I have a string in the following format.
"ad60 ad60-12 ad60 12-ad60"
now I want to find the matches only where "ad60" is written.
I started off with
/\bad60\b/i
but regex finds that '-' is not part of the character string. which returns 4 matches.
I tried many things but they all either return 4 matches or return nothing.
Any kind of help would be appreciated.
You can use:
var s = "ad60 ad60-12 ad60 12-ad60";
var r = s.replace(/(^|\s)ad60(?=\s|$)/g, "$1##");
//=> ## ad60-12 ## 12-ad60
this should find any word starting from letter
/[a-z][a-z\d]+/i
ADDED
if you wish to find any words which are included ad60 string try this
/[a-z\d\-\_]*ad60[a-z\d\-\_]*/i
but words separated by space will not get in result
"ad60 ad60-12 ad60 12-ad60".match(/(^ad60\s+)|(\s+ad60$)|(\s+ad60\s+)|(^ad60$)/ig);
will result in ["ad60 ", " ad60 "]
then you can trim the white spaces in matched elements.
Related
I have a string of text with HTML line breaks. Some of the <br> immediately follow a number between two delimiters «...» and some do not.
Here's the string:
var str = ("«1»<br>«2»some text<br>«3»<br>«4»more text<br>«5»<br>«6»even more text<br>");
I’m looking for a conditional regex that’ll remove the number and delimiters (ex. «1») as well as the line break itself without removing all of the line breaks in the string.
So for instance, at the beginning of my example string, when the script encounters »<br> it’ll remove everything between and including the first « to the left, to »<br> (ex. «1»<br>). However it would not remove «2»some text<br>.
I’ve had some help removing the entire number/delimiters (ex. «1») using the following:
var regex = new RegExp(UsedKeys.join('|'), 'g');
var nextStr = str.replace(/«[^»]*»/g, " ");
I sure hope that makes sense.
Just to be super clear, when the string is rendered in a browser, I’d like to go from this…
«1»
«2»some text
«3»
«4»more text
«5»
«6»even more text
To this…
«2»some text
«4»more text
«6»even more text
Many thanks!
Maybe I'm missing a subtlety here, if so I apologize. But it seems that you can just replace with the regex: /«\d+»<br>/g. This will replace all occurrences of a number between « & » followed by <br>
var str = "«1»<br>«2»some text<br>«3»<br>«4»more text<br>«5»<br>«6»even more text<br>"
var newStr = str.replace(/«\d+»<br>/g, '')
console.log(newStr)
To match letters and digits you can use \w instead of \d
var str = "«a»<br>«b»some text<br>«hel»<br>«4»more text<br>«5»<br>«6»even more text<br>"
var newStr = str.replace(/«\w+?»<br>/g, '')
console.log(newStr)
This snippet assumes that the input within the brackets will always be a number but I think it solves the problem you're trying to solve.
const str = "«1»<br>«2»some text<br>«3»<br>«4»more text<br>«5»<br>«6»even more text<br>";
console.log(str.replace(/(«(\d+)»<br>)/g, ""));
/(«(\d+)»<br>)/g
«(\d+)» Will match any brackets containing 1 or more digits in a row
If you would prefer to match alphanumeric you could use «(\w+)» or for any characters including symbols you could use «([^»]+)»
<br> Will match a line break
//g Matches globally so that it can find every instance of the substring
Basically we are only removing the bracketed numbers if they are immediately followed by a line break.
I'm trying to make a JS regex that returns example in all of these input strings, which may have question marks in them.
example?after?after
example?after
example?
example
example would be a dynamic string so I can't literally match the string 'example'.
'example?after?after'.match(/(.*?)\?/)[1]
'example?after'.match(/(.*?)\?/)[1]
'example?'.match(/(.*?)\?/)[1]
'example'.match(/(.*?)\?/)[1]
The first 3 above return example as expected but the last one gives an error. How can I modify the regex to return the expected result?
You may use
/^[^?]+/
See the regex demo
The ^ matches the start of string and [^?]+ matches one or more symbols other than ?.
var regex = /^[^?]+/;
var strs =[ 'example?after?after', 'example?after', 'example?', 'example'];
for (var s of strs)
{
console.log(s + " => " + s.match(regex));
}
I am trying to match everything after (but not including!) the last occurrence of a string in JavaScript.
The search, for example, is:
[quote="user1"]this is the first quote[/quote]\n[quote="user2"]this is the 2nd quote and some url https://www.google.com/[/quote]\nThis is all the text I\'m wirting about myself.\n\nLook at me ma. Javascript.
Edit: I'm looking to match everything after the last quote block. So I was trying to match everything after the last occurrence of "quote]" ? Idk if this is the best solution but its what i've been trying.
I'll be honest, i suck at this Regex stuff.. here is what i've been trying with the results..
regex = /(quote\].+)(.*)/ig; // Returns null
regex = /.+((quote\]).+)$/ig // Returns null
regex = /( .* (quote\]) .*)$/ig // Returns null
I have made a JSfiddle for anyone to have a play with here:
https://jsfiddle.net/au4bpk0e/
One option would be to match everything up until the last [/quote], and then get anything following it. (example)
/.*\[\/quote\](.*)$/i
This works since .* is inherently greedy, and it will match every up until the last \[\/quote\].
Based on the string you provided, this would be the first capturing group match:
\nThis is all the text I\'m wirting about myself.\n\nLook at me ma. Javascript.
But since your string contains new lines, and . doesn't match newlines, you could use [\s\S] in place of . in order to match anything.
Updated Example
/[\s\S]*\[\/quote\]([\s\S]*)$/i
You could also avoid regex and use the .lastIndexOf() method along with .slice():
Updated Example
var match = '[\/quote]';
var textAfterLastQuote = str.slice(str.lastIndexOf(match) + match.length);
document.getElementById('res').innerHTML = "Results: " + textAfterLastQuote;
Alternatively, you could also use .split() and then get the last value in the array:
Updated Example
var textAfterLastQuote = str.split('[\/quote]').pop();
document.getElementById('res').innerHTML = "Results: " + textAfterLastQuote;
I have this string
"/mp3/mysong.mp3"
I need to do make this string look like this with javascript.
"/mp3/myusername/mysong.mp3"
My guess would be to find second occurrence of "/", then append "myusername/" there or prepend "/myusername" but I'm not sure how to do this in javascript.
Just capture the characters upto the second / symbol and store it into a group. Then replace the matched characters with the characters inside group 1 plus the string /myusername
Regex:
^(\/[^\/]*)
Replacement string:
$1/myusername
DEMO
> var r = "/mp3/mysong.mp3"
undefined
> r.replace(/^(\/[^\/]*)/, "$1/myusername")
'/mp3/myusername/mysong.mp3'
OR
Use a lookahead.
> r.replace(/(?=\/[^/]*$)/, "/myusername")
'/mp3/myusername/mysong.mp3'
This (?=\/[^/]*$) matches a boundary which was just before to the last / symbol. Replacing the matched boundary with /myusername will give you the desired result.
This works -
> "/mp3/mysong.mp3".replace(/(.*?\/)(\w+\.\w+)/, "$1myusername\/$2")
"/mp3/myusername/mysong.mp3"
Demo and explanation of the regex here
use this :
var str = "/mp3/mysong.mp3";
var res = str.replace(/(.*?\/){2}/g, "$1myusername/");
console.log(res);
this will insert the text myusername after the 2nd / .
I need to split an email address and take out the first character and the first character after the '#'
I can do this as follows:
'bar#foo'.split('#').map(function(a){ return a.charAt(0); }).join('')
--> bf
Now I was wondering if it can be done using a regex match, something like this
'bar#foo'.match(/^(\w).*?#(\w)/).join('')
--> bar#fbf
Not really what I want, but I'm sure I miss something here! Any suggestions ?
Why use a regex for this? just use indexOf to get the char at any given position:
var addr = 'foo#bar';
console.log(addr[0], addr[addr.indexOf('#')+1])
To ensure your code works on all browsers, you might want to use charAt instead of []:
console.log(addr.charAt(0), addr.charAt(addr.indexOf('#')+1));
Either way, It'll work just fine, and This is undeniably the fastest approach
If you are going to persist, and choose a regex, then you should realize that the match method returns an array containing 3 strings, in your case:
/^(\w).*?#(\w)/
["the whole match",//start of string + first char + .*?# + first string after #
"groupw 1 \w",//first char
"group 2 \w"//first char after #
]
So addr.match(/^(\w).*?#(\w)/).slice(1).join('') is probably what you want.
If I understand correctly, you are quite close. Just don't join everything returned by match because the first element is the entire matched string.
'bar#foo'.match(/^(\w).*?#(\w)/).splice(1).join('')
--> bf
Using regex:
matched="",
'abc#xyz'.replace(/(?:^|#)(\w)/g, function($0, $1) { matched += $1; return $0; });
console.log(matched);
// ax
The regex match function returns an array of all matches, where the first one is the 'full text' of the match, followed by every sub-group. In your case, it returns this:
bar#f
b
f
To get rid of the first item (the full match), use slice:
'bar#foo'.match(/^(\w).*?#(\w)/).slice(1).join('\r')
Use String.prototype.replace with regular expression:
'bar#foo'.replace(/^(\w).*#(\w).*$/, '$1$2'); // "bf"
Or using RegEx
^([a-zA-Z0-9])[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+#([a-zA-Z0-9-])[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$
Fiddle