我可以从嵌套模板访问顶级模板变量吗？
假设我有一个带有这样的嵌套子模板的模板。游乐场链接
package main
import (
"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 main
import (
"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 here
Person name is: {{.SecondName}} {{.FirstName}}
{{else}}
Person name is: {{.FirstName}} {{.SecondName}}
{{end}}
{{end}}
`
怎么做？是否可以？
交互式爱情3回答
-
阿晨1998
您可以做的一件事是使用模板函数将传递到子模板的变量与父模板中的变量“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))然后在模板中您将执行以下操作：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
对于上述复杂的解决方案，更好的解决方案是停止尝试使用顶级配置选项，而是将其编写为模板函数，并将配置变量放在函数闭包中{{- 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另一种选择是将整个切换编写为字符串函数，即：{{- define "person"}}Person name is: {{SwitchNames .FirstName .SecondName}}{{end}}并SwitchNames作为字符串函数...Funcs(template.FuncMap{ "SwitchNames": func(first, second string) string { if switchNames { return second + " " + first } return first + " " + second },})...可以更干净或更干净，具体取决于实际的复杂性https://play.golang.org/p/UPB3NIpzw0N -
蝴蝶刀刀
我最终做的是添加一个单独的结构配置并将其复制到各处。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 = cfor _, 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，但是该怎么办https://play.golang.org/p/f95i4me8XLP
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