query-string pick 方法返回过滤器中的第一个匹配项和损坏的 URL

Question

I am trying to clean a URL and only keep some parameters that I need in order to parse them so I was using pick method providing it the url, and the filter which is a regex test method here I am testing to check if the key in the query parameter matches the regular expression

const groupRegex = new RegExp('^(GRP_)[a-zA-Z0-9/-]','g');
export const parseGroups= (url:string)=>{
    let pickedURL = qs.pick(url,(key,value)=>groupRegex.test(key));
    console.log(pickedURL);
}
var url=`http://localhost:3000/tat?GRP_Bob[]=SW&GRP_sa[]=QW&GRP_sa[]=AA&projects[]=MP,PM&releases[]=2021.4,2022.1`
parseGroups(url)

for example http://localhost:3000/tat?GRP_Bob[]=SW&GRP_sa[]=QW&GRP_sa[]=AA&projects[]=MP,PM&releases[]=2021.4,2022.1 it should return http://localhost:3000/tat?GRP_Bob=SW&GRP_sa=QW&GRP_sa=AA yet it only tests for the first request parameter only and logs http://localhost:3000/tat?GRP_Bob%5B%5D=SW

I am trying to clean the url from any other parameters that doesn't match my regular expression so I can parse the URL and extract the object so it can be like this for example

{
       GRP_Bob:["SW"],
       GRP_sa:["QW","AA"]
 }

Instead of having other parameters parsed also which are not necessary. I know I can just parse the url normally, and then loop on the returned query object, and remove any key that doesn't match the regex, but is there anything wrong I am doing in the above snippet?

UPDATE: I changed the filter function to be (key,value)=>key.startsWith('GRP_'))

export const parseGroups= (url:string)=>{
    let pickedURL = qs.pick(url,(key,value)=>key.startsWith('GRP_'));
    console.log(pickedURL);
    let parsedURL = qs.parseUrl(pickedURL)
    console.log(parsedURL.query)
}
var url=`http://localhost:3000/tat?GRP_Bob[]=SW&GRP_sa[]=QW&GRP_sa[]=AA&projects[]=MP,PM&releases[]=2021.4,2022.1`
parseGroups(url)

and the pickedURL logged this http://localhost:3000/tat?GRP_Bob%5B%5D=SW&GRP_sa%5B%5D=QW&GRP_sa%5B%5D=AA which is likely to be correct. it came out like that

GRP_Bob[]: "SW"
GRP_sa[]: (2) ['QW', 'AA']

So I am confused actually what's going on with the regular expression approach, and why the keys in the second approach have [] in it?

Answer 1

Ah yeh. This one is a rare gotcha and totally unexpected every time I see it. RegExp actually has state. See Why does JavaScript's RegExp maintain state between calls? and Why does Javascript's regex.exec() not always return the same value? .

In your case, you don't need the g flag, so removing that should also fix your problem since that makes the regex stateless.