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0

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Javascript: includes return false while matching partial array element

Very simple example here:

var u = []
u.push("https://cloudinary.com/products/media_optimizer/web-performance-guide#get-started")

u.includes("https://cloudinary.com/products/media_optimizer/web-performance-guide") //return false

What am I doing wrong here?

about 4 years ago · Juan Pablo Isaza
3 answers
Answer question

0

Iterate over the array, checking each element.

const search = (arr, fragment) => arr.some(v => v.includes(fragment));

const u = [
  "https://cloudinary.com/products/media_optimizer/web-performance-guide#get-started"
];

const found = search(u, "https://cloudinary.com/products/media_optimizer/web-performance-guide");

console.log(found);

Reference:

  • .some()
about 4 years ago · Juan Pablo Isaza Report

0

The value you are searching for must be an exact match, and you are not searching for the ending #get-started

about 4 years ago · Juan Pablo Isaza Report

0

You are searching the array for an exact match to your string. You can use .includes(...) on the string in the array which will return true.

u[0].includes("https://cloudinary.com/products/media_optimizer/web-performance-guide")
about 4 years ago · Juan Pablo Isaza Report
Answer question
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