go - 没有足够的参数来调用" html / template"。必须

时间:2016-03-09 23:27:12

标签: html go arguments render template-engine

我在Golang中编写了一个包装器函数,用于从多个文件中呈现模板,如下所示:

func RenderTemplate(w http.ResponseWriter, data interface{}, tmpl... string) {
    cwd, _ := os.Getwd()
    for _,file:=range tmpl{
        file=filepath.Join(cwd,"./view/"+file+".html")
    }
    t, err := template.ParseFiles(tmpl...)
    if err != nil {
        http.Error(w, err.Error(), http.StatusInternalServerError)
        return
    }
    templates:=template.Must(t)
    err = templates.Execute(w, data)
    if err != nil {
        http.Error(w, err.Error(), http.StatusInternalServerError)
    }
}

当我从main函数运行服务器时,控制台会抛出错误:

not enough arguments in call to "html/template".Must

如果我这样写:

templates,err:=template.Must(t)

它也会抛出同样的错误,加上:

assignment count mismatch: 2 = 1

我打算将此函数用于服务器中的路由处理程序:

func IndexHandler(w http.ResponseWriter, r *http.Request) {
    files:=[]string{"base","index"}
    util.RenderTemplate(w, nil, files...)
}

index.html使用模板嵌套

base.html延伸

base.html模板:

{{define "base"}}
<!DOCTYPE html>
<html>
<head>
    <meta charget="utf-8">
    <title>{{template "title".}}</title>
    <script type="text/javascript" src="https://code.jquery.com/jquery-2.2.0.min.js"></script>
    <script type="text/javascript" src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
    <script type="text/javascript" src="/js/isotope.pkgd.min.js"></script>
    <link rel="stylesheet" type="text/css" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">
    <link rel="stylesheet" type="text/css" href="https://maxcdn.bootstrapcdn.com/font-awesome/4.5.0/css/font-awesome.min.css">
    <link rel="stylesheet" type="text/css" href="/css/style.css">
</head>
<body>
    {{template "index".}}
</body>
</html>
{{end}}

index.html模板:

<!DOCTYPE html>
<html>
{{define "title"}}Homepage{{end}}
<body>
    {{define "index"}}
    <div class="wrapper">
        <div class="page-content">
            <div class="container">
                <div class="left">
                    <img src="../public/images/img_landing_page_mac.png">
                </div>
                <div class="right">
                    <h2 style="font-size: 33px; letter-spacing: 5px">Organize <br>Modern Knowledge<br> for Mankind</h2>
                    <p style="font-size: 20px;margin-top: 35px;letter-spacing: 4px">Consume, Colect and Revisit <br>Knowledge at Your Fingertips</p>
                    <a href="#" style="margin-top: 80px;display: inline-block;margin-left: -17px"><img src="../public/images/btn_get_chrome.png"></a>
                </div>
            </div>
        </div>
    </div>
    {{end}}
</body>
</html>

我错过了什么吗?我检查了"html/template".Must的原型并没有得到发生的事情

2 个答案:

答案 0 :(得分:1)

您不需要调用ParseFiles和Must,您可以调用其中一个

func RenderTemplate(w http.ResponseWriter, data interface{}, tmpl... string) {
    cwd, _ := os.Getwd()
    for _,file:=range tmpl{
        file=filepath.Join(cwd,"./view/"+file+".html")
    }
    t, err := template.ParseFiles(tmpl...)
    if err != nil {
        http.Error(w, err.Error(), http.StatusInternalServerError)
        return
    }

    err = t.Execute(w, data)
    if err != nil {
        http.Error(w, err.Error(), http.StatusInternalServerError)
    }
}

我相信上面的功能应该做你想要的......

答案 1 :(得分:1)

template.Must()有这个签名:

func Must(t *Template, err error) *Template 

Must()的论据是&#34;巧合&#34;与ParseFiles()ParseGlob()的返回值相同,因此您可以在Must()中使用这些函数,如果错误为非零,则会产生恐慌。所以你可以说

t := template.Must(template.ParseFiles(....))

并且不关心错误检查。这只是一个便利函数,类似于整个标准库中的所有其他Must()函数,例如regexp.MustCompile()

Must()的实施非常简单:

func Must(t *Template, err error) *Template {
    if err != nil {
            panic(err)
    }
    return t
}

请参阅https://golang.org/src/text/template/helper.go?s=576:619#L11