下面是我的代码:// swagger:route GET /user_list get_user//// Get a user profile// Parameters:// user_id: userParam// Consumes:// - application/json// Produces:// - application/json//// Responses:// 200: UserResponse在我的请求模型中,我添加了如下参数:// swagger:parameters userParamtype UserRequest struct { // aaaa // in: query UserId string `json:"user_id"`}生成不可靠的json时出现错误。panic: assignment to entry in nil mapgoroutine 1 [running]:github.com/go-swagger/go-swagger/scan.(*setOpParams).Parse(0xc428986a20, 0xc426439f30, 0x1, 0x1, 0x0, 0x0) /home/sotsys-056/go_10/src/github.com/go-swagger/go-swagger/scan/route_params.go:137 +0x635github.com/go-swagger/go-swagger/scan.(*tagParser).Parse(0xc426e873d0, 0xc426439f30, 0x1, 0x1, 0x0, 0xc426439f30) /home/sotsys-056/go_10/src/github.com/go-swagger/go-swagger/scan/scanner.go:526 +0x52当我删除参数部分时,它工作正常,我不会生成swagger json,但是带有参数我无法生成它。我也尝试过这些方法。// Parameters:// - user_id: [in:query required:true type:string] Description goes herparameters: - name: user_id in: query description: this argument is a string schema: type: string
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慕码人2483693
TA贡献1860条经验 获得超9个赞
您的示例不完整。但是,使用您已有的代码,您的代码应该看起来像最简单的形式:
// swagger:parameters GetUser
type GetUser struct {
// UserId that identifies a user.
//
// in: query
UserId string `json:"user_id"`
}
// Returns a user.
// swagger:response UserResponse
type UserResponse struct {
// in: body
Body struct {}
}
// swagger:route GET /user_list GetUser
//
// Get a user profile.
//
// Responses:
// 200: UserResponse
router.Get("/user_list", handler.GetUser)
注意我们有三件事:
请求对象
响应对象
同时使用请求和响应对象的路由
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