從 object 的拼接數組中更新數據

Question

所以，我已經拼接了數據，我想將它的 object 從disabled = true更新為disabled = false .. 我一直在尋找另一個答案，但找不到..

任何建議表示贊賞。 謝謝

這是我的下拉菜單

const newDrop = [
  { key: "1", label: "Monthly GMV", sublabel: "Max.1" },
  { key: "2", label: "AOV", sublabel: "Max.1" },
  { key: "3", label: "Monthly Order", sublabel: "Max.1" },
  { key: "4", label: "Last Purchase", sublabel: "Max.1" },
  { key: "5", label: "Has Purchase" },
  { key: "6", label: "Has Purchase Spesific Product" },
  { key: "7", label: "Located In", sublabel: "Max.1" },
];

選擇 function：選擇下拉列表時，將添加新屬性disabled:true

newInput({ key }) {
    const {
      changeFormAttribute,
      form: { selectedDrop, newDrop, data },
    } = this.props;

    itemIndex = newDrop.map((itm) => {
      let x = itm.key === key;

      return x && itm.key !== "6" && itm.key !== "5" ? { ...itm, disabled: true } : { ...itm };
    });
    
    changeFormAttribute({
      newDrop: itemIndex,
    });
  }

刪除function disabled:false這是onClick function，點擊時應該拼接數據，並檢查是否禁用拼接數據disabled === true 拼接的工作正常，但沒有將屬性更改為disabled = false

 deleteInput(index) {
    const {
      changeFormAttribute,
      form: { selectedDrop },
    } = this.props;
  
    let selected = selectedDrop || [];
    selected.splice(index, 1).map((e) => {return e.disabled === true ? false:true })

    changeFormAttribute({ selectedDrop: selected });
  }

Answer 1

您的splice function 沒有返回元素。 您可能需要將其更改為

 selected.splice(index, 1).map((e) => {
     e.disabled === true ? false:true
     return e; 
 })