I'm having a problem in searching in an array. I wanted to check if a certain string exists in one of the elements, example
Array:
["4_1", "4_2", "4_3"]
I will check if the string "4" exists in one of the variables.
Thank you!
3 Answers
You could use a loop for it.
var arr = ["4_1", "4_2", "4_3"];
search = RegExp(4),
result = arr.some(search.test.bind(search));
document.write(result);4 Comments
mplungjan
I do not fully grasp the need for the complexity of your code. Why be case insensitive on a number and why the bind?
Praveen Kumar Purushothaman
The
some is really good. Yes, why case insensitive on a number?
Nina Scholz
@PraveenKumar, it was for future reference ... ;)
Nina Scholz
@mplungjan, try it without bind, it does not work outside of a function around.
The Easiest way is to use Array.prototype.join && indexOf methods
["4_1", "4_2", "4_3"].join("").indexOf("4")
Update
According to @t.niese's comment this answer may cause wrong result, for example if you are looking for 14 it will return 2 - what is wrong because your array doesnt contain element which begins on 14. In this case better to use @Nina's answer or you can join it in another way
["4_1", "4_2", "4_3"].join(" ").indexOf("14") // -1
2 Comments
t.niese
The performance might not be the best for larger arrays, and will also have a problem, if the OP decides to use a string with more then on char as
needle , because it will then give a false positive for ["4_1", "4_2", "4_3"].join("").indexOf("14").t.niese
For all
.join it will be possible to find a failure case: ["4_1", "4_2", "4_3"].join(separator).indexOf("1"+separator+"4") will show a result. If the OP really only looks for numbers, then it will work. But because the question itself is only a generic [...]I wanted to check if a certain string exists in one of the elements[...] then under this conditions a .join will never be a good solution.You can actually make a loop and check the indexOf to be not -1:
["4_1", "4_2", "4_3"].forEach(function (element, index, array) {
if (element.indexOf(4) != -1)
alert("Found in the element #" + index + ", at position " + element.indexOf(4));
});
answered May 4, 2016 at 8:51
Praveen Kumar Purushothaman
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STRING.indexOf(needle)in a loop ?