Question

我不确定标题中的解释是否很好，但这是一个示例：

我有两个数组：我想把arr2中的每个字符串都放在arr1中，看是否有匹配项。

let arr1 = ["Apple","Ape","Orange","Apricot","Pineapple"]
let arr2 = ["Ap","Or"]

我想找回以下数组（以任何顺序）：

["Apple","Ape","Orange","Apricot"]

我找到了.includes（），但它仅搜索完全匹配项。

Answer 1

不需要库，只需filter遍历单词中是否包含第二个数组的子字符串中的some，就可以let arr1 = ["Apple","Ape","Orange","Apricot","Pineapple"] let arr2 = ["Ap","Or"] console.log( arr1.filter((word) => arr2.some(substr => word.includes(substr))) );：

 WITH cte1
     AS (SELECT id,sdate,edate from t
     union all
     select c.id,DATEADD(DAY, 1, c.sdate),c.edate from cte1 c where DATEADD(DAY, 1, c.sdate)<=c.edate
        ) 
SELECT sdate,count(id) as total FROM   cte1 
group by sdate
OPTION (MAXRECURSION 0)

Answer 2

传统方法是使用两次迭代来检查indexOf()

let arr1 = ["Apple","Ape","Orange","Apricot","Pineapple"]
let arr2 = ["Ap","Or"]

for(var i in arr1){
  for(var j in arr2){
     if(arr1[i].indexOf(arr2[j]) == 0){
       console.log(arr1[i]);
       break;
     }
  }
}

确定一个字符串是否存在于数组中，从另一个数组中的另一组字符串开始

2 个答案: