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RODRIGO MARTINEZ
5,912 PointsphoneNumbers is array of 10 digit phone numbers, where the first three digits, in parentheses, are area codes.
Can't get it right. Please help. Excercise " phoneNumbers is array of 10 digit phone numbers, where the first three digits, in parentheses, are area codes. Using reduce, return the total phone numbers with a 503 area code. Store the total in the variable numberOf503."
const phoneNumbers = ["(503) 123-4567", "(646) 123-4567", "(503) 987-6543", "(503) 234-5678", "(212) 123-4567", "(416) 123-4567"];
let numberOf503;
// numberOf503 should be: 3
// Write your code below
numberOf503 =
phoneNumbers.reduce((count, number) => {
if (number.substring(0,4) === '(503)') {
return count +1;
}
return count;
});
console.log(numberOf503);
const phoneNumbers = ["(503) 123-4567", "(646) 123-4567", "(503) 987-6543", "(503) 234-5678", "(212) 123-4567", "(416) 123-4567"];
let numberOf503;
// numberOf503 should be: 3
// Write your code below
numberOf503 =
phoneNumbers.reduce((count, number) => {
if (number.substring(0,4) === '(503)') {
return count +1;
}
return count;
});
console.log(numberOf503);
3 Answers
Steven Parker
231,272 PointsYou're pretty close, but:
- you need an explicit starting value of 0 to begin the "count" with (otherwise it starts with a string)
- the "substring" needs to have 5 characters instead of 4 to match what it is being compared with
RODRIGO MARTINEZ
5,912 PointsSteve, Thanks a lot for the help. I thought that the substring would count the characters from 0 like with the index numbers of an array. Should i always start the count from "0"? In case i 'd set an explicit starting value of "1", would that means that it starts counting from the secnd element in the array? Thanks again
Steven Parker
231,272 PointsYou don't always need an explicit starting value, but in this case you need to because you're counting the items instead of combining their values. And starting a count a 1 would make it too big at the end.
RODRIGO MARTINEZ
5,912 Pointsthanks again man. You're awesome. Very generous. Thanks for beeing always willing to help.