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Default usage output to stdout 

If the command has not set an output explicitly everything will go to
stderr.  This makes a lot of sense, but is a huge PITA for people who
want to be able to grep the help output.  It is very common for users to
want to do

command --help | grep flag

This patch fixes that by default help output (but not error output like
an invalid command) to stdout instead of defaulting to stderr.
This commit is contained in:
Eric Paris 2015-03-18 11:16:12 -04:00 committed by bep
parent 6264dc67e1
commit 6de96b849c
1 changed files with 11 additions and 3 deletions
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14
command.go
View File

@ -79,7 +79,7 @@ func (c *Command) SetArgs(a []string) {
c.args = a
}
func (c *Command) Out() io.Writer {
func (c *Command) getOut(def io.Writer) io.Writer {
if c.output != nil {
return *c.output
}
@ -87,10 +87,18 @@ func (c *Command) Out() io.Writer {
if c.HasParent() {
return c.parent.Out()
} else {
return os.Stderr
return def
}
}
func (c *Command) Out() io.Writer {
return c.getOut(os.Stderr)
}
func (c *Command) getOutOrStdout() io.Writer {
return c.getOut(os.Stdout)
}
// SetOutput sets the destination for usage and error messages.
// If output is nil, os.Stderr is used.
func (c *Command) SetOutput(output io.Writer) {
@ -658,7 +666,7 @@ func (c *Command) Usage() error {
// by the default HelpFunc in the commander
func (c *Command) Help() error {
c.mergePersistentFlags()
err := tmpl(c.Out(), c.HelpTemplate(), c)
err := tmpl(c.getOutOrStdout(), c.HelpTemplate(), c)
return err
}