REBOL解析规则
问题描述：
这个伟大的工程（感谢苏南达的建议How can I parse [a/b] ? syntax error in Rebol?）：REBOL解析规则
attribute: copy []
class: copy []
definition-rule: compose/deep [some [set class word! 'is 'defined 'by
[some [copy attribute to (to-lit-word "/") thru (to-lit-word "/") ]]
copy attribute to end]]
parse [Customer is defined by First Name/Last Name/Email] definition-rule
，但我现在需要添加一些附加指令（追加输出类），它不工作了：
attribute: copy []
class: copy []
definition-rule: compose/deep [some [set class word! (append output class) 'is 'defined 'by
[some [copy attribute to (to-lit-word "/") thru (to-lit-word "/") ]]
copy attribute to end]]
parse [Customer is defined by First Name/Last Name/Email] definition-rule
答
的这里的问题是，组成是吃所有表达式在括号中。你很高兴它吃（以点亮的词“/”），但是你认真不希望它吃（附加输出类），因为这是为在解析方言。
有可能是一个聪明的做法，但这应该工作：删除通过执行点燃字解析规则之外工作撰写 ...
attribute: copy []
class: copy []
output: copy ""
fs: to-lit-word "/" ;; define a forward slash lit-word
definition-rule: [
some [set class word! (append output class) 'is 'defined 'by [
some [copy attribute to fs thru fs]
]
copy attribute to end]
]
parse [Customer is defined by First Name/Last Name/Email] definition-rule
== true
我不能完全肯定你正在尝试使用此代码做的，但你想在年底提取的属性集，然后再考虑这种变化：
attribute: copy []
attributes: copy []
class: copy []
output: copy ""
fs: to-lit-word "/" ;; define a forward slash lit-word
definition-rule: [
some [set class word! (append output class) 'is 'defined 'by [
some [copy attribute to fs thru fs (append/only attributes attribute)]
]
copy attribute to end (append/only attributes attribute)]
]
parse [Customer is defined by First Name/Last Name/Email] definition-rule
print ["==" class mold attributes]
== Customer [[First Name] [Last Name] [Email]]
+0
非常感谢，我需要在第二个例子中这样做的： 属性：复制[] 类：复制[] FS：要点亮的词“/” 输出：复制“” 定义规则：[ 一些[套课文！ （追加输出连接属性“;”）] [ ] 复制属性结束（追加输出属性）] [ ] 解析[客户由名/姓/电子邮件定义] definition-rule 探针输出 – 2009-09-22 20:02:46
答
我重新发布代码注释，因为它是不可读，W我想要做的正是这样的帽子：
attribute: copy []
class: copy []
fs: to-lit-word "/"
output: copy ""
definition-rule: [
some [set class word! (append output join class "|") 'is 'defined 'by [
some [copy attribute to fs thru fs (append output join attribute ";")]
]
copy attribute to end (append output attribute)]
]
parse [Customer is defined by First Name/Last Name/Email] definition-rule
probe output
好吧，让我们看看是否再次回答Sunanda。我不太明白为什么一个教程提出问题，但我准备犯错。 – 2009-09-22 04:56:02
我很乐意在任何有REBOL问题的论坛上回答有关REBOL的问题（如果我知道答案的话）。到目前为止，REBOL社区对*的使用非常少 - 主要是RebolTutorial提出的问题，还有几个人（主要是我到目前为止）的回应。如果这给任何人造成问题，那么Reboltutorial可以切换到REBOL特定的论坛之一以获得他/她的Q＆A支持。 – Sunanda 2009-09-22 13:57:59