我可以将数组的所有元素作为单独的字符串吗

Question

正如标题所说，我想将数组的每个元素都作为自己的字符串，以便检查某个其他数组是否包括数组1中的任何字符串。

我觉得我一直在努力，我不确定这是否可能，但我可能很愚蠢。如有任何帮助，我们将不胜感激！

const v = items.map((value) => {
if(value.tags.includes('weapon')) return value.id
})
const results = v.filter(e => {
return e !== undefined
})
const c = Object.keys((challengerItems)).map((key) => {return key})
const o = Object.keys((opposerItems)).map((key) => {return key})
const va = results.filter(element => {
return element
});
if(challengerItems === undefined) return message.reply(``❓` <@${message.author.id}> does not have any weapons! Duel canceled!`)
if(opposerItems === undefined) return message.reply(``❓` <@${opposer.id}> does not have any weapons! Duel canceled!`)
if(!c.includes(CHECK IF USER HAS ANY WEAPONS)) return message.reply(``❓` <@${message.author.id}> does not have any weapons! Duel canceled!`) 
if(!o.includes(CHECK IF OTHER USER HAS ANY WEAPONS)) return message.reply(``❓` <@${opposer.id}> does not have any weapons! Duel canceled!`) 

Answer 1

使用map将每个元素转换为字符串(我使用toString()，如果适合，可以使用JSON.stringify(

使用find和indexOf定位阵列2中包含在阵列1 中的项目

var arr1 = [2, 17, 'B'];
var arr2 = ["18", "17", "dog"];
// convert arr1 elements to string
arr1 = arr1.map(item => item.toString())
console.log(arr2.find(item => arr1.indexOf(item) > -1))