Hi Tamás,
Have you actually tried that and gotten it to work? It does not compile
for me but this does (note method call vs. property reference):
slog.SetDefault(slog.New(myHandler{Handler:slog.Default().Handler()}))
However, when delegating the Handle() method it seems to cause an infinite
loop:
func (m MyHandler) Handle(ctx context.Context, r slog.Record) error {
return m.Handler.Handle(ctx, r)
}
See https://goplay.tools/snippet/qw07m0YflLd
I know about this because just this past weekend I was trying to write a
TeeHandler to output the default to the screen and JSON to a file just this
past weekend and ran into an infinite loop problem with the default handler.
I tried my best to figure out why it needed to be structured the way it was
in that it seems to call itself recursively. I wanted to post a question to
this list to see if there was a workaround, or if not to see if there might
be interest in allowing it to work, but I could not get my head around it so
eventually gave up and just used the TextHandler instead.
Shame though. It would be nice to be able to reuse the default handler but
AFACT it is not possible (though if I am wrong I would love for someone to
show me how to get it to work.)
-Mike
On Monday, August 28, 2023 at 12:50:50 PM UTC-4 Tamás Gulácsi wrote:
slog.SetDefault(slog.New(myHandler{Handler:slog.Default().Handler}))
vl...@mailbox.org a következőt írta (2023. augusztus 28., hétfő, 15:06:37
UTC+2):
Hi,
When reading trough the log/slog documentation, it seems one can create
a logger with a different handler, which is either NewTextHandler or
NewJSONHandler.
Why can't I configure the defaultHandler? Let's say I want my logger to
behave exactly like the defaultHandler, but output to a logfile or
Stdout instead.
The defaultHandler's output is different compared to the NewTextHandler:
slog.Info("ok"), gives me:
INFO ok
The NextTextHandler gives me:
level=INFO msg="ok"
Regards,
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