Question

这是我的目标：

var obj = {
  "flag": false,
  "text": "good text",
  "someArray": [
    {
      "questionId": 11,
      "text": "qwe",
      "TypeId": "B",
      "source": "oooooo"
    },
    {
      "questionId": 12,
      "text": "zxc",
      "TypeId": "A",
      "source": "pppppp"
    },
    {
      "questionId": 13,
      "text": "asd",
      "TypeId": "D",
      "source": "cccccc"
    }
  ]
}

我尝试做的是从someArray属性TypeId和source中删除对象。我怎么能用lodash做到这一点？我正在尝试使用_.pick，但我不知道如何在someArray内使用它。

Answer 1

您希望使用与SORT FIELDS=COPY OUTFIL REMOVECC, HEADER1=(1:C'COUNTRY_CODE',13:C',',14:C'CHECK_SEQ_NUMBER',30:C',', 31:C'FORM_INDICATOR') OUTREC BUILD=(1:1,2,3:C',',4:4,5,9:C',',10:10,1,80:X)相反的_.pick，并通过使用_.omit

映射来返回没有给定属性的对象

＆＃13;

_.map

＆＃13;

var obj = {
  "flag": false,
  "text": "good text",
  "someArray": [{
    "questionId": 11,
    "text": "qwe",
    "TypeId": "B",
    "source": "oooooo"
  }, {
    "questionId": 12,
    "text": "zxc",
    "TypeId": "A",
    "source": "pppppp"
  }, {
    "questionId": 13,
    "text": "asd",
    "TypeId": "D",
    "source": "cccccc"
  }]
}

obj.someArray = _.map(obj.someArray, o => _.omit(o, ['TypeId', 'source']));

console.log(obj)

＆＃13;

如何使用lodash从对象中删除属性？

1 个答案: