Shell 脚本获取 XML 标记的值

Question

I have the XML data below.

<java:Parameter>
       <java:Name>flagCount</java:Name>
       <java:Value>Y</java:Value>
</java:Parameter>
<java:Parameter>
       <java:Name>returnCode</java:Name>
       <java:Value>001</java:Value>
</java:Parameter>

I want to get only Value of Name "returnCode" (Value is 001).

I try to do by command below, but does't work.

echo -e 'cat //java:Value/text()' | xmllint --shell test_check_main.xml

XPath error : Undefined namespace prefix
xmlXPathEval: evaluation failed
//java:Value/text(): no such node

How to implement by shell scripting?

Answer 1

First fix the xml:

xml=$( cat <<'EOF'
<java:Parameter>
   <java:Name>flagCount</java:Name>
   <java:Value>Y</java:Value>
</java:Parameter>
<java:Parameter>
   <java:Name>returnCode</java:Name>
   <java:Value>001</java:Value>
</java:Parameter>
EOF
)
xml2="<java:root xmlns:java=\"http://fakens.com\">""$xml""</java:root>"

Then you can get it with this xpath statement:

echo "$xml2" | xmllint --xpath '//*[local-name()="Parameter"]/*[local-name()="Name" and contains(text(),"returnCode")]/parent::*[namespace-uri()="http://fakens.com"]/*[local-name()="Value"]/text()' -

It is a bit round-about and annoying, but it is correct.

Answer 2

I hope this helps you :)

cat yourXML.xml | grep -A 1 'returnCode' | tail -1 | grep -oP '(?<=<java:Value>).*' | cut -d '<' -f1

There can be multiple ways of doing this but currently i came up with this one.

Answer 3

another way...

grep returnCode file -A1 | tail -1 | cut -d\\< -f2 | cut -d> -f2