英文:
How to generate interface implementations in VS Code for Go?
问题
在VSCode中，如何为接口生成实现代码？
比如，我有以下接口：
type ServerInterface interface {
// 为设备设置值
SetSomethingForDeviceById(ctx echo.Context, id int64) error
}
如何生成实现该接口的方法？
英文:
In VSCode, how do I generate the implementation for an interface?
Say, I have this interface:
type ServerInterface interface {
// Set value for a device
SetSomethingForDeviceById(ctx echo.Context, id int64) error
}
How do I generate methods that implement it?
答案1
得分: 5
VScode支持使用Go扩展生成接口。
以下是操作步骤：
首先，你需要定义一个结构体：
type ApiServer struct {}
然后，使用Ctrl-Shift-P，找到命令：“Go generate interface stubs”
接下来，输入类似以下格式的内容：接收者名称、类型、接口名称：
s ReceiverType package.InterfaceName
按下回车键，缺失的方法将会生成：
package api
import "github.com/labstack/echo/v4"
// 为设备设置值
func (s ApiServer) SetSomethingForDeviceById(ctx echo.Context, id int64) error {
panic("not implemented")
}
@clément-jean 补充道：
这个命令依赖于 https://github.com/josharian/impl：在生成代码之前，你需要先安装它。
英文:
VScode supports interface generation with the Go extension.
Here's how you do it:
First, you start with defining your struct:
type ApiServer struct {}
Now, use Ctrl-Shift-P, and find this command: "Go generate interface stubs"
Now type something like this: receiver name, type, interface name:
> s ReceiverType package.InterfaceName
Hit Enter. Missing methods are generated:
package api
import "github.com/labstack/echo/v4"
// Set value for a device
func (s ApiServer) SetSomethingForDeviceById(ctx echo.Context, id int64) error {
panic("not implemented")
}
@clément-jean added that:
> this command depends on https://github.com/josharian/impl: you need to install it before being able to generate the code.
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