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原始对象↓

let obj1 = [{name:'蔬菜-1',id:'123'},
            {name:'蔬菜-2',id:'122'},
            {name:'新鲜蔬菜',id:'112'},
            {name:'劣质苹果',id:'114'},
            {name:'苹果-新增',id:'115'},
            {name:'苹果-优质',id:'116'},
            {name:'香蕉',id:'106'},
            {name:'梨子',id:'156'}]

期待的效果↓

let result = [{name:'蔬菜',id:'123'},
              {name:'苹果',id:'114'},
              {name:'香蕉',id:'106'},
              {name:'梨子',id:'156'}]

也就是如果name里面有相同的字符,改造后的name就是那个相同的字符,其余的去掉,然后id的话就用在前的第一个就行,没有相同的话就一起遍历到result里面,这该怎么写呢

不知道我的问题描述是否清晰,还请各位大神指教,小弟感激不尽!!


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1 Answer

这个简单,遍历一遍用map存起来,最后转数组就行,给你写个伪代码

const result = new Map()
obj1.forEach(item => {
  const name = item.name.split('-')[0]
  if (!result.has(name)) {
    result.set(name, { name, id: item.id })
  }
})
return [result.values()]

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
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