如何在Go中使用各种元素解析巨大的XML文件？

How can you parse a huge XML file that's having various elements (i.e. not same element repeated multiple times).

Example:

<stuff>
    <header>...</header>
    <item>...</item>
    ...
    <item>...</item>
    <something>...</sometihng>
</stuff>

I want to write a script in Go that would allow me to split this file in multiple smaller files with specific amount of tags per file. All examples on how to parse XML with Go seems to rely on knowing the elements that you have in the file.

Can the file be parsed without knowing that? Something like for each element in XML no matter what element is there (header, item, something, etc...)

Use the standard xml Decoder.

Call Token to read tokens one by one. When a start element of interest is found, call DecodeElement to decode the element to a Go value.

Here's a sketch of how to use the decoder:

d := xml.NewDecoder(r)
for {
    t, tokenErr := d.Token()
    if tokenErr != nil {
        if tokenErr == io.EOF {
           break
        }
        // handle error
    }
    switch t := t.(type) {
    case xml.StartElement:
        if t.Name.Space == "foo" && t.Name.Local == "bar" {
            var b bar
            if err := d.DecodeElement(&b, &t); err != nil {
                // handle error
            }
            // do something with b
        }
    }
}

This isn't so much a limit of Go as a limit of xml. XML elements only make sense according to their schema (which predefines what elements are in other elements).

You should look to SAX parses, something like https://github.com/kokardy/saxlike

you can also check following library which has been tested with big xml files. It has written to address the performance issue in go default xml package.

xml stream parser