可能看起來很傻，但我沒有找到任何不會訴諸其他功能的東西，所以......
ar1 = [[1, 10, "Item1", 50], [1, 10, "Item1", 50],[1, 10, "Item3", 50]];
let ar2 = ar1.filter(e => e[2] === "Item1")[0];//To get only of the filtered records
它回傳：
[1, 10, "Item1", 50]
但要讓它回傳：
[[1, 10, "Item1", 50]]
謝謝！
uj5u.com熱心網友回復：
也許是這樣：
function lfunko() {
const a = [[1, 10, "Item1", 50], [1, 10, "Item1", 50],[1, 10, "Item3", 50]];
Logger.log(a.filter(r => r[2] == "Item1").filter((r,i) => i == 1));
}
Execution log
12:42:24 PM Notice Execution started
12:42:23 PM Info [[1.0, 10.0, Item1, 50.0]]
12:42:25 PM Notice Execution completed
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