duanliang9288 2015-06-05 09:04 采纳率: 100%
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Golang:将URL作为GET参数传递

I want to get an URL as a get aparameter

ex: example.com?domain=site.come?a=val&b=val

the problem when i use

query := r.URL.Query()
domain := query.Get("domain") 

to get the domain name it give just domain=site.come?a=val

I think because when the r.URL.Query() meet & it consider it as a new parameter

does anyone know how can I solve this problem

Thank you in advance.

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1条回答 默认 最新

  • dongtanliefang8765 2015-06-05 09:13
    关注

    You need to URL-Encode your query string, like this:

    package main
    
    import (
        "fmt"
        "net/url"
    )
    
    func main() {
        query := make(url.Values)
        query.Add("domain", "example.com?foo=bar")
    
        fmt.Println(query.Encode())
    }
    

    Which outputs domain=example.com%3Ffoo%3Dbar.

    You can set that string as a RawQuery of an url.URL value and if you then access the query like you did, it will have the correct value.

    If the URL is correctly encoded then you should be able to run the following code with your URL value and get the correct result:

    package main
    
    import (
        "fmt"
        "net/url"
    )
    
    func main() {
        query := make(url.Values)
        query.Add("domain", "example.com?foo=bar&abc=123&jkl=qwe")
    
        url := &url.URL{RawQuery: query.Encode(), Host: "domain.com", Scheme: "http"}
        fmt.Println(url.String())
    
        abc := url.Query().Get("domain")
        fmt.Println(abc)
    }
    

    This prints:

    http://domain.com?domain=example.com%3Ffoo%3Dbar%26abc%3D123%26jkl%3Dqwe
    

    (the complete URI with the encoded parameter called "domain")

    example.com?foo=bar&abc=123&jkl=qwe
    

    (the decoded value of said parameter)

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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