text/template: template.New(name).ParseFiles(file) not behaving as expected

[udf2457]

Go version

1.22

Output of go env in your module/workspace:

Not applicable for this

What did you do?

t, err := template.New(name).ParseFiles(file)
if err != nil {
  log.Error("failed to load template", "err", err.Error())
  os.Exit(1)
}

then elsewhere ...

t.Execute(...)

What did you see happen?

On Execute() an error was raised:

name: \"name\" is an incomplete or empty template"

What did you expect to see?

Execute() should execute as expected.

Instead, I had to change template.New(name).ParseFiles(file) to template.ParseFiles(file)

I therefore suggest:

1. The preferred option would be chaining template.New(name).ParseFiles(file) works, as reasonably expected
2. It errors out if chained

The present situation is unacceptable because you do not discover the problem until Execute() time and when you do, the error message is not helpful in diagnosing the cause.

[AlexanderYastrebov]

It works when name == file (t, err := template.New("test.tmpl").ParseFiles("test.tmpl")) and fails when not (t, err := template.New("test").ParseFiles("test.tmpl"))

The documentation https://pkg.go.dev/text/template#Template.ParseFiles has somewhat vague explanation of the behaviour.

[AlexanderYastrebov]

See previous #11247

[AlexanderYastrebov]

Since this works (as documented):

func main() {
	t, err := template.New("whatever").ParseFiles("test.tmpl")
	if err != nil {
		log.Fatalf("Failed to load template: %v", err)
	}

	err = t.ExecuteTemplate(os.Stdout, "test.tmpl", nil)
	if err != nil {
		log.Fatalf("Failed to execute template: %v", err)
	}
}

it might be impossible to change the behaviour of template.New("foo").ParseFiles("bar") to return error earlier due to backwards compatibility.

[seankhliao]

given:

Since the templates created by ParseFiles are named by the base names of the argument files,

I think this is unambiguously working as intended.

it "works" when the name matches simply because it's overwritten.
note that it's also possible to do New().Parse().ParseFiles()