编码特殊字符 [英] Encoding Special Characters

问题描述

我正在寻找一种方法来将基本ASCII字符集（1-127）中的所有字符更改为等效的扩展代码。代码表示为&＃[code];

例如，度数符号将变为&＃176;

有没有办法在序列化或编码过程中执行此操作？

谢谢！< br> Brian

解决方案

您可以实现自己的XmlWriter（可以通过扩展现有的XmlWriter实现来完成），也可以使用XSLT和样式表复制除了将编码更改为US-ASCII之外的所有内容：

< xsl：stylesheet

version = " 1.0"

xmlns： xsl = " http：//www.w3.org/1999/XSL/Transform" >

< xsl：output method = " xml" < font style ="color：red"> encoding = " US-ASCII" />

< xsl：template match = " @ * | node（）" >

< xsl：copy >

< xsl：apply-templates 选择 = " @ * | node（）" />

< / xsl：copy >

< / xsl：template >

< / xsl：stylesheet >

I'm looking for a way to change all characters not in the basic ASCII character set (1-127) to their equivalent extended code.  The codes are represented as &#[code]; 

For example, a degree symbol would become &#176;

Is there a way to do this as part of a serialization or encoding process?

Thanks!
Brian

解决方案

You can either implement your own XmlWriter (can be done by extending an existing XmlWriter implementation) that does that or you can use XSLT and a stylesheet copying everything but changing the encoding to US-ASCII:

<xsl:stylesheet

version="1.0"

xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output method="xml" encoding="US-ASCII"/>

<xsl:template match="@* | node()">

<xsl:copy>

<xsl:apply-templates select="@* | node()"/>

</xsl:copy>

</xsl:template>

</xsl:stylesheet>

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