I am trying to filter out words that include zz or ZZ but can't quite get it to work. I am trying to use includes() within the filter() method. I'm not sure if the || operand is the correct choice here.
function removeZZ (str){
let splitStr = str.split(" ")
let noBuzz = splitStr.filter(word => !word.includes("zz") ||!word.includes("ZZ") )
return noBuzz.join(" ")
}
You can use word.toLowerCase().includes
const str = "wed zzAA ZZSS 34f34f"
function removeZZ(str) {
return str
.split(" ")
.filter(word => !word.toLowerCase().includes("zz"))
.join(" ");
}
console.log(removeZZ(str))With your approach you need to use && instead of ||
const str = "wed zzAA ZZSS 34f34f"
function removeZZ(str) {
let splitStr = str.split(" ")
let noBuzz = splitStr.filter(word => !word.includes("zz") && !word.includes("ZZ"))
return noBuzz.join(" ")
}
console.log(removeZZ(str))Or simply you can make each word lowercase and check if it includes 'zz'. Try the following example.
function removeZZ (str){
let splitStr = str.split(" ")
let noBuzz = splitStr.filter(word => !word.toLowercase().include('zz'))
return noBuzz.join(" ")
}
Your logic is a bit inverted by using the || instead of && operator.
Instead you could write this:
let noBuzz = splitStr.filter(word => !word.includes("zz") && !word.includes("ZZ") )
Or use an Array of blocked words instead:
let noBuzz = splitStr.filter(word => !["zz", "ZZ"].includes(word))
Or, as they are the same and only the case is different, try a Regex:
let noBuzz = splitStr.filter(word => !word.match(/zz/i))