Shell脚本提取递归xml标签

Question

I have an XML file of form:

...
<element1>
<element2>
<group1>
<tag1>value</tag1>
<tag2>value</tag2>
</group1>
<group1>
<tag1>value</tag1>
<tag2>value</tag2>
</group1>
<element2>
...

I used

sed -n '/\<group1\>/,\<\/group1>/p' filename

to extract all content of group1 tags, including all childs. This is exactly what I want.

<group1>
<tag1>value</tag1>
<tag2>value</tag2>
</group1>
<group1>
<tag1>value</tag1>
<tag2>value</tag2>
</group1>

However, if the input XML is of form

...
<element1>
<element2>
<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>
<element3>
<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>
...

And I tried to extract following content

<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>
<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>

The sed command above just returns:

<group2>
     <group2>value</group2>

It understands the stop pattern </group2> and do no more extraction. I'm quite confused here. Why doesn't it continue extracting the next <group2> , as in <group1> case. Is there any way to make it work with sed? and any other alternatives?

Answer 1

您可以像这样更改sed

sed -n '/\<group1\>/,/^<\/group1>/p' filename  | grep -v 'element3'

Answer 2

Far better to use XPath with a command line xpath interpreter, like xpath, xmlstarlet, my xidel or xmllint.

All group elements on the 3rd level:

/elememt1/*/group1

All group elements that do not contain a group2:

//group2[not(group2)]

Answer 3

Some like this?

awk '/^<group2>/,/^<\/group2>/' file
<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>
<group2>
     <group2>value</group2>
     <otherTag>value</otherTag>
</group2>

This works if there are different spacing on the tag, if all is adjusted to the left, it will not work