在 XML/Xpath 中转义引号的特定问题 [英] Specific Problem with escaping quotes in XML/Xpath
问题描述
我对一些包含引号的入站消息有特定问题(例如)'s-Hertogenbosch，我需要转义第一个单引号，因为我无法在数据库..谁能告诉我解决方案如何忽略该报价?..或者简单地说，除了第一条和最后一条外，我如何在每条入站消息中动态转义不必要的引号.
I have specific problem with some inbound messages that contains quotes(for example) 's-Hertogenbosch and i need to escape the first single quote because i can't inject whole node-set in the DataBase..can anyone tell me solution how can i ignore that quote?..or briefly how at all can i dynamically escape unnecessary quotes in every inbound message except first and last one.
我尝试了 translate($InboundMessage, ' " ', "(blankspace)" ) 并且它工作正常但它删除了所有引号..我需要保留第一个和最后一个..例如我的入站消息是
I tryed with translate($InboundMessage, ' " ', "(blankspace)" ) and it worked fine but it removed all quotes..i need to keep first and last one..for example my Inbound message is
'去年我访问了's-Hertogenbosch'..那个单引号有害，因为在输出中我得到了'去年我访问了'而没有城市名称.
'Last year i visited 's-Hertogenbosch '..that single quote makes harm because on the output i got 'Last year i visited' without the name of the city.
解决方案?
问候.
推荐答案
试试这个表达式....
Try this expression....
<xsl:value-of select='replace($InboundMessage, "(.)'+(.)", "$1$2")'/>
或者，如果你想为双引号做...
Or, if you want to do it for double quotes...
<xsl:value-of select="replace($InboundMessage, '(.)"+(.)', '$1$2')"/>
如果您只能使用 XSLT1.0 和 XPATH 1.0，那么您必须进行一些字符串操作
If you can only use XSLT1.0 and XPATH 1.0, then you have to do some string manipluation
<xsl:value-of select='concat(substring($InboundMessage, 1, 1), translate(substring($InboundMessage, 2, string-length($InboundMessage) - 2), "'", ""), substring($InboundMessage, string-length($InboundMessage)))'/>
或者，根据您对构建 SQL 的评论，您可以替换所有撇号，并将外部撇号放回您构建的 SQL 字符串中
Alternatively, given your comment about building SQL, you could just replace all apostrophes, and put the outer apostrophes back in the SQL string you build
<xsl:value-of select='concat("INSERT INTO ", $Destination, " (", $Column, ") VALUES ('", translate($InboundMessage, "'", ""), "')")'/>
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