js reduce()方法配合 RegExp正则模糊匹配

最新推荐文章于 2024-01-25 19:57:52 发布

原创最新推荐文章于 2024-01-25 19:57:52 发布 · 513 阅读

0 ·

CC 4.0 BY-SA版权

js基础理论同时被 2 个专栏收录

26 篇文章

订阅专栏

笔记

16 篇文章

订阅专栏

本文深入探讨了JavaScript数组中的reduce方法使用技巧，通过实例详细解释如何利用reduce方法实现数组对象去重，特别关注于合并相同examid的项并确保answer属性值不重复。文章提供了完整的代码示例和相关资源链接，帮助读者掌握reduce方法的高级应用。

这里提供一个小需求，需要将 examid 相同的项合并，并将此项中的 answer 属性值合起来且不重复
具体数据如下：

state = {
	arr: [
              { examid: '1052', stb: '1', optionid: '4199', answer: 'A' },
              { examid: '1265', stb: '1', optionid: '5051', answer: 'A' },
              { examid: '970', stb: '1', optionid: '3871', answer: 'A' },
              { examid: '875', stb: '1', optionid: '3491', answer: 'A' },
              { examid: '1174', stb: '1', optionid: '4687', answer: 'A' },
              { examid: '1017', stb: '1', optionid: '4059', answer: 'AB' },
              { examid: '1017', stb: '1', optionid: '4060', answer: 'B' },
              { examid: '1096', stb: '1', optionid: '4375', answer: 'AB' },
              { examid: '1096', stb: '1', optionid: '4376', answer: 'B' },
              { examid: '856', stb: '1', optionid: '3421', answer: 'AB' },
              { examid: '856', stb: '1', optionid: '3422', answer: 'B' },
              { examid: '1059', stb: '1', optionid: '4227', answer: 'AB' },
              { examid: '1059', stb: '1', optionid: '4228', answer: 'B' },
              { examid: '1039', stb: '1', optionid: '4147', answer: 'AB' },
              { examid: '1039', stb: '1', optionid: '4148', answer: 'c' }
          ]
}

public xxx = ()=>{
    console.log(this.state.arr,'arr');
    let newObj = {}
    let newArr = this.state.arr.reduce((item: any,currentValue) => {
    
      // 先判断是否重复
      if (newObj[currentValue.examid]) {

		// 重复，则判断 answer 属性中是否已有当前项的 answer 值，通过正则模糊匹配
        let answer = currentValue.answer
        let reg = new RegExp(answer)
        if (!item[newObj[currentValue.examid]-1].answer.match(reg)) {
          item[newObj[currentValue.examid]-1] = {...item[newObj[currentValue.examid]-1], answer: item[newObj[currentValue.examid]-1].answer + currentValue.answer}
        }
      } else {
		
		// 未重复，则将当前项加入到数组中
        newObj[currentValue.examid] = item.push(currentValue)
      }
      return item
    },[])
    console.log(newArr,'newArr');
  }