import subpackages with golang, by importing parent dir?
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0
down vote
favorite
I have this main.go file:
package main
import "one/entities/bar"
func main(){
}
I have this dir structure:
When I build my project with:
go install main
I get this compilation error:
src/main/main.go:3:8: no Go files in
/home/oleg/codes/oresoftware/oredoc/test/builds/go/src/one/entities/bar
that error makes sense. Is there some way to import all the subpackages from within the one/entities/bar
directory?
Something like this:
package main
import bar "one/entities/bar/*"
func main(){
}
(using some sort of * syntax and importing all subpackages in the bar namespace).
Ultimately I am trying do something like this:
package main
import (
"log"
"one/entities/bar"
)
func main(){
v := bar.Get.Basic.Req.Headers{}
log.Fatal(v)
}
go
add a comment |
up vote
0
down vote
favorite
I have this main.go file:
package main
import "one/entities/bar"
func main(){
}
I have this dir structure:
When I build my project with:
go install main
I get this compilation error:
src/main/main.go:3:8: no Go files in
/home/oleg/codes/oresoftware/oredoc/test/builds/go/src/one/entities/bar
that error makes sense. Is there some way to import all the subpackages from within the one/entities/bar
directory?
Something like this:
package main
import bar "one/entities/bar/*"
func main(){
}
(using some sort of * syntax and importing all subpackages in the bar namespace).
Ultimately I am trying do something like this:
package main
import (
"log"
"one/entities/bar"
)
func main(){
v := bar.Get.Basic.Req.Headers{}
log.Fatal(v)
}
go
Import statements imports apackage
. As long as the code inside the directory is of the same package, it is imported. You then can access it bypackagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?
– Chen A.
Nov 17 at 22:16
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have this main.go file:
package main
import "one/entities/bar"
func main(){
}
I have this dir structure:
When I build my project with:
go install main
I get this compilation error:
src/main/main.go:3:8: no Go files in
/home/oleg/codes/oresoftware/oredoc/test/builds/go/src/one/entities/bar
that error makes sense. Is there some way to import all the subpackages from within the one/entities/bar
directory?
Something like this:
package main
import bar "one/entities/bar/*"
func main(){
}
(using some sort of * syntax and importing all subpackages in the bar namespace).
Ultimately I am trying do something like this:
package main
import (
"log"
"one/entities/bar"
)
func main(){
v := bar.Get.Basic.Req.Headers{}
log.Fatal(v)
}
go
I have this main.go file:
package main
import "one/entities/bar"
func main(){
}
I have this dir structure:
When I build my project with:
go install main
I get this compilation error:
src/main/main.go:3:8: no Go files in
/home/oleg/codes/oresoftware/oredoc/test/builds/go/src/one/entities/bar
that error makes sense. Is there some way to import all the subpackages from within the one/entities/bar
directory?
Something like this:
package main
import bar "one/entities/bar/*"
func main(){
}
(using some sort of * syntax and importing all subpackages in the bar namespace).
Ultimately I am trying do something like this:
package main
import (
"log"
"one/entities/bar"
)
func main(){
v := bar.Get.Basic.Req.Headers{}
log.Fatal(v)
}
go
go
edited Nov 17 at 22:12
asked Nov 17 at 22:01
Alexander Mills
17k30146287
17k30146287
Import statements imports apackage
. As long as the code inside the directory is of the same package, it is imported. You then can access it bypackagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?
– Chen A.
Nov 17 at 22:16
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17
add a comment |
Import statements imports apackage
. As long as the code inside the directory is of the same package, it is imported. You then can access it bypackagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?
– Chen A.
Nov 17 at 22:16
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17
Import statements imports a
package
. As long as the code inside the directory is of the same package, it is imported. You then can access it by packagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?– Chen A.
Nov 17 at 22:16
Import statements imports a
package
. As long as the code inside the directory is of the same package, it is imported. You then can access it by packagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?– Chen A.
Nov 17 at 22:16
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can't do such thing, according to the specs of import
:
The import names an identifier (PackageName) to be used for access and an
ImportPath that specifies the package to be imported.
ImportDecl = "import" ( ImportSpec | "(" { ImportSpec ";" } ")" ) .
ImportSpec = [ "." | PackageName ] ImportPath .
ImportPath = string_lit .
The import syntax requires a package name or path of a package. So no wildcard import.
This statement will clarify the why;
The PackageName is used in qualified identifiers to access exported
identifiers of the package within the importing source file.
You must specify in which package the element you are referring to exists. e.g, bar.Get.Basic.Req.Headers
. This can't be determined without an explicit import.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can't do such thing, according to the specs of import
:
The import names an identifier (PackageName) to be used for access and an
ImportPath that specifies the package to be imported.
ImportDecl = "import" ( ImportSpec | "(" { ImportSpec ";" } ")" ) .
ImportSpec = [ "." | PackageName ] ImportPath .
ImportPath = string_lit .
The import syntax requires a package name or path of a package. So no wildcard import.
This statement will clarify the why;
The PackageName is used in qualified identifiers to access exported
identifiers of the package within the importing source file.
You must specify in which package the element you are referring to exists. e.g, bar.Get.Basic.Req.Headers
. This can't be determined without an explicit import.
add a comment |
up vote
1
down vote
accepted
You can't do such thing, according to the specs of import
:
The import names an identifier (PackageName) to be used for access and an
ImportPath that specifies the package to be imported.
ImportDecl = "import" ( ImportSpec | "(" { ImportSpec ";" } ")" ) .
ImportSpec = [ "." | PackageName ] ImportPath .
ImportPath = string_lit .
The import syntax requires a package name or path of a package. So no wildcard import.
This statement will clarify the why;
The PackageName is used in qualified identifiers to access exported
identifiers of the package within the importing source file.
You must specify in which package the element you are referring to exists. e.g, bar.Get.Basic.Req.Headers
. This can't be determined without an explicit import.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can't do such thing, according to the specs of import
:
The import names an identifier (PackageName) to be used for access and an
ImportPath that specifies the package to be imported.
ImportDecl = "import" ( ImportSpec | "(" { ImportSpec ";" } ")" ) .
ImportSpec = [ "." | PackageName ] ImportPath .
ImportPath = string_lit .
The import syntax requires a package name or path of a package. So no wildcard import.
This statement will clarify the why;
The PackageName is used in qualified identifiers to access exported
identifiers of the package within the importing source file.
You must specify in which package the element you are referring to exists. e.g, bar.Get.Basic.Req.Headers
. This can't be determined without an explicit import.
You can't do such thing, according to the specs of import
:
The import names an identifier (PackageName) to be used for access and an
ImportPath that specifies the package to be imported.
ImportDecl = "import" ( ImportSpec | "(" { ImportSpec ";" } ")" ) .
ImportSpec = [ "." | PackageName ] ImportPath .
ImportPath = string_lit .
The import syntax requires a package name or path of a package. So no wildcard import.
This statement will clarify the why;
The PackageName is used in qualified identifiers to access exported
identifiers of the package within the importing source file.
You must specify in which package the element you are referring to exists. e.g, bar.Get.Basic.Req.Headers
. This can't be determined without an explicit import.
answered Nov 17 at 22:24
Chen A.
5,09721531
5,09721531
add a comment |
add a comment |
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Import statements imports a
package
. As long as the code inside the directory is of the same package, it is imported. You then can access it bypackagename.<func/var>
. Why would you subdivide single package to many? Is it necessary?– Chen A.
Nov 17 at 22:16
@ChenA. yeah in the bar folder, I am going to have to generate a type alias that points to each branch in the folder tree, it sux, but what I have do I guess.
– Alexander Mills
Nov 17 at 22:17