");
var spParen = spEq[1].split("(");
return spParen[0];
}
Regex (demo)
Here is some sample regex you could use
/[^=]+=([^(]+).*/g
This says
[^=]+ - One or more character which is not an =
= - The = itself
( - creates a matching group so you can access your match in code
[^(]+ - One or more character which is not a (
) - closes the matching group
.* - Matches the rest of the line
the /g on the end tells it to perform the match on all lines.
Using look around you can search for string preceded by = and followed by ( as following.
Regex: (?<==)[A-Z ]+(?=\()
Explanation:
(?<==) checks if [A-Z ] is preceded by an =.
[A-Z ]+ matches your pattern.
(?=\() checks if matched pattern is followed by a (.
Regex101 Demo
var str = "234234= AWORDHERE('sdf.'aa')";
var regexp = /.*=\s+(\w+)\(.*\)/g;
var match = regexp.exec(str);
alert( match[1] );
I made my solution for this just a little more general than you asked for, but I don't think it takes much more time to execute. I didn't measure. If you need greater efficiency than this provides, comment and I or someone else can help you with that.
Here's what I did, using the command prompt of node:
> var s = "234234= AWORDHERE('sdf.'aa')"
undefined
> var a = s.match(/(\w+)=\s*(\w+)\s*\(.*/)
undefined
> a
[ '234234= AWORDHERE(\'sdf.\'aa\')',
'234234',
'AWORDHERE',
index: 0,
input: '234234= AWORDHERE(\'sdf.\'aa\')' ]
>
As you can see, this matches the number before the = in a[1], and it matches the AWORDHERE name as you requested in a[2]. This will work with any number (including zero) spaces before and/or after the =.
Related
I'm trying to execute regex replace after match char, example 3674802/3 or 637884-ORG
The id can become one of them, in that case, how can I use regex replace to match to remove after the match?
Input var id = 3674802/3 or 637884-ORG;
Expected Output 3674802 or 637884
You could use sbustring method to take part of string only till '/' OR '-':
var input = "3674802/3";
var output = input.substr(0, input.indexOf('/'));
var input = "637884-ORG";
var output = input.substr(0, input.indexOf('-'));
var input = "3674802/3";
if (input.indexOf('/') > -1)
{
input = input.substr(0, input.indexOf('/'));
}
console.log(input);
var input = "637884-ORG";
if (input.indexOf('-') > -1)
{
input = input.substr(0, input.indexOf('-'));
}
console.log(input);
You can use a regex with a lookahead assertion
/(\d+)(?=[/-])/g
var id = "3674802/3"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())
id = "637884-ORG"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())
You don't need Regex for this. Regex is far more powerful than what you need.
You get away with the String's substring and indexOf methods.
indexOf takes in a character/substring and returns an integer. The integer represents what character position the character/substring starts at.
substring takes in a starting position and ending position, and returns the new string from the start to the end.
If are having trouble getting these to work; then, feel free to ask for more clarification.
You can use the following script:
var str = '3674802/3 or 637884-ORG';
var id = str.replace(/(\d+)[-\/](?:\d+|[A-Z]+)/g, '$1');
Details concerning the regex:
(\d+) - A seuence of digits, the 1st capturing group.
[-\/] - Either a minus or a slash. Because / are regex delimiters,
it must be escaped with a backslash.
(?: - Start of a non-capturing group, a "container" for alternatives.
\d+ - First alternative - a sequence of digits.
| - Alternative separator.
[A-Z]+ - Second alternative - a sequence of letters.
) - End of the non-capturing group.
g - global option.
The expression to replace with: $1 - replace the whole finding with
the first capturing group.
Thanks To everyone who responded to my question, was really helpful to resolve my issue.
Here is My answer that I built:
var str = ['8484683*ORG','7488575/2','647658-ORG'];
for(i=0;i<str.length;i++){
var regRep = /((\/\/[^\/]+)?\/.*)|(\-.*)|(\*.*)/;
var txt = str[i].replace(regRep,"");
console.log(txt);
}
I'm trying to extract out a group of words from a larger string/cookie that are separated by hyphens. I would like to replace the hyphens with a space and set to a variable. Javascript or jQuery.
As an example, the larger string has a name and value like this within it:
facility=34222%7CConner-Department-Store;
(notice the leading "C")
So first, I need to match()/find facility=34222%7CConner-Department-Store; with regex. Then break it down to "Conner Department Store"
var cookie = document.cookie;
var facilityValue = cookie.match( REGEX ); ??
var test = "store=874635%7Csomethingelse;facility=34222%7CConner-Department-Store;store=874635%7Csomethingelse;";
var test2 = test.replace(/^(.*)facility=([^;]+)(.*)$/, function(matchedString, match1, match2, match3){
return decodeURIComponent(match2);
});
console.log( test2 );
console.log( test2.split('|')[1].replace(/[-]/g, ' ') );
If I understood it correctly, you want to make a phrase by getting all the words between hyphens and disallowing two successive Uppercase letters in a word, so I'd prefer using Regex in that case.
This is a Regex solution, that works dynamically with any cookies in the same format and extract the wanted sentence from it:
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Demo:
var str = "facility=34222%7CConner-Department-Store;";
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Explanation:
Use this Regex (/([A-Z][a-z]+)-?/g to match the words between -.
Replace any - occurence in the matched words.
Then just join these matches array with white space.
Ok,
first, you should decode this string as follows:
var str = "facility=34222%7CConner-Department-Store;"
var decoded = decodeURIComponent(str);
// decoded = "facility=34222|Conner-Department-Store;"
Then you have multiple possibilities to split up this string.
The easiest way is to use substring()
var solution1 = decoded.substring(decoded.indexOf('|') + 1, decoded.length)
// solution1 = "Conner-Department-Store;"
solution1 = solution1.replace('-', ' ');
// solution1 = "Conner Department Store;"
As you can see, substring(arg1, arg2) returns the string, starting at index arg1 and ending at index arg2. See Full Documentation here
If you want to cut the last ; just set decoded.length - 1 as arg2 in the snippet above.
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1)
//returns "Conner-Department-Store"
or all above in just one line:
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1).replace('-', ' ')
If you want still to use a regular Expression to retrieve (perhaps more) data out of the string, you could use something similar to this snippet:
var solution2 = "";
var regEx= /([A-Za-z]*)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/;
if (regEx.test(decoded)) {
solution2 = decoded.match(regEx);
/* returns
[0:"facility=34222|Conner-Department-Store",
1:"facility",
2:"34222",
3:"Conner-Department-Store",
index:0,
input:"facility=34222|Conner-Department-Store;"
length:4] */
solution2 = solution2[3].replace('-', ' ');
// "Conner Department Store"
}
I have applied some rules for the regex to work, feel free to modify them according your needs.
facility can be any Word built with alphabetical characters lower and uppercase (no other chars) at any length
= needs to be the char =
34222 can be any number but no other characters
| needs to be the char |
Conner-Department-Store can be any characters except one of the following (reserved delimiters): :/?#[]#;,'
Hope this helps :)
edit: to find only the part
facility=34222%7CConner-Department-Store; just modify the regex to
match facility= instead of ([A-z]*)=:
/(facility)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/
You can use cookies.js, a mini framework from MDN (Mozilla Developer Network).
Simply include the cookies.js file in your application, and write:
docCookies.getItem("Connor Department Store");
I am trying to create a regex that can extract all words from a given string that only contain alphanumeric characters.
Yes
yes absolutely
#no
*NotThis
orThis--
Good *Bad*
1ThisIsOkay2 ButNotThis2)
Words that should have been extracted: Yes, yes, absolutely, Good, 1ThisIsOkay2
Here is the work I have done thus far:
/(?:^|\b)[a-zA-Z0-9]+(?=\b|$)/g
I had found this expression that works in Ruby ( with some tweaking ) but I have not been able to convert it to Javascript regex.
Use /(?:^|\s)\w+(?!\S)/g to match 1 or more word chars in between start of string/whitespace and another whitespace or end of string:
var s = "Yes\nyes absolutely\n#no\n*NotThis\norThis-- \nGood *Bad*\n1ThisIsOkay2 ButNotThis2)";
var re = /(?:^|\s)\w+(?!\S)/g;
var res = s.match(re).map(function(m) {
return m.trim();
});
console.log(res);
Or another variation:
var s = "Yes\nyes absolutely\n#no\n*NotThis\norThis-- \nGood *Bad*\n1ThisIsOkay2 ButNotThis2)";
var re = /(?:^|\s)(\w+)(?!\S)/g;
var res = [];
while ((m=re.exec(s)) !== null) {
res.push(m[1]);
}
console.log(res);
Pattern details:
(?:^|\s) - either start of string or whitespace (consumed, that is why trim() is necessary in Snippet 1)
\w+ - 1 or more word chars (in Snippet 2, captured into Group 1 used to populate the resulting array)
(?!\S) - negative lookahead failing the match if the word chars are not followed with non-whitespace.
You can do that (where s is your string) to match all the words:
var m = s.split(/\s+/).filter(function(i) { return !/\W/.test(i); });
If you want to proceed to a replacement, you can do that:
var res = s.split(/(\s+)/).map(function(i) { return i.replace(/^\w+$/, "#");}).join('');
I need a regex in Javascript that would allow me to match an order number in two different formats of order URL:
The URLs:
http://store.apple.com/vieworder/1003123464/test#test.com
http://store.apple.com/vieworder/W411234368/test#test.com/AOS-A=
M-104121
The first one will always be all numbers, and the second one will always start with a W, followed by just numbers.
I need to be able to use a single regex to return these matches:
1003123464
W411234368
This is what I've tried so far:
/(vieworder\/)(.*?)(?=\/)/g
RegExr link
That allows me to match:
vieworder/1003123464
vieworder/W411234368
but I'd like it to not include the first capture group.
I know I could then run the result through a string.replace('vieworder/'), but it'd be cool to be able to do this in just one command.
Use your expression without grouping vieworder
vieworder\/(.*?)(?=\/)
DEMO
var string = 'http://store.apple.com/vieworder/1003123464/test#test.com http://store.apple.com/vieworder/W411234368/test#test.com/AOS-A=M-104121';
var myRegEx = /vieworder\/(.*?)(?=\/)/g;
var index = 1;
var matches = [];
var match;
while (match = myRegEx.exec(string)) {
matches.push(match[index]);
}
console.log(matches);
Use replace instead of match since js won't support lookbehinds. You could use capturing groups and exec method to print the chars present inside a particular group.
> var s1 = 'http://store.apple.com/vieworder/1003123464/test#test.com'
undefined
> var s2 = 'http://store.apple.com/vieworder/W411234368/test#test.com/AOS-A='
undefined
> s1.replace(/^.*?vieworder\/|\/.*/g, '')
'1003123464'
> s2.replace(/^.*?vieworder\/|\/.*/g, '')
'W411234368'
OR
> s1.replace(/^.*?\bvieworder\/([^\/]*)\/.*/g, '$1')
'1003123464'
I'd suggest
W?\d+
That ought to translate to "one or zero W and one or more digits".
I'm attempting to get any words starting with #, such as in "#word", but only get the "word" value.
My sample text is:
#bob asodija qwwiq qwe #john #cat asdasd#qeqwe
My current regex is:
/\B#(\w+)/gi
This works perfectly, except that "#" is still being captured. The output of this match is:
"#bob"
"#john"
"#cat"
I've tried setting the # in a back reference, but its still including the # in the results.
/\B(?:#)(\w+)/gi
You want to use the match array returned from exec
var teststr = '#bob asodija qwwiq qwe #john #cat asdasd#qeqwe';
var exp = /\B#(\w+)/gi;
var match = exp.exec(teststr);
while(match != null){
alert(match[1]); // match 1 = 1st group captured
match = exp.exec(teststr);
}
Here's a neat trick using the String.replace method, which can take a function as the replacement.
var matches = [];
var str = "#bob asodija qwwiq qwe #john #cat asdasd#qeqwe";
str.replace( /\B#(\w+)/g, function( all, firstCaptureGroup ) {
matches.push( firstCaptureGroup );
});
console.log( matches ); //["bob", "john", "cat"]
Here is a better solution without additional calculations except of regular expression:
(?<=\B#)(\w+)