Question

我想创建一个可以与ID匹配的终端，但不完全匹配。虽然ID是

terminal ID         : '^'?('a'..'z'|'A'..'Z'|'_') ('a'..'z'|'A'..'Z'|'_'|'0'..'9')*;

我想要定义的终端是

terminal TYPE: (('a'..'z'|'A'..'Z')?('a'..'z'|'A'..'Z'|'_'|'0'..'9')*)?

因为TYPE可以匹配ID我收到RULE_ID错误，在这种情况下我该怎么办？

______EDIT__________

Domainmodel :
    (elements+=XType)*;

terminal TYPE: ('a'..'z'|'A'..'Z')('a'..'z'|'A'..'Z'|'0'..'9'|'_')*;

MyID: 
    TYPE | ID
;

XType:
    DataType | Entity;

DataType:
    'datatype' name=MyID;

Entity:
    'entity' name=MyID ('extends' superType=[Entity])? '{'
        (features+=Feature)*
    '}';

Feature:
    (many?='many')? name=MyID ':' type=[XType];

模型（基于博客示例）

datatype String

entity Blog {
    title: String
    title2: String
    many posts: Post
    many Posts: Post
}

entity HasAuthor {
    author: String
}

entity Post extends HasAuthor {
    title: String
    content: String
    many comments: Comment
}

entity Comment extends HasAuthor {
    content: String
}

Answer 1

您可以引入数据类型规则

MyID: ID | TYPE;

MyID的值转换器，并在您使用ID的地方使用它

或者您忘记输入并检查验证器内的限制范围

Domainmodel :
    (elements+=XType)*;

terminal TYPE: ('a'..'z'|'A'..'Z')('a'..'z'|'A'..'Z'|'0'..'9'|'_')*;

MyID: 
    TYPE | ID
;

XType:
    DataType | Entity;

DataType:
    'datatype' name=MyID;

Entity:
    'entity' name=MyID ('extends' superType=[Entity|MyID])? '{'
        (features+=Feature)*
    '}';

Feature:
    (many?='many')? name=MyID ':' type=[XType|MyID];

在xtext中定义作为ID子集的终端

1 个答案: