假設我有一個帶有這樣的嵌套子模板的模板。游樂場鏈接package mainimport ( "os" "text/template")type Person struct { FirstName string SecondName string}type Document struct { DocName string People []Person}const document = `Document name: {{.DocName}}{{range $person:=.People}}{{template "person" $person}}{{end}}{{- define "person"}}Person name is: {{.FirstName}} {{.SecondName}}{{end}}`func main() { d := Document{ DocName: "first try", People: []Person{ {"Brian", "Kernighan"}, {"Dennis", "Ritchie"}, }, } t := template.Must(template.New("document").Parse(document)) err := t.Execute(os.Stdout, d) if err != nil { panic(err) }}package mainimport ( "os" "text/template")type Person struct { FirstName string SecondName string}type Document struct { DocName string People []Person}const document = `Document name: {{.DocName}}{{range $person:=.People}}{{template "person" $person}}{{end}}{{- define "person"}}Person name is: {{.FirstName}} {{.SecondName}}{{end}}`func main() { d := Document{ DocName: "first try", People: []Person{ {"Brian", "Kernighan"}, {"Dennis", "Ritchie"}, }, } t := template.Must(template.New("document").Parse(document)) err := t.Execute(os.Stdout, d) if err != nil { panic(err) }}type Person struct { FirstName string SecondName string}type Document struct { DocName string People []Person SwitchNameOrder bool}const document = `Document name: {{.DocName}}{{range $person:=.People}}{{template "person" $person}}{{end}}{{- define "person"}}{{if $.SwitchNameOrder}} // <---- panic herePerson name is: {{.SecondName}} {{.FirstName}}{{else}}Person name is: {{.FirstName}} {{.SecondName}}{{end}}{{end}}`怎么做？是否可以？
3 回答
阿晨1998
TA貢獻2037條經驗 獲得超6個贊
您可以做的一件事是使用模板函數將傳遞到子模板的變量與父模板中的變量“合并”。
type Person struct {
FirstName string
SecondName string
}
type Document struct {
DocName string
People []Person
SwitchNameOrder bool
}
func personWithDocument(p Person, d Document) interface{} {
return struct {
Person
Document Document
}{p, d}
}
t := template.Must(template.New("document").Funcs(template.FuncMap{
"personWithDocument": personWithDocument,
}).Parse(document))
然后在模板中您將執行以下操作：
const document = `
Document name: {{.DocName}}
{{range $person:=.People}}
{{template "person" (personWithDocument $person $) }}
{{end}}
{{- define "person"}}
{{if .Document.SwitchNameOrder}}
Person name is: {{.SecondName}} {{.FirstName}}
{{else}}
Person name is: {{.FirstName}} {{.SecondName}}
{{end}}
{{end}}
`
https://play.golang.org/p/YorPsMdr9g_H
收到一只叮咚
TA貢獻1821條經驗 獲得超5個贊
對于上述復雜的解決方案，更好的解決方案是停止嘗試使用頂級配置選項，而是將其編寫為模板函數，并將配置變量放在函數閉包中
{{- define "person"}}
{{if SwitchNameOrder}}
Person name is: {{.SecondName}} {{.FirstName}}
{{else}}
Person name is: {{.FirstName}} {{.SecondName}}
{{end}}
{{end}}
和
t := template.Must(template.New("document").Funcs(template.FuncMap{
"SwitchNameOrder": func() bool {
return switchNames // variable sits in closure
},
}).Parse(document))
https://play.golang.org/p/O6QHtmxweOi
另一種選擇是將整個切換編寫為字符串函數，即：
{{- define "person"}}
Person name is: {{SwitchNames .FirstName .SecondName}}
{{end}}
并SwitchNames作為字符串函數
...Funcs(template.FuncMap{
"SwitchNames": func(first, second string) string {
if switchNames {
return second + " " + first
}
return first + " " + second
},
})...
可以更干凈或更干凈，具體取決于實際的復雜性
https://play.golang.org/p/UPB3NIpzw0N
蝴蝶刀刀
TA貢獻1801條經驗 獲得超8個贊
我最終做的是添加一個單獨的結構配置并將其復制到各處。那是，
type Config struct {
SwitchNameOrder bool
}
type Person struct {
FirstName string
SecondName string
Config Config
// this could also be a pointer,
// but I don't want to deal with nils, so let's copy
}
type Document struct {
DocName string
People []Person
Config Config
}
和
c := Config{SwitchNameOrder: true}
d.Config = c
for _, p := range d.People {
p.Config = c
}
然后在模板中使用它
{{- define "person"}}
{{if .Config.SwitchNameOrder}}
Person name is: {{.SecondName}} {{.FirstName}}
{{else}}
Person name is: {{.FirstName}} {{.SecondName}}
{{end}}
{{end}}
很丑，但是該怎么辦
https://play.golang.org/p/f95i4me8XLP
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