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    <p>The issue you're seeing is because of the nested for loop. You are essentially doing a cross-compare: for every item in i1, you are comparing it to every item in i2 (remember that j starts again at 0 every time i advances... the two loops don't run in parallel).</p> <p>Since I understand from the comments below that you want to be able to compare one array to the other, even if the items in each are in a different order, I've edited my original suggestion. Note that the snippet below does not normalize differences in case between the two arrays... don't know if that's a concern. Also note that it only compares i1 against i2... not both i1 to i2 and i2 to i1, which would make the task a little more challenging.</p> <pre><code>function testItems(i1, i2) { var newArray = []; var newArray2 = []; for (var i = 0; i &lt; i1.length; i++) { var found = false; for (var j = 0; j &lt; i2.length; j++) { if (i1[i] == i2[j]) found = true; } if (found) { newArray.push(i1[i]) } else { newArray2.push(i1[i]) } } } </code></pre> <p>As an alternative, you could consider using a hash table to index i1/i2, but since the example of strings in your comment include spaces and I don't know if you're using any javascript helper libraries, it's probably best to stick with the nested for loops. The snippet also makes no attempt to weed out duplicates.</p> <p>Another optimization you might consider is that your newArray and newArray2 arrays contain their own length property, so you don't need to pass the count to your HTML writer. When the writer receives the arrays, it can ask each one for the .length property to know how large each one is.</p>
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    1. POWhy is my nested for loop not working as I expected?
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    1. USJarret Hardie
    1. USJarret Hardie
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    1. POWhy is my nested for loop not working as I expected?
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    1. COHi Jarret, Thanks for your wonderful answer. There's a problem, I want to compare it word by word regardless of their order. Example: 1) Upper Deck New Jersey Nets Vince Carter 2) NBA All-Star Vinyl New Jersey Nets The code above will return mismatch.. Sorry for not mentioning this.
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      1. USRaymond G.
    2. COAh, gotcha! No worries... in that case, your nest for-loop approach is very close. I've edited the code above to reflect a way to tweak what you had that might meet your needs.
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      1. USJarret Hardie
    3. COYou could add a `break` if a match has been found.
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