Regexp to explode url-CodePudding

I have a string url like "home/products/product_name_1/details/some_options" And i want to parse it into array with Regexp to ["home", "products","product","details","some"]

So the rule is "split by words if backslash, but if the word have underscores - take only that part that comes before first underscore"

JavaScript equivalent for this regex is

str.split("/").map(item => item.indexOf("_") > -1 ? item.split("_")[0] : item)

Please help!

CodePudding user response：

you can use this pattern

(?<!\w)[^/_]

results

['home', 'products', 'product', 'details', 'some']

python code

import re
str="home/products/product_name_1/details/some_options"

re.findall('(?<!\w)[^/_] ',str)

['home', 'products', 'product', 'details', 'some']

CodePudding user response：

Try this:

input = ["home/products/product_name_1/details/some_options",
    "company/products/cars_all/details/black_color",
    "public/places/1_cities/disctricts/1234_something"]

let pattern = /([a-zA-Z\d]*)(?:\/|_.*?(?:\/|$))/gmi

input.forEach(el => {
    let matches = el.matchAll(pattern)
    for (const match of matches) {
        console.log(match[1]);
    }
})

Remove \d from the regex pattern if you dont want digits in the url. I have used matchAll here, matchAll returns a iterator, use that to get each match object, inside which the first element is the full match, and the second elemnt(index: 1) is the required group.

/([a-zA-Z\d]*)(?:\/|_.*?(?:\/|$))/gmi

/
([a-zA-Z\d]*)         capture group to match letters and digits
(?:\/|_.*?(?:\/|$))   non capture group to match '/' or '_' and everything till another '/' or end of the line is found 
/gmi

You can test this regex here: https://regex101.com/r/B5Bo74/1

CodePudding user response：

You can use:

\b[^\W_]

\b A word boundary to prevent a partial match
[^\W_] Match 1 word characters except for _

See a regex demo.

const s = "home/products/product_name_1/details/some_options";
const regex = /\b[^\W_] /g;
console.log(s.match(regex));

If there has to be a leading / or the start of the string before the match, you can use an alternation (?:^|\/) and use a capture group for the values that you want to keep:

const s = "home/products/product_name_1/details/some_options";
const regex = /(?:^|\/)([^\W_] )/g;
console.log(Array.from(s.matchAll(regex), m => m[1]));

CodePudding user response：

Given input:

string "home/products/product_name_1/details/some_options"

Expected output:

array ["home", "products", "product", "details", "some"]
Note: ignore/exclude name, 1, options (because word occurs after 1st underscore).

Task:

split URI by slash into a set of path-segments (words)
(if the path-segment or word contains underscores) remove the part after first underscore

Regex to match

With a regex \/|_\w you could match the URL-path separator (slash) and excluded word-part (every word after an underscore).

Then use this regex

either as separator to split the string into its parts（excluding the regex matches): e.g. in JS split(/\/|_\w /)
or as search-pattern in replace to prepare a string that can be easily split: e.g. in JS replaceAll(/\/|_\w /g, ',') to obtain a CSV row which can be easily split by comma `split(',')

Beware: The regular-expression itself (flavor) and functions to apply it depend on your environment/regex-engine and script-/programming-language.

Regex applied in Javascript

split by regex

For example in Javascript use url.split(/\/|_\w*/) where:

/pattern/: everything inside the slashes is the regex-pattern
\/: a c slash (URL-path-separator)
|: the alternate junction, interpreted as boolean OR
_\w*: zero or more (*) word-characters (w, i.e. letter from alphabet, numeric digit or underscore) following an underscore

replace into CSV, then split and exclude (map,replace,filter)

The CSV containing path-segments can be split by comma and resulting parts (path-segments) can be filtered or replaced to exclude unwanted sub-parts.

using:

replaceAll to transform to CSV or remove empty strings. Note: global flag required when calling replaceAll with regex
map to remove unwanted parts after underscore
filter(s => s) to filter out empty strings

const url = "home/products/product_name_1/details/some_options";

// step by step
let pathSegments = url.split('/');
console.log('pathSegments:', pathSegments);
let firstWordsInSegments = pathSegments.map(s => s.replaceAll(/_\w*/g,''));
console.log(firstWordsInSegments);

// replace to obtain CSV and then split
let csv = "home/products/product_name_1/details/some_options/_/home".replaceAll(/\/|_\w /g, ',');
console.log('csv:', csv);
let parts = csv.split(',');
console.log('parts:', parts); // contains empty parts
let nonEmptyParts = parts.filter(s => s);
console.log('nonEmptyParts:', nonEmptyParts); // filtered out empty parts

Bonus Tip

Try your regex online (e.g. regex101 or regexplanet). See the demo on regex101.

CodePudding user response：

You could split the url with this regex

(_\w*) |(\/)

This matches the /, _name_1 and _options.

BUT depending what you are trying to to, or which language do you use, there are way better options to do this.

CodePudding user response：

You can try a pattern like \/([^\/_] ){1,} (assuming that the path starts with '/' and the components are separated by '/'); depending on language you might get an array or iterator that will give the components.

CodePudding user response：

Try ^[[:alpha:]] |(?<=\/)[[:alpha:]] or ^[a-zA-Z] |(?<=\/)[a-zA-Z] if [[:alpha:]] is not supported , it matches one or more characters on the beginning or after slash until first non char.