Bash 脚本：使用 xmllint 将 XML 元素放入数组
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Bash script: Get XML elements into array using xmllint
提问by tzippy
This is a follow up question to this post.
这是这篇文章的后续问题 。
I want to end up with an array, containing all the <description>elements of the xml.
我想最终得到一个数组，其中包含<description>xml 的所有元素。
array[0] = "<![CDATA[A title for .... <br />]]>"
array[1] = "<![CDATA[A title for .... <br />]]>"
...
...
file.xml:
<item>
<description><![CDATA[A title for the URLs<br /><br />
http://www.foobar.com/foo/bar
<br />http://bar.com/foo
<br />http://myurl.com/foo
<br />http://desiredURL.com/files/ddd
<br />http://asdasd.com/onefile/g.html
<br />http://second.com/link
<br />]]></description>
</item>
</item>
<description> ...</description>
<item>
回答by édouard Lopez
A Bashsolution could be
一个Bash解决方案可能是
let itemsCount=$(xmllint --xpath 'count(//item/description)' /tmp/so.xml)
declare -a description=( )
for (( i=1; i <= $itemsCount; i++ )); do
description[$i]="$(xmllint --xpath '//item['$i']/description' /tmp/so.xml)"
done
echo ${description[@]}
Disclaimer
免责声明
Consider that bashmay not be the right tool. XSLT/XPath could give you direct access to the content of the descriptionelement as describe in previous answer. For instance:
考虑到这bash可能不是正确的工具。XSLT/XPath 可以让您直接访问description元素的内容，如先前答案中所述。例如：
xmllint --xpath '//item/description/text()' /tmp/so.xml
Return every <description>content
返回所有<description>内容
