Finding a specific string in array of strings

Question

In an array of strings, a loop returns false if it finds that one of the strings is not what we are looking for.

If it doesn't find an unmatching string, the array is correct and it should return true. It keeps returning false even when the array has no "mistakes"

I have tried using indexOf, for loop and while loops, all of them unsuccessful.

function brackets() {
    var testArr = ['()', '{}', '()']

    /* Method 1 --- returns false even when the parenthesis are ok, I guess
    it's because the indexOf only searches for the first element that matches 
    the criteria */

    if (testArr.indexOf("()") == -1 || testArr.indexOf("{}") == -1 ||
        testArr.indexOf("[]") == -1) {
        return false
    } else {
        return true
    }

    /* Method 2 --- for loop. Same story, returns false, even when all   
    testArr[i] === any of the cases and none of them is !==, it behaves as if it was 
    false. I'm not sure why */

    for (i = 0; i < testArr.length; i++) {
        if (testArr[i] !== "()" || testArr[i] !== "{}" || testArr[i] !== "[]") {
            return false
        }
    }
    return true
}

brackets()

testArr.indexOf("[]") clearly fails, and as it's a logical OR it returns false if any of them fail ? — adeneo
– adeneo, Commented May 24, 2015 at 14:24

Luca Colonnello · Accepted Answer · 2015-05-24 15:29:09Z

4

In the second method you can use an AND operator to solve this problem.

function brackets() {
var testArr = ['()', '{}', '()'];

/* Method 2 --- for loop. Same story, returns false, even when all   
testArr[i] === any of the cases and none of them is !==, it behaves as if it was 
false. I'm not sure why */

for (i = 0; i < testArr.length; i++) {
    if (testArr[i] !== "()" && testArr[i] !== "{}"  && testArr[i] !== "[]") {
        return false;
    }
}
return true;
}

brackets();

answered May 24, 2015 at 15:29

Luca Colonnello

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Comments

victor175 · Accepted Answer · 2015-05-24 14:24:56Z

0

Change your array to : var testArr = ['()', '{}', '[]']

Since you do

 if (testArr.indexOf("()") == -1 || testArr.indexOf("{}") == -1 ||
        testArr.indexOf("[]") == -1)

Then even if one of these parenthesis is absent from the array, the condition will return false.

answered May 24, 2015 at 14:24

victor175

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1 Comment

Bergi Over a year ago

@vsl0610: but the array in your example clearly has that mistake that it doesn't contain []? If you have problems with "an array that contains no mistakes and still returns false", then you should provide code that demonstrates that case.

Mistergreen · Accepted Answer · 2015-05-24 15:16:01Z

0

You've created an array with

var testArr = ['()', '{}', '()']

so if you need to test the string you need to test the 2-d array

if (testArr[0].indexOf("()") == -1

edited May 24, 2015 at 15:16

answered May 24, 2015 at 14:57

Mistergreen

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Comments

brroshan · Accepted Answer · 2015-05-24 17:27:45Z

0

This should do it:

var testArr = ['()', '{}', '()'];

if(testArr.some(function(e){ return /(){}/g.test(e) })){
   console.log("found");
} else {
   console.log("not found");   
}

Finding all instances of "()" and "{}"

https://jsfiddle.net/dhoh1932/1/

answered May 24, 2015 at 17:27

brroshan

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