如何從JavaScript中的數組對象中刪除特定的鍵和值？

Question

在這里，我在這個arrayData中有arrayData數組對象。我有多個對象，我想從該數組對象中刪除索引並鍵入鍵值。如何從此arrayData中刪除？

arrayData : [
  0: {
       index: 0
       is_required: true
       name: "vmvdnksl"
       type: "LONG_TEXT"
     }
  1: {
       index: 1
       is_required: true
       name: "dsvnlk"
       type: "MULTIPLE_SELECTORS"
     }
   ]

刪除索引和類型后，我想要這種類型的結果

 arrayData : [
  0: {
       is_required: true
       name: "vmvdnksl"
     }
  1: {
       is_required: true
       name: "dsvnlk"
     }
   ]

Answer 1

您可以使用rest參數。 當您有很多要保留的鑰匙並只取其中幾個鑰匙時，它將非常方便。

 const arrayData= [{index: 0,is_required: true,name: "vmvdnksl",type: "LONG_TEXT"},{index: 1,is_required: true,name: "dsvnlk",type: "MULTIPLE_SELECTORS"}]; const result = arrayData.map(({type,index,...rest}) => ({...rest})); console.log(result); 

Answer 2

您可以使用delete關鍵字從對象中刪除屬性。

 var arrayData = [ 0: { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, 1: { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } ]; for (var i in arrayData) { for (var j in arrayData[i]) { if (j === 'index' || j === 'type') { delete arrayData[i][j]; } } } console.log(arrayData); 

Answer 3

您可以使用Array.map（）和destructuring此任務：

 const arrayData = [ { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } ]; let res = arrayData.map(({is_required, name}) => ({is_required, name})); console.log(res); 

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我更喜歡永遠不要更改原始數據，但是在您需要這樣做的情況下，您可以這樣做（或使用delete如其他人所示）：

 const arrayData = [ { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } ]; let res = arrayData.forEach( ({is_required, name}, idx, arr) => arr[idx] = ({is_required, name}) ); console.log(arrayData); 

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Answer 4

嘗試使用JavaScript的delete運算符 ，如以下代碼示例所示：

for(let o in arrayData){
    delete arrayData[o].index;
    delete arrayData[o].type
}

Answer 5

對於數組，您可以使用map()函數

  var arrayData = [ { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } ], mappedArrayData = arrayData.map(({is_required, name}) => { return {is_required, name}; }) console.log(mappedArrayData); 

對於對象，請使用delete運算符。

 var arrayData = { 0: { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, 1: { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } }; for (let key in arrayData) { delete arrayData[key].index; delete arrayData[key].type; } console.log(arrayData); 

Answer 6

您可以簡單地使用Array.map()僅顯示必需的屬性：

 const arrayData= [ { index: 0, is_required: true, name: "vmvdnksl", type: "LONG_TEXT" }, { index: 1, is_required: true, name: "dsvnlk", type: "MULTIPLE_SELECTORS" } ]; const result = arrayData.map(({is_required, name}) => ({is_required, name})); console.log(result);