如何取消选择嵌套的 JSON 对象？

Question

目前我的代码是console.log("body:", body.readResults)。但我不想要 t 和 r 的值。如何取消选择它们并获得其余部分？

"readResults": [
   {
       "id": "Channel1.Device1.Coil",
       "s": true,
       "r": "",
       "v": true,
       "t": 1553586722927
    }
]

Answer 1

你可以轻松做到：

function cleanMyResult(result) {
    const cleanResult = Object.assign({}, result); // this will preserve your original object, but it costs some performance
    delete cleanResult.t;
    delete cleanResult.r;
    return cleanResult;
}

const cleanedResults = body.readResults.map(cleanMyResult);
console.log(cleanedResults);

请同时查看 SO 上很多已经很好的答案：

• Filter object properties by key in ES6
• Simplest way to copy JS object and filter out certain properties
• How do I remove a property from a JavaScript object?