簡體   English   中英

如何從 arrays 列表中獲取特定數組

[英]How to get a specific array from a list of arrays

我想根據代碼得到一個特定的數組,數組是這樣的:

const arr = [
  [
    "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=11f18ac1-0476-4a1e-ada6-09e6566abc19",
    1595619171842,
    "0b7ad06f-7776-4bab-a8c6-53fd5fd5bd9b"
  ],
  [
    "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=b64c143d-e817-434f-bf6f-0bd0e8d9e7b5",
    1595619171844,
    "2f44a130-71d9-47ce-b5d5-04587c3c81fc"
  ],
  [
    "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=71dc5d26-75f4-4141-905e-074b0705eac4",
    1595619171845,
    "d7eb2a05-1f5a-48dd-b7ac-f3b071499d00"
  ],
  [
    "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=d3645614-0ea3-4d17-80ab-57c6c6525fab",
    1595619171846,
    "940fb9a7-6fdd-4f8b-a808-26a9c60114bf"
  ]
];

如何獲取代碼為“d7eb2a05-1f5a-48dd-b7ac-f3b071499d00”的數組? 我正在使用reduce來獲取更新的圖像,但現在我不知道!

Array#find返回數組的第一個元素,該數組對給定的 function 返回 true。

const specificArray = arr.find(subArray => {
     return subArray[2] === "d7eb2a05-1f5a-48dd-b7ac-f3b071499d00";
}

Array#find (帶有解構)最適合此目的。

const res = arr.find(([,,code])=>code==="d7eb2a05-1f5a-48dd-b7ac-f3b071499d00");

 const arr = [ [ "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=11f18ac1-0476-4a1e-ada6-09e6566abc19", 1595619171842, "0b7ad06f-7776-4bab-a8c6-53fd5fd5bd9b" ], [ "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=b64c143d-e817-434f-bf6f-0bd0e8d9e7b5", 1595619171844, "2f44a130-71d9-47ce-b5d5-04587c3c81fc" ], [ "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=71dc5d26-75f4-4141-905e-074b0705eac4", 1595619171845, "d7eb2a05-1f5a-48dd-b7ac-f3b071499d00" ], [ "https://firebasestorage.googleapis.com/v0/b/ziro-a…=media&token=d3645614-0ea3-4d17-80ab-57c6c6525fab", 1595619171846, "940fb9a7-6fdd-4f8b-a808-26a9c60114bf" ] ]; const res = arr.find(([,,code])=>code==="d7eb2a05-1f5a-48dd-b7ac-f3b071499d00"); console.log(res);

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM