筛选不包含其他数组中的字符串的字符串？

Question

I want to filter an array of urls by excluding ones that contain any substrings that are in a blacklist array.

const urls = [
'http://example.com/people/chuck', 
'http://example.com/goats/sam', 
'http://example.com/goats/billy', 
'http://example.com/goats/linda', 
'http://example.com/cows/mary', 
'http://example.com/cows/betty', 
'http://example.com/people/betty']

const blacklist = ['cows', 'goats']

let cleanUrls = [];

I can do this with for-loops but I want to find a clean/concise way using filter and/or reduce.

If I didn't need to loop over x number of blacklist items:

cleanUrls = urls.filter( url => !url.includes(blacklist[0]) )                  
                .filter( url => !url.includes(blacklist[1]) ) 

I also don't want to just iterate through the blacklist with a forEach or map because I want to immediately stop if a particular url matches any blacklist entry.

Plain JS please. Thank you. :)

Answer 1

You can use filter() like this:

const cleanUrls = urls.filter(u => blacklist.every(s => !u.includes(s)));

Or

const cleanUrls = urls.filter(u => !blacklist.some(s => u.includes(s)));

Description:

• .includes() will check the existence of a sub-string withing another string.
• .every() will check whether none of the strings of blacklist array exists in a particular URL.
• .some() will check whether any of the string of blacklist array exists in a particular URL.
• .filter() will select only those URL's which will pass the test.

Demo:

 const urls = [ 'http://example.com/people/chuck', 'http://example.com/goats/sam', 'http://example.com/goats/billy', 'http://example.com/goats/linda', 'http://example.com/cows/mary', 'http://example.com/cows/betty', 'http://example.com/people/betty' ]; const blacklist = ['cows', 'goats']; const cleanUrls = urls.filter(u => blacklist.every(s => !u.includes(s))); console.log(cleanUrls); 

Docs:

• Array.prototype.filter()
• Array.prototype.includes()
• Array.prototype.some()

Answer 2

You can do this with filter() , some() and includes() methods.

 const urls = ['http://example.com/people/chuck', 'http://example.com/goats/sam', 'http://example.com/goats/billy', 'http://example.com/goats/linda', 'http://example.com/cows/mary', 'http://example.com/cows/betty', 'http://example.com/people/betty']; const blacklist = ['cows', 'goats'] const result = urls.filter(url => !blacklist.some(e => url.includes(e))) console.log(result) 

Answer 3

It might be easuer to turn the blacklist i to a Regex:

 const block = new RegExp(blacklist.join("|"));

Them its as simple as:

 cleanUrls = urls.filter(url => !url.match(block));

Answer 4

You can use forEach & indexOf to check if the element from second array is present in the url

 const urls = [ 'http://example.com/people/chuck', 'http://example.com/goats/sam', 'http://example.com/goats/billy', 'http://example.com/goats/linda', 'http://example.com/cows/mary', 'http://example.com/cows/betty', 'http://example.com/people/betty' ] const blacklist = ['cows', 'goats']; var newArray = []; blacklist.forEach(function(item) { if (newArray.length === 0) { newArray = urls.filter(function(items) { return (items.indexOf(item) === -1) }) } else { newArray = newArray.filter(function(items) { return (items.indexOf(item) === -1) }) } }); console.log(newArray) 

Answer 5

I also don't want to just iterate through the blacklist with a forEach or map because I want to immediately stop if a particular url matches any blacklist entry.

The function stop searching if an element from the blacklist appears.

Recursive way to solve this problem:

 const urls = [ 'http://example.com/people/chuck', 'http://example.com/goats/sam', 'http://example.com/goats/billy', 'http://example.com/goats/linda', 'http://example.com/cows/mary', 'http://example.com/cows/betty', 'http://example.com/people/betty'] const blacklist = ['cows', 'goats'] function filter([url, ...urls] =[], blackList=[], filteredUrls =[] ){ return !url || blacklist.some(el => url.includes(el)) ? filteredUrls : filter(urls, blackList, [...filteredUrls, url]) } console.log(filter(urls, blacklist)); 

Answer 6

Others have answered the question. Providing two different scenarios.

 const urls = [ 'http://example.com/people/chuck', 'http://example.com/goats/billy', 'http://example.com/aagoatsss/billyyy', 'http://example.com/people/betty' ] const blacklist = ['cows', 'goats'] //1st approach. It will also block 'http://example.com/aagoatsss/billyyy' let cleanUrls1 = urls.filter( url => blacklist.find( blockWord => url.indexOf(blockWord) != -1 //Url contains Block Word ) === undefined //url donot contain any block word. ) console.log(cleanUrls1) //2nd approach. It will pass 'http://example.com/aagoatsss/billyyy' let cleanUrls2 = urls.filter( url => blacklist.find( blockWord => url.split('/').indexOf(blockWord) != -1 //Url contains Block Word ) === undefined //url donot contain any block word. ) console.log(cleanUrls2)