Use "M" shorthand for messages >= one million #4659
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Implements [Unread messages count] Abandon "k" (= thousand) when it reaches million+ thelounge#4615
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You didn't edit the test. |
Implements [Unread messages count] Abandon "k" (= thousand) when it reaches million+ thelounge#4615
brettgilio
changed the title
brettgilio
changed the title
Done. |
brunnre8
added
Meta: Internal
xPaw
reviewed
Feb 15, 2023
| @@ -6,10 +6,15 @@ describe("roundBadgeNumber helper", function () { | |||
| expect(roundBadgeNumber(123)).to.equal("123"); | |||
| }); | |||
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| it("should return numbers above 999 in thousands", function () { | |||
| it("should return numbers between 1000 and 999999 with a 'k' suffix", function () { | |||
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add a few more expects, e.g. 999999 is in fact 999k
xPaw
reviewed
Feb 15, 2023
| @@ -3,5 +3,9 @@ export default (count: number) => { | |||
| return count.toString(); | |||
| } | |||
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| return (count / 1000).toFixed(2).slice(0, -1) + "k"; | |||
| const suffixes = ["", "k", "M"]; | |||
| const magnitudeIndex = Math.min(Math.floor(Math.log10(count) / 3), suffixes.length - 1); | |||
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While I understand the use case here, I think it's simpler to just have two if cases to handle M/k.
if(count >= 1000000) M
if(count >= 1000) k
count
xPaw
reviewed
Feb 15, 2023
| expect(roundBadgeNumber(1000)).to.be.equal("1.0k"); | ||
| }); | ||
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| it("should return numbers above 999999 with a 'M' suffix", function () { | ||
| expect(roundBadgeNumber(1000000)).to.be.equal("1.0M"); | ||
| expect(roundBadgeNumber(1234567)).to.be.equal("1.2M"); |
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Please add a test for much higher magnitude number (with current code it would validate that it magnitudeIndex works` as expected.
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Implements [Unread messages count] Abandon "k" (= thousand) when it reaches million+ #4615