这个 XSLT 2.0 转换:
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="text()">
<xsl:variable name="vLines" select="tokenize(., '\n\r?')"/>
<xsl:for-each select="$vLines">
<xsl:value-of select=
"if(matches(.,':\s*(\d+(\.\d+)?)'))
then
(replace(., '^.*?:\s*(\d+(\.\d+)?).*$','$1'),'
')
else
if(matches(.,',\s*\d+(\.\d+)?'))
then
(replace(., '^.*?,\s*(\d+(\.\d+)?).*$','$1'),'
')
else
'null
'
"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
应用于以下 XML 文档时:
<t>OK - load average: 0.34,0.02,0.34
USERS OK - 0 users currently logged in,100,200
HTTP WARNING: HTTP/1.1 403 Forbidden
PING OK - Packet loss = 0%
DISK OK - free space: / 34285 MB (82% inode=94%):</t>
产生想要的正确结果:
0.34
100
null
null
null
单个 XPath 2.0 表达式解决方案:
for $vLine in
('OK - load average: 0.34,0.02,0.34'
, 'USERS OK - 0 users currently logged in,100,200'
,'HTTP WARNING: HTTP/1.1 403 Forbidden'
, 'PING OK - Packet loss = 0%'
, 'DISK OK - free space: / 34285 MB (82% inode=94%):'
)
return
if(matches($vLine,':\s*(\d+(\.\d+)?)'))
then
(replace($vLine, '^.*?:\s*(\d+(\.\d+)?).*$','$1'),'
')
else
if(matches($vLine,',\s*\d+(\.\d+)?'))
then
(replace($vLine, '^.*?,\s*(\d+(\.\d+)?).*$','$1'),'
')
else
'null
'