使用重写规则的 Scala XML 转换
Scala XML transformation with Rewrite rules
我有一个 XML 模板,其中预定义了一些字段。我想使用 RewriteRules
.
基于它的模板和新的 Value
值构建新的 XML
例如。
模板:
val template = <xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value></Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value></Value>
</LastName>
</Person>
</Persons>
</xml>
case class Person(fullName: String, lastName: String)
val persons = Seq(Person("John Smith", "Smith"), Person("Bob Saver", "Saver"))
输出应该是:
<xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>John Smith</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Smith</Value>
</LastName>
</Person>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>Bob Saver</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Saver</Value>
</LastName>
</Person>
</Persons>
</xml>
可以用RewriteRules
吗?
您不需要 RewriteRules
用于此目的。您可以在 xml 模板中定义变量。
scala> def template(id: String = "defaultValue can be here") = <someXml>{id}</someXml>
template: (id: String)scala.xml.Elem
scala> template("person")
res4: scala.xml.Elem = <someXml>person</someXml>
scala> template("person2")
res5: scala.xml.Elem = <someXml>person2</someXml>
否则
Scala - replace xml element with specific text
我有一个 XML 模板,其中预定义了一些字段。我想使用 RewriteRules
.
Value
值构建新的 XML
例如。 模板:
val template = <xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value></Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value></Value>
</LastName>
</Person>
</Persons>
</xml>
case class Person(fullName: String, lastName: String)
val persons = Seq(Person("John Smith", "Smith"), Person("Bob Saver", "Saver"))
输出应该是:
<xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>John Smith</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Smith</Value>
</LastName>
</Person>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>Bob Saver</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Saver</Value>
</LastName>
</Person>
</Persons>
</xml>
可以用RewriteRules
吗?
您不需要 RewriteRules
用于此目的。您可以在 xml 模板中定义变量。
scala> def template(id: String = "defaultValue can be here") = <someXml>{id}</someXml>
template: (id: String)scala.xml.Elem
scala> template("person")
res4: scala.xml.Elem = <someXml>person</someXml>
scala> template("person2")
res5: scala.xml.Elem = <someXml>person2</someXml>
否则 Scala - replace xml element with specific text