On Tue, Jan 25, 2022 at 11:35 AM Victor Giordano <vitucho3...@gmail.com> wrote:
>
> Hi all! I'm dealing with some multi valued column in postgres, and i have
> stomp into a bizzare behavoir in `fmt.Sscanf`
>
> Please take a look at the following snippet, i also leave my question within
> the code and this link to the playground so you can actually try it and see
> with you own eyes.
>
> ```
> package main
>
> import "fmt"
>
> func main() {
> var a int
> var b string
> input := "(1,beto)" // doesn't like the parenthesis
>
> _, err := fmt.Sscanf(input, "(%d,%s)", &a, &b)
> if err != nil {
> fmt.Println(err)
> } else {
> fmt.Println(a, b)
> }
>
> input = "1,beto" // this works
> _, err = fmt.Sscanf(input, "%d,%s", &a, &b)
> if err != nil {
> fmt.Println(err)
> } else {
> fmt.Println(a, b)
> }
>
> input = "beto,1" // doesn't like an string at first place, ¿¿WHAT i
> do?!?!
> _, err = fmt.Sscanf(input, "%s,%d", &b, &a)
> if err != nil {
> fmt.Println(err)
> } else {
> fmt.Println(a, b)
> }
> }
> ```
Using %s in fmt.Sscanf will read all characters up to a space or
newline. This is documented at https://pkg.go.dev/fmt which says
"Input processed by verbs is implicitly space-delimited: the
implementation of every verb except %c starts by discarding leading
spaces from the remaining input, and the %s verb (and %v reading into
a string) stops consuming input at the first space or newline
character."
Basically fmt.Sscanf is not the right tool for the problem you are
trying to solve.
Ian
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