英文:
Using struct method in Golang template
问题
在Go模板中，结构体方法通常与公共结构体属性以相同的方式调用，但在这种情况下却不起作用：http://play.golang.org/p/xV86xwJnjA
{{with index . 0}}{{.FirstName}} {{.LastName}} is {{.SquareAge}} years old.{{end}}
错误信息：
执行“person”时出错，错误位置为<.SquareAge>: SquareAge不是main.Person结构体类型的字段
在{{$person}}的例子中也存在同样的问题：
{{$person := index . 0}}{{$person.FirstName}} {{$person.LastName}} is{{$person.SquareAge}} years old.
相比之下，以下代码可以正常工作：
{{range .}}{{.FirstName}} {{.LastName}} is {{.SquareAge}} years old.{{end}}
如何在{{with}}和{{$person}}的例子中调用SquareAge()方法？
英文:
Struct methods in Go templates are usually called same way as public struct properties but in this case it just doesn't work: http://play.golang.org/p/xV86xwJnjA
{{with index . 0}}{{.FirstName}} {{.LastName}} is {{.SquareAge}} years old.{{end}}
Error:
executing "person" at <.SquareAge>: SquareAge is not a fieldof struct type main.Person
Same problem with:
{{$person := index . 0}}{{$person.FirstName}} {{$person.LastName}} is{{$person.SquareAge}} years old.
In constrast, this works:
{{range .}}{{.FirstName}} {{.LastName}} is {{.SquareAge}} years old.{{end}}
How to call SquareAge() method in {{with}} and {{$person}} examples?
答案1
得分: 13
如先前在https://stackoverflow.com/questions/10200178/call-a-method-from-a-go-template中所回答的，该方法的定义如下：
func (p *Person) SquareAge() int {return p.Age * p.Age}
该方法仅适用于类型*Person。
由于在SquareAge方法中不会改变Person对象，你可以将接收器从p *Person更改为p Person，这样它就可以与你之前的切片一起使用。
或者，如果你将以下代码：
var people = []Person{{"John", "Smith", 22},{"Alice", "Smith", 25},{"Bob", "Baker", 24},}
替换为：
var people = []*Person{{"John", "Smith", 22},{"Alice", "Smith", 25},{"Bob", "Baker", 24},}
它也会起作用。
可工作示例1：http://play.golang.org/p/NzWupgl8Km
可工作示例2：http://play.golang.org/p/lN5ySpbQw1
英文:
As previously answered in https://stackoverflow.com/questions/10200178/call-a-method-from-a-go-template, the method as defined by
func (p *Person) SquareAge() int {return p.Age * p.Age}
is only available on type *Person.
Since you don't mutate the Person object in the SquareAge method, you could just change the receiver from p *Person to p Person, and it would work with your previous slice.
Alternatively, if you replace
var people = []Person{{"John", "Smith", 22},{"Alice", "Smith", 25},{"Bob", "Baker", 24},}
with
var people = []*Person{{"John", "Smith", 22},{"Alice", "Smith", 25},{"Bob", "Baker", 24},}
It'll work as well.
Working example #1: http://play.golang.org/p/NzWupgl8Km
Working example #2: http://play.golang.org/p/lN5ySpbQw1
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