我有这个dtd: http://fast-code.sourceforge.net/template.dtd
但是当我在xml中包含时，我得到警告:
没有为文档检测到语法约束(DTD或XML模式)。
xml是:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE templates PUBLIC "//UNKNOWN/" "http://fast-code.sourceforge.net/template.dtd">
<templates>
<template type="INSTANCE_OF_CLASS">
<description>Used to Create instance of class</description>
<variation>asasa</variation>
<variation-field>asasa</variation-field>
<class-pattern>asasa</class-pattern>
<getter-setter>setter</getter-setter>
<allowed-file-extensions>java</allowed-file-extensions>
<number-required-classes>1</number-required-classes>
<allow-multiple-variation>false</allow-multiple-variation>
<template-body>
<![CDATA[
// Creating new instance of ${class_name}
final ${class_name} ${instance} = new ${class_name}();
#foreach ($field in ${fields})
${instance}.${field.setter}(${field.value});
#end
]]>
</template-body>
</template>
</templates>
编辑:我改变了xml，我现在得到这个错误:
元素类型“template”的内容必须匹配“(description,variation?，variation-field?,允许
multiple-variation ?,阶级结构? getter-setter ? allowed-file-extensions吗?阈值-
类?模板体内)”。
答:
下面每个DTD部分的注释。更多信息请参考官方规范。
<!
DOCTYPE ----------------------------------------- correct
templates --------------------------------------- correct Name matches root element.
PUBLIC ------------------------------------------ correct Accessing external subset via URL.
"//UNKNOWN/" ------------------------------------ invalid? Seems useless, wrong, out-of-place.
Safely replaceable by DTD URL in next line.
"http://fast-code.sourceforge.net/template.dtd" - invalid URL is currently broken.
>
简单的解释:
一个非常基本的DTD看起来就像这里的第二行:
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE nameOfYourRootElement>
<nameOfYourRootElement>
</nameOfYourRootElement>
详细解释:
dtd用于建立商定的数据格式并验证此类数据的接收。它们定义了XML文档的结构，包括:
法律要素清单
特殊字符
字符串
还有更多
E.g.
<!DOCTYPE nameOfYourRootElement
[
<!ELEMENT nameOfYourRootElement (nameOfChildElement1,nameOfChildElement2)>
<!ELEMENT nameOfChildElement1 (#PCDATA)>
<!ELEMENT nameOfChildElement2 (#PCDATA)>
<!ENTITY nbsp " ">
<!ENTITY author "Your Author Name">
]>
Meaning of above lines...
Line 1) Root element defined as "nameOfYourRootElement"
Line 2) Start of element definitions
Line 3) Root element children defined as "nameOfYourRootElement1" and "nameOfYourRootElement2"
Line 4) Child element, which is defined as data type #PCDATA
Line 5) Child element, which is defined as data type #PCDATA
Line 6) Expand instances of to   when document is parsed by XML parser
Line 7) Expand instances of &author; to Your Author Name when document is parsed by XML parser
Line 8) End of definitions
我真的不能说为什么会出现“No grammar constraints…”警告，但是我可以在Eclipse中通过完全删除DOCTYPE声明来引起它。当我把声明放回去并再次验证时，我得到了这个错误消息:
元素类型template的内容
必须匹配
”(描述+、变异? variation-field ?, allow-multiple-variation ?,阶级结构?,getter-setter ?, allowed-file-extensions ?,模板体内+)。
我相信这是正确的(“number-required-classes”元素是不允许的)。
我知道这是旧的，但我通过同样的问题，并在春季文档中找到了解决方案，以下xml配置已经为我解决了这个问题。
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/tx
<!-- THIS IS THE LINE THAT SOLVE MY PROBLEM -->
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">
在我把上面的行建议在这个论坛的主题，我有同样的警告信息，并把这个…
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xml>
它给了我以下警告信息…
The content of element type "template" must match "
(description,variation?,variation-field?,allow- multiple-variation?,class-
pattern?,getter-setter?,allowed-file-extensions?,number-required-
classes?,template-body)".
因此，请尝试使用我的XML配置中的建议行。
