Is it possible to pass a string of go code into go run instead of go run /some/path/script.go? I tried:
echo "some awesome go code here" | go run
But does not work. Thanks.
I don't think that there is such an option. At least not with the standard *g compilers or
go run.
You can try using gccgo as GCC supports reading from stdin.
Since I thought that this would be a useful thing to have, I wrote a relatively small Python script that achieves what I think you want. I called it go-script, and here are some usage examples:
# Assuming that test.go is a valid go file including package and imports
$ go-script --no-package < test.go
# Runs code from stdin, importing 'fmt' and wrapping it in a func main(){}
$ echo 'fmt.Println("test")' | go-script --import fmt --main
$ echo 'fmt.Println("test")' | go-script -ifmt -m
Help:
Usage: go-script [options]
Options:
-h, --help show this help message and exit
-i PACKAGE, --import=PACKAGE
Import package of given name
-p, --no-package Don't specify 'package main' (enabled by default)
-m, --main Wrap input in a func main() {} block
-d, --debug Print the generated Go code instead of running it.
The source (also available as a gist):
#!/usr/bin/env python
from __future__ import print_function
from optparse import OptionParser
import os
import sys
parser = OptionParser()
parser.add_option("-i", "--import", dest="imports", action="append", default=[],
help="Import package of given name", metavar="PACKAGE")
parser.add_option("-p", "--no-package", dest="package", action="store_false", default=True,
help="Don't specify 'package main' (enabled by default)")
parser.add_option("-m", "--main", dest="main", action="store_true", default=False,
help="Wrap input in a func main() {} block")
parser.add_option("-d", "--debug", dest="debug", action="store_true", default=False,
help="Print the generated Go code instead of running it.")
(options, args) = parser.parse_args()
stdin = ""
for line in sys.stdin.readlines():
stdin += "%s\n" % line
out = ""
if options.package:
out += "package main\n\n"
for package in options.imports:
out += "import \"%s\"\n" % package
out += "\n"
if options.main:
out += "func main() {\n%s\n}\n" % stdin
else:
out += stdin
if options.debug:
print(out)
else:
tmpfile = "%s%s" % (os.environ["TMPDIR"], "script.go")
f = open(tmpfile, 'w')
print(out, file=f)
f.close()
os.execlp("go", "", "run", tmpfile)
This works
cat <<EOF | tee /tmp/blah.go | go run /tmp/blah.go
package main
import "fmt"
func main() {
fmt.Println("Hello, World!")
}
EOF
If you want to not have to open a file and edit it first. Although I wouldn't find this super practical for every day use.
Related
I wrote a bash script to automate download my Go program and install to my linux machine. I can via the curl to download but when my Go program prompt the user input, it go to infinite loop. The error shows EOF error.
Do anyone have any idea about it?
Install.sh
#!/bin/bash
set -e
set -o noglob
curl https://coderkk.net/ReadInput -o ReadInput
chmod a+x ReadInput
./ReadInput
ReadInput.go
package main
import (
"bufio"
"fmt"
"os"
"strings"
)
func main() {
var text string
reader := bufio.NewReader(os.Stdin)
for {
fmt.Print("Please enter you name here : ")
text, _ = reader.ReadString('\n')
text = strings.Replace(text, "\n", "", -1)
if text != "" {
break
}
}
fmt.Printf("Hi %s\n", text)
}
The problem is not related to Go - although you shouldn't ignore any potential errors from reader.ReadString - it is due to how you are invoking your downloaded script.
The pipe takes the output of curl and turns that into the stdin for sh. This is fine but your script requires stdin from the terminal - which is now lost.
Change your invocation method to something like this:
sh -c "$(curl -sfL https://coderkk.net/install.sh)"
this will pass the contents of your script to the sh interpreter but preserve the user's terminal stdin.
How do I read string argument include & in Go for example this link
$ ./main
https://www.youtube.com/watch?v=G3PvTWRIhZA&list=PLQVvvaa0QuDeF3hP0wQoSxpkqgRcgxMqX
without use double quotation (")
$ ./main
"https://www.youtube.com/watch?v=G3PvTWRIhZA&list=PLQVvvaa0QuDeF3hP0wQoSxpkqgRcgxMqX"
main.go
package main
import (
"os"
"fmt"
)
func main() {
link := os.Args[1]
fmt.Println(link)
}
$ go build main.go
$ ./main
https://www.youtube.com/watch?v=G3PvTWRIhZA&list=PLQVvvaa0QuDeF3hP0wQoSxpkqgRcgxMqX
output will be
https://www.youtube.com/watch?v=G3PvTWRIhZA
Both #JimB and #Adrian are correct, the & needs to be escaped.
If you absolutely must find a workaround, you could opt to not use a command-line argument and rather read input instead to bypass need for escaping.
Example:
package main
import (
"fmt"
)
func main() {
var input string
fmt.Scan(&input)
fmt.Println(input)
}
input:
$ ./main
https://www.youtube.com/watch?v=G3PvTWRIhZA&list=PLQVvvaa0QuDeF3hP0wQoSxpkqgRcgxMqX
output:
https://www.youtube.com/watch?v=G3PvTWRIhZA&list=PLQVvvaa0QuDeF3hP0wQoSxpkqgRcgxMqX
I have the file playground.go:
package main
import (
"fmt"
)
var GitCommit string
func main() {
fmt.Printf("Hello world, version: %s\n", GitCommit)
}
why the behavior of the build ldflag -X depends on specifying the source filename?:
# CASE 1
$ go build -ldflags "-X main.GitCommit=abc" && ./go_playground
Hello world, version:
$ go tool nm go_playground
52a900 D main.GitCommit
# CASE 2
$ go build -ldflags "-X main.GitCommit=abc" playground.go && ./playground
Hello world, version: abc
$ go tool nm playground
5242f0 D main.GitCommit
4c5678 R main.GitCommit.str
EDIT: it seems that the problem happens when case 1 is executed from a symlinked directory; when case 1 is executed from the original directory, the variable is passed. I'm not sure if this is expected, or it's a bug.
I want to escape a restricted shell spawning a bash shell via Go. In other words, I want to do this but using Go:
python -c 'import pty; pty.spawn("/bin/bash")'
I am totally new to Go. I have tried this (following the answer in this question Go: How to spawn a bash shell) but nothing happens:
package main
import "os"
import "os/exec"
func main() {
shell := exec.Command("/bin/bash")
shell.Stdout = os.Stdout
shell.Stdin = os.Stdin
shell.Stderr = os.Stderr
shell.Run()
}
Also if I add the fmt.Println("hello") line at the end of the main function nothing is printed
UPDATE
Maybe I did not expalined well. What I am trying to achieve it's to spawn a shell gotten a restricted shell. This is what I did:
Listener:
nc.traditional -l -p 8080 -e /bin/bash
Connects to listener: And I exec the code here
nc.traditional localhost 8080 -v
Your program works fine for me. I put some error checking in and an extra print statement which should make what is happening clearer. You are getting an interactive shell, it just looks exactly like your previous shell.
$ go run Go/shell.go
$ # whoa another shell
$ exit
exit
exiting
$ # back again
$
Here is the revised program
package main
import (
"fmt"
"log"
"os"
"os/exec"
)
func main() {
shell := exec.Command("/bin/bash")
shell.Stdout = os.Stdout
shell.Stdin = os.Stdin
shell.Stderr = os.Stderr
err := shell.Run()
if err != nil {
log.Fatalf("command failed: %v", err)
}
fmt.Printf("exiting\n")
}
I'm passing argument from console. There are some flags too. Like:
go run test.go "-IP=10.10.10.10" "-db=kite" "-wv=45" "-cv=75" "A = value1" "B = value2" "C = 100" "D := ((A-B)/A)*C" "D ?"
Here, -IP, -db, -wv, -wc these four are flags and others are passing as normal argument as I know.
Number of flags can be variable.
How can I know how many flags are passed to my program. In this case 4 flags are passed.
If you use the standard flag package to parse command-line flags, you can call the NFlag function to get the number of flags:
package main
import "fmt"
import "flag"
func main() {
flag.Bool("a", true, "A value");
flag.Bool("b", true, "B value");
flag.Parse();
fmt.Println(flag.NFlag())
}
Test:
$ go run test.go
0
$ go run test.go -a
1
$ go run test.go -a -b
2