simple regex /:[a-z]+/ not working as expected in javascript - javascript

Below is a very simple regex code, which works correctly in php and ruby, but not in JS. Plead help me get it working:
var r = /:[a-z]+/
var s = '/a/:b/c/:d'
var m = r.exec(s)
// now m is [":b"]
// it should be [":b", ":d"]
// because that's what i get in ruby and php

Using RegExp.exec() with g (global) modifier is meant to be used inside a loop for getting all matches.
var str = '/a/:b/c/:d'
var re = /:[a-z]+/g
var matches;
while (matches = re.exec(str)) {
// In array form, match is now your next match..
}
You can also use the String.match() method here.
var s = '/a/:b/c/:d',
m = s.match(/:[a-z]+/g);
console.log(m); //=> [ ':b', ':d' ]

var r = /:[a-z]+/g; // i put the g tag here because it needs to match all occurrences
var s = '/a/:b/c/:d';
var m = s.match(r);
console.log(m); // [':b',':d']
I used match because it returns all the matches in an array where as with exec you would have to loop through like the other examples.

Related

reg expression on OS 10 vs OS 11 [duplicate]

I am looking for an alternative for this:
(?<=\.\d\d)\d
(Match third digit after a period.)
I'm aware I can solve it by using other methods, but I have to use a regular expression and more importantly I have to use replace on the string, without adding a callback.
Turn the lookbehind in a consuming pattern and use a capturing group:
And use it as shown below:
var s = "some string.005";
var rx = /\.\d\d(\d)/;
var m = s.match(/\.\d\d(\d)/);
if (m) {
console.log(m[1]);
}
Or, to get all matches:
const s = "some string.005 some string.006";
const rx = /\.\d\d(\d)/g;
let result = [], m;
while (m = rx.exec(s)) {
result.push(m[1]);
}
console.log( result );
An example with matchAll:
const result = Array.from(s.matchAll(rx), x=>x[1]);
EDIT:
To remove the 3 from the str.123 using your current specifications, use the same capturing approach: capture what you need and restore the captured text in the result using the $n backreference(s) in the replacement pattern, and just match what you need to remove.
var s = "str.123";
var rx = /(\.\d\d)\d/;
var res = s.replace(rx, "$1");
console.log(res);

How to String include after character in nodejs, JavaScript

I want to do this in node.js
example.js
var str = "a#universe.dev";
var n = str.includes("b#universe.dev");
console.log(n);
but with restriction, so it can search for that string only after the character in this example # so if the new search string would be c#universe.dev it would still find it as the same string and outputs true because it's same "domain" and what's before the character in this example everything before # would be ignored.
Hope someone can help, please
Look into String.prototype.endsWith: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/endsWith
First, you need to get the end of the first string.
var ending = "#" + str.split("#").reverse()[0];
I split your string by the # character, so that something like "abc#def#ghi" becomes the array ["abc", "def", "ghi"]. I get the last match by reversing the array and grabbing the first element, but there are multiple ways of doing this. I add the separator character back to the beginning.
Then, check whether your new string ends the same:
var n = str.endsWith(ending);
console.log(n);
var str = "a#universe.dev";
var str2 = 'c#universe.dev';
str = str.split('#');
str2 = str2.split('#');
console.log(str[1] ===str2[1]);
With split you can split string based on the # character. and then check for the element on position 1, which will always be the string after #.
Declare the function
function stringIncludeAfterCharacter(s1, s2, c) {
return s1.substr(s1.indexOf(c)) === s2.substr(s2.indexOf(c));
}
then use it
console.log(stringIncludeAfterCharacter('a#universe.dev', 'b#universe.dev', '#' ));
var str = "a#universe.dev";
var n = str.includes(str.split('#')[1]);
console.log(n);
Another way !
var str = "a#universe.dev";
var n = str.indexOf(("b#universe.dev").split('#')[1]) > -1;
console.log(n);

How to extract string in regex

I have string in this format:
var a="input_[2][invoiceNO]";
I want to extract "invoiceNo" string. I've tried:
var a="input_[2][invoiceNO]";
var patt = new RegExp('\[(.*?)\]');
var res = patt.exec(a);
However, I get the following output:
Array [ "[2]", "2" ]
I want to extract only invoiceNo from the string.
Note: Input start can be any string and in place of number 2 it can be any number.
I would check if the [...] before the necessary [InvoiceNo] contains digits and is preceded with _ with this regex:
/_\[\d+\]\s*\[([^\]]+)\]/g
Explanation:
_ - Match underscore
\[\d+\] - Match [1234]-like substring
\s* - Optional spaces
\[([^\]]+)\] - The [some_invoice_123]-like substring
You can even use this regex to find invoice numbers inside larger texts.
The value is in capture group 1 (see m[1] below).
Sample code:
var re = /_\[\d+\]\s*\[([^\]]+)\]/g;
var str = 'input_[2][invoiceNO]';
while ((m = re.exec(str)) !== null) {
alert(m[1]);
}
You can use this regex:
/\[(\w{2,})\]/
and grab captured group #1 from resulting array of String.match function.
var str = 'input_[2][invoiceNO]'
var m = str.match(/\[(\w{2,})\]/);
//=> ["[invoiceNO]", "invoiceNO"]
PS: You can also use negative lookahead to grab same string:
var m = str.match(/\[(\w+)\](?!\[)/);
var a="input_[2][invoiceNO]";
var patt = new RegExp('\[(.*?)\]$');
var res = patt.exec(a);
Try this:
var a="input_[2][invoiceNO]";
var patt = new RegExp(/\]\[(.*)\]/);
var res = patt.exec(a)[1];
console.log(res);
Output:
invoiceNO
You could use something like so: \[([^[]+)\]$. This will extract the content within the last set of brackets. Example available here.
Use the greediness of .*
var a="input_[2][invoiceNO]";
var patt = new RegExp('.*\[(.*?)\]');
var res = patt.exec(a);

Match all entries { * }

I need to match all entries like { * } for string (using Javascript). How can I do it?
Here is an input: "#SD#{date};{time};{lat1};{lat2};{lon1};{lon2};{speed};{course};{height};{sats}\r\n";
thanks in advance
Using regular expression maybe:
var r = /{.*?}/g;
var s = "#SD#{date};{time};{lat1};{lat2};{lon1};{lon2};{speed};{course};{height};{sats}\r\n";
var matches = s.match(r);
Or if you want to strip those {} in results, you can run an .exec() loop to get a captured data only:
var r = /{(.*?)}/g;
var s = "#SD#{date};{time};{lat1};{lat2};{lon1};{lon2};{speed};{course};{height};{sats}\r\n";
var matches = [];
var match = null;
while(match = r.exec(s)) {
matches.push(match[1]);
}
You can try with the following regular expression:
var str = "#SD#{date};{time};{lat1};{lat2};{lon1};{lon2};{speed};{course};{height};{sats}\r\n".
var m = str.match(/{.*?}/g);
You'll find all the matches inside m.
Further references: http://www.w3schools.com/jsref/jsref_match.asp
For just the text within the brackets, you should be able to use this pattern:
var pattern = new RegExp("([^{|^}]+)(?=})", "g");
var testString = "#SD#{date};{time};{lat1};{lat2};{lon1};{lon2};{speed};{course};{height};{sats}\r\n";
var bracketTextArray = testString.match(pattern);
The bracketTextArray variable will be an array that contains the following values:
"date", "time", "lat1", "lat2", "lon1", "lon2", "speed", "course", "height", "sats"
The regex matches every occurrence of one or more of any character except { and } (that's this part: ([^{|^}]+)), that are immediately followed by a } character (that's this part: (?=})). The "g" in the RegExp definition makes the pattern "greedy", so that it will find all occurrences.

split string to two substrings from nth occerence of a charactor jquery

i have a string like this .
var url="http://localhost/elephanti2/chaink/stores/stores_ajax_page/5/b.BusinessName/asc/1/11"
i want to get substrings
http://localhost/elephanti2/chaink/stores/stores_ajax_page
and
5/b.BusinessName/asc/1/11
i want to split string from the 7 th slash and make the two sub-strings
how to do this ??,
i looked for split()
but in this case if i use it i have to con-cat the sub-strings and make what i want . is there a easy way ??
try this one:
var url="http://localhost/elephanti2/chaink/stores/stores_ajax_page/5/b.BusinessName/asc/1/11";
var parts = url.split('/');
var p1 = parts.slice(0,6).join('/');
var p2 = parts.slice(7).join('/');
alert(p1);
alert(p2);
p1 should get the first part and p2 is the second part
You can try this regex. Generally if your url pattern always follow this structure, it will work.
var pattern = /(\w+:\/\/(\w+\/){5})/i;
var url = "http://localhost/elephanti2/chaink/stores/stores_ajax_page/5/b.BusinessName/asc/1/11";
var result = url.split(pattern);
alert(result[1]);
alert(result[3]);
Try this :
var str = 'http://localhost/elephanti2/chaink/stores/stores_ajax_page/5/b.BusinessName/asc/1/11',
delimiter = '/',
start = 7,
tokens = str.split(delimiter).slice(start),
result = tokens.join(delimiter);
var match = str.match(/([^\/]*\/){5}/)[0];
Find this fiddle

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