英文:
How to update field value in a Go template
问题
我有以下情况，我正在将包含映射的结构体传递给模板：
package mainimport ("log""os""text/template")var fns = template.FuncMap{"plus1": func(x int) int {return x + 1},}type codec struct {Names map[string]stringCount int}func main() {a := map[string]string{"one": "1","two": "2","three": "3"}t := template.Must(template.New("abc").Funcs(fns).Parse(`{{$l := len .Names}}{{range $k, $v := .Names}}{{if ne (plus1 $.Count) $l}}{{$k}} {{$v}} {{end}}{{end}}.`))err := t.Execute(os.Stdout, codec{a, 0})if err != nil {log.Println(err)}}
我想要增加codec的Count字段，以便我知道我已经看到了多少个映射项。
英文:
I have the following case, where I am passing a struct containing a map to a template:
package mainimport ("log""os""text/template")var fns = template.FuncMap{"plus1": func(x int) int {return x + 1},}type codec struct {Names map[string]stringCount int}func main() {a := map[string]string{"one": "1","two": "2","three": "3"}t := template.Must(template.New("abc").Funcs(fns).Parse(`{{$l := len .Names}}{{range $k, $v := .Names}}{{if ne (plus1 $.Count) $l}}{{$k}} {{$v}} {{end}}{{end}}.`))err := t.Execute(os.Stdout, codec{a, 0})if err != nil {log.Println(err)}}
I would like to increment the Count field of codec so that I can know how many items of the map I've seen.
答案1
得分: 2
一种解决方案是将plus1函数作为闭包，直接作用于codec的值：
// 首先创建一个codec实例
c := codec {a, 0}
// 现在将函数定义为闭包，并引用c
fns := template.FuncMap{
"plus1": func() int {
c.Count++
return c.Count
},
}
// 现在在模板中不需要传递任何参数给它
t := template.Must(template.New("abc").Funcs(fns).Parse({{$l := len .Names}}{{range $k, $v := .Names}}{{if ne (plus1) $l}}{{$k}} {{$v}} {{end}}{{end}}.))
输出结果为：
one 1 three 3
我猜这就是你的目标？并且该值在执行结束时保留在c中。
英文:
One solution is to make the plus1 function a closure that acts directly on the value of the codec:
// first create a codec instancec := codec {a, 0}// now define the function as a closure with a reference to cfns := template.FuncMap{"plus1": func() int {c.Count++return c.Count},}// now we don't need to pass anything to it in the templatet := template.Must(template.New("abc").Funcs(fns).Parse(`{{$l := len .Names}}{{range $k, $v := .Names}}{{if ne (plus1) $l}}{{$k}} {{$v}} {{end}}{{end}}.`))
The output was:
one 1 three 3
which I'm guessing is what you were aiming for? And the value is retained in c at the end of execution.
答案2
得分: 1
你可以在你的结构体上简单地定义一个方法：
type codec struct {Names map[string]stringCount int}func (c *codec) IncAndGet() int {c.Count++return c.Count}
从模板中调用它：
c := &codec{Count: 2}t := template.Must(template.New("").Parse(`{{.IncAndGet}} {{.IncAndGet}}`))t.Execute(os.Stdout, c)
输出结果（在Go Playground上尝试）：
3 4
请注意，为了使其工作，该方法需要一个指针接收器（func (c *codec) IncAndGet()），并且你必须将指针传递给Template.Execute()（在我们的示例中，c是一个指针：c := &codec{Count: 2}）。
如果你不想要任何结果，只想计数，可以将其定义为具有string返回类型并返回空字符串""：
func (c *codec) Inc() string {c.Count++return ""}
英文:
You can simply define a method on your struct:
type codec struct {Names map[string]stringCount int}func (c *codec) IncAndGet() int {c.Count++return c.Count}
Calling it from a template:
c := &codec{Count: 2}t := template.Must(template.New("").Parse(`{{.IncAndGet}} {{.IncAndGet}}`))t.Execute(os.Stdout, c)
Output (try it on the <kbd>Go Playground</kbd>):
3 4
Note that for this to work, the method needs a pointer receiver (func (c *codec) IncAndGet()) and you have to pass a pointer to Template.Execute() (c is a pointer in our example: c := &codec{Count: 2}).
If you don't want any result just counting, define it to have a string return type and return the empty string "":
func (c *codec) Inc() string {c.Count++return ""}
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