XSLT Atrtibute价值 [英] XSLT Atrtibute Value

问题描述

我正在寻找以下问题的解决方案：

我生成的XML文档就像


< .... >

< struct name =" number" type =" PartNumber"> 100< / struct>

< struct name =" identity" type =" PartIdentity"> 1< / struct>

< enum name =" ex_identity" type =" .....">< / enum>

< ....>


现在我想要使用XSLT文件创建一个新的结构化XML文件，例如


< ....>

< number sort =" struct" type =" PartNumber"> 100< / struct>

< identity sort =" struct" type =" PartIdentity"> 1< / struct>

< ex_identity sort =" enum" type =" ........


我的问题是，我无法读出该属性的价值

"命名"所以我不能

创建它的新元素。此外，我需要为新元素创建其他

属性，或者我可以轻松地复制它们（？）。

有人有任何解决方案吗？ PLZ帮我搜索了很长时间。


Big THX parvus

Hi, I′m searching a solution for the following problem:
I′ve a generated XML Document like

<....>
<struct name="number" type="PartNumber">100</struct>
<struct name="identity" type="PartIdentity">1</struct>
<enum name="ex_identity" type="....."></enum>
<....>

Now i want to use a XSLT file to create a new structured XML file like

<....>
<number sort="struct" type="PartNumber">100</struct>
<identity sort="struct" type="PartIdentity">1</struct>
<ex_identity sort="enum" type="........

My problem ist, that i couldn′t read out the value of the attribute
"name" so i can′t
create a new element of it. Furthermore i need to create the other
attributes for the new element or maybe i can copy them easily(?). Does
anybody have any solution? PLZ help me im searching for a long time.

Big THX parvus

推荐答案

parvus写道：

[...]

parvus wrote:
[...]

现在我想使用XSLT文件创建一个新的结构化XML文件，如

< ....>
< number sort =" struct" type =" PartNumber"> 100< / struct>
< identity sort =" struct" type =" PartIdentity"> 1< / struct>
< ex_identity sort =" enum" type =" ........

Now i want to use a XSLT file to create a new structured XML file like

<....>
<number sort="struct" type="PartNumber">100</struct>
<identity sort="struct" type="PartIdentity">1</struct>
<ex_identity sort="enum" type="........

[...]

使用枚举是一个好主意。但上面的格式不是很好。

也许你想要别的东西？

[...]
Using enum is a good idea. But the above is not well formed XML.
Maybe you want something else?

在文章< 11 ******* ***************@z14g2000cwz.googlegroups .com> ;,

parvus< pa ****** @ web.de>写道：

In article <11**********************@z14g2000cwz.googlegroups .com>,
parvus <pa******@web.de> wrote:

< struct name =" number"类型= QUOT; PARTNUMBER"> 100℃; /结构>
< number sort =" struct" type =" PartNumber"> 100< / struct>

<struct name="number" type="PartNumber">100</struct> <number sort="struct" type="PartNumber">100</struct>

你想要类似的东西


< xsl：element名称= QUOT; {@名称}"> ......


- Richard

You want something like

<xsl:element name="{@name}"> ...

-- Richard

我正在搜索以下问题的解决方案：
我生成的XML文档如

< ....>
< struct name =" number" type =" PartNumber"> 100< / struct>
< struct name =" identity" type =" PartIdentity"> 1< / struct>
< enum name =" ex_identity" type =" .....">< / enum>
< ....>

现在我想使用XSLT文件创建一个新的结构化XML文件，如

< ....>
< number sort =" struct" type =" PartNumber"> 100< / struct>
< identity sort =" struct" type =" PartIdentity"> 1< / struct>
< ex_identity sort =" enum" type =" ........

我的问题是，我无法读出属性
name的值。所以我不能创建它的新元素。


为什么你不能？ < element name =" {@ name}" />不工作？


此外我需要为新元素创建其他属性，或者我可以轻松复制它们（？）。有没有人有任何解决方案？ PLZ帮我搜索了很长时间。

Hi, I′m searching a solution for the following problem:
I′ve a generated XML Document like

<....>
<struct name="number" type="PartNumber">100</struct>
<struct name="identity" type="PartIdentity">1</struct>
<enum name="ex_identity" type="....."></enum>
<....>

Now i want to use a XSLT file to create a new structured XML file like

<....>
<number sort="struct" type="PartNumber">100</struct>
<identity sort="struct" type="PartIdentity">1</struct>
<ex_identity sort="enum" type="........

My problem ist, that i couldn′t read out the value of the attribute
"name" so i can′t
create a new element of it.
Why couldn''t you? Does <element name="{@name}"/> not work?

Furthermore i need to create the other attributes for the new element or maybe i can copy them easily(?). Does
anybody have any solution? PLZ help me im searching for a long time.

< attribute name =" sort">< value-of select =" local-name（ ）" />< / attribute>


Soren

<attribute name="sort"><value-of select="local-name()"/></attribute>

Soren

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