在Strings.join GoLang中使用Separator

问题描述

我正在建立一个URL前面的URL

I am building a URL with the front of the url

var SearchUrl = "https://www.example.org/3/search/movie?query="

然后形成保存要搜索关键字的数据。

Then form data which holds the keywords to search for.

var MovieSearch []string = r.Form["GetSearchKey"]  

以及保存api键的url的第三部分

And the third part of the url which holds the api key

var apiKey = "&api_key=######"

我是使用 ArrayToString（）来解析表单输入数据

I am using the ArrayToString() to parse the form input data

func ArrayToString(array []string) string{
    str := strings.Join(array, "+")
    return str 
}

然后构建这样的网址，

And then building the url like this,

var SearchUrl = "https://api.example.org/3/search/movie?query="
var MovieSearch []string = r.Form["GetSearchKey"]  
var apiKey = "&api_key=########"
UrlBuild := []string {SearchUrl, ArrayToString(MovieSearch), apiKey}
OUTPUT_STRING := ArrayToString(UrlBuild)

一个词的URL输出完美。网址如下所示：

The output of the URL with one word works perfectly. The URL looks like this,

https://api.example.org/3/search/movie?query=+bad+&api_key=#####

当我在输入字段中添加第二个字时， url看起来像这样，

When I add a second word to the input field the url looks like this,

https://api.example.org/3/search/movie?query=bad santa&api_key=e######

我需要关键字在它们之间没有空格，就像该网址无效。

I need the keywords to not have a space between them like in the url that does not work.

这是我当前基于Cedmundo的答案进行的尝试

This is my current attempt based on the answer from Cedmundo

func ArrayToQuery(values []string) string {
    return url.QueryEscape(strings.Join(values, " "))
}

func searchHandler(w http.ResponseWriter, r *http.Request) {
  display(w, "search", &Page{Title: "Search"})
   fmt.Println("method:", r.Method) 
        r.ParseForm()

var MovieSearch = r.Form["GetSearchKey"]  

var SearchKeys = ArrayToQuery(MovieSearch)

params := fmt.Sprintf("?query=%s&api_key=eagaggagagagagagag", url.QueryEscape(SearchKeys))
perform := url.URL{
    Scheme:     "https",  
    Host:       "api.example.org",
    Path:       "3/search/movie",
    RawQuery:   params,
}


fmt.Println(perform) // <- Calls .String()
}

错误

./main.go:63: url.QueryEscape undefined (type string has no field or method QueryEscape)
./main.go:75: url.QueryEscape undefined (type string has no field or method QueryEscape)
./main.go:76: url.URL undefined (type string has no field or method URL)
./main.go:94: undefined: OUTPUT_STRING
./main.go:96: undefined: OUTPUT_STRING

推荐答案

正常情况下，应该使用url包的值。

Normally, one should use url package's Values.

下面是一个例子，这是我认为你想要的， a href =http://play.golang.org/p/9gC6NNW--_ =nofollow>在玩
无论是简单的main还是http.HandlerFunc形式：

Here's an example, that does what I think you want, on playBoth a simple main, and in http.HandlerFunc form:

package main

import "fmt"
import "net/url"
import "net/http"

func main() {
    baseURL := "https://www.example.org/3/search/movie"
    v := url.Values{}
    v.Set("query", "this is a value")
    perform := baseURL + "?" + v.Encode()
    fmt.Println("Perform:", perform)
}

func formHandler(w http.ResponseWriter, r *http.Request) {
    baseURL := "https://www.example.org/3/search/movie"
    v := url.Values{}

    v.Set("query", r.Form.Get("GetSearchKey")) // take GetSearchKey from submitted form
    v.Set("api_ley", "YOURKEY") // whatever your api key is

    perform := baseURL + "?" + v.Encode() // put it all together
    fmt.Println("Perform:", perform) // do something with it
}

输出：
执行：https：//www.example.org/3/search/movie？query = this +是+ a +值

请注意，这些值是否会放入查询字符串中，以便您正确转义。

Notice how the values are put in to query string, properly escaped, for you.

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