文本/模板问题Parse()与ParseFiles()

Question

我正在尝试使用text/template包做一些简单的工作.模板顶部给出的样本就是我正在使用的样本.

如何编写'解析'文件以便template.ParseFiles()正确读取和执行它？

package main

import (
    "text/template"
    "os"
)

type Inventory struct {
    Material string
    Count    uint
}

func main() {
    sweaters := Inventory{"wool", 17}
    tmpl, err := template.New("test").Parse("{{.Count}} items are made of {{.Material}}")
    // tmpl, err := template.New("test").ParseFiles("file.txt")

    if err != nil { panic(err) }
    err = tmpl.Execute(os.Stdout, sweaters)
    if err != nil { panic(err) }
}

/*
Contents of file.txt:
{{.Count}} items are made of {{.Material}}

Error thrown:
panic: template: test:1: "test" is an incomplete or empty template

goroutine 1 [running]:
main.main()
    /tmp/templates/t.go:19 +0x21a

goroutine 2 [syscall]:
created by runtime.main
    /var/tmp/portage/dev-lang/go-1.0.1/work/go/src/pkg/runtime/proc.c:221
*/

我在这里的golang操场上发布了这段代码的副本

编辑#1:我一直在研究这个问题...因为它Execute()是实际抛出异常的方法,而不是ParseFiles()部分,我检查了方法定义:

// Execute applies a parsed template to the specified data object,
// and writes the output to wr.
func (t *Template) Execute(wr io.Writer, data interface{}) (err error) {
    defer errRecover(&err)
    value := reflect.ValueOf(data)
    state := &state{
        tmpl: t,
        wr:   wr,
        line: 1,
        vars: []variable{{"$", value}},
    }
    if t.Tree == nil || t.Root == nil {
        state.errorf("%q is an incomplete or empty template", t.name)
    }
    state.walk(value, t.Root)
    return
}

所以,在预感中,我将t.Tree的值转换为内联的"非文件"样式,tmpl是:&parse.Tree{Name:"test", Root:(*parse.ListNode)(0xf840030700), funcs:[]map[string]interface {}(nil), lex:(*parse.lexer)(nil), token:[2]parse.item{parse.item{typ:6, val:""}, parse.item{typ:9, val:"{{"}}, peekCount:1, vars:[]string(nil)}当运行时ParseFiles(),tmpl是:(*parse.Tree)(nil).我觉得很奇怪,一个是解除引用,一个值是一个指针.这可能有助于解决这个谜团

Answer 1

sweaters := Inventory{"wool", 17}
tmpl, err := template.ParseFiles("file.txt")
if err != nil {
    panic(err)
}
err = tmpl.ExecuteTemplate(os.Stdout, "file.txt", sweaters)
if err != nil {
    panic(err)
}

如果您有许多文件,可以使用ParseGlob:

tmpl, err := template.ParseGlob("*.txt")
if err != nil {
    panic(err)
}
err = tmpl.ExecuteTemplate(os.Stdout, "file.txt", sweaters)
if err != nil {
    panic(err)
}
err = tmpl.ExecuteTemplate(os.Stdout, "file2.txt", sweaters)
if err != nil {
    panic(err)
}

Answer 2

Go模板parseFiles中有一个小技巧。

func parseFiles(t *Template, filenames ...string) (*Template, error) {
    if len(filenames) == 0 {
        // Not really a problem, but be consistent.
        return nil, fmt.Errorf("template: no files named in call to ParseFiles")
    }
    for _, filename := range filenames {
        b, err := ioutil.ReadFile(filename)
        if err != nil {
            return nil, err
        }
        s := string(b)
        name := filepath.Base(filename)
        // First template becomes return value if not already defined,
        // and we use that one for subsequent New calls to associate
        // all the templates together. Also, if this file has the same name
        // as t, this file becomes the contents of t, so
        //  t, err := New(name).Funcs(xxx).ParseFiles(name)
        // works. Otherwise we create a new template associated with t.
        var tmpl *Template
        if t == nil {
            t = New(name)
        }
        if name == t.Name() {
            tmpl = t
        } else {
            tmpl = t.New(name)
        }
        _, err = tmpl.Parse(s)
        if err != nil {
            return nil, err
        }
    }
    return t, nil
}

仅具有相同名称的模板将被重用，否则创建新的模板。作为您的样本：

tmpl, err := template.New("test").ParseFiles("file.txt")

tmpl是名为“ test”的模板，并关联了另一个名为“ file.txt”的模板，您在“ test”模板上调用Execute，此模板是空模板，因此引发错误“ test是不完整或空模板”。

当您将模板名称更改为file.txt时，它起作用了

tmpl, err := template.New("file.txt").ParseFiles("file.txt")