我需要从一个XML文件中的几个XML块中检索一个以上的值。如何使用xmllint做到这一点?
我注意到此解决方案(xml_grep get attribute from element)并尝试扩展它。不幸的是,到目前为止没有任何运气。
[
{
"operation": "shift",
"spec": {
"*": {
"orderNumber": "[&1].orderNumber",
"items": {
"*": {
"itemQuantity": "[&1].items[].itemQuantity",
"item": {
"external_id": "[&1].items[].external"
}
}
}
}
}
}
]
示例XML文件:
xmllint --xpath 'string(//identity/@name @placeofbirth @photo)' file.xml
想要输出
<eid>
<identity>
<name>Menten</name>
<firstname>Kasper</firstname>
<middlenames>Marie J</middlenames>
<nationality>Belg</nationality>
<placeofbirth>Sint-Truiden</placeofbirth>
<photo>base64-string</photo>
</identity>
<identity>
<name>Herbal</name>
<firstname>Jane</firstname>
<middlenames>Helena</middlenames>
<nationality>Frans</nationality>
<placeofbirth>Paris</placeofbirth>
<photo>notavailable</photo>
</identity>
</eid>
答案 0 :(得分:1)
一种实现方法是
# Read xml into variable
xmlStr=$(cat test.xml)
# Count identity nodes
nodeCount=$(echo "$xmlStr" | xmllint --xpath "count(//identity)" -)
# Iterate the nodeset by index
for i in $(seq 1 $nodeCount);do
echo "$xmlStr" | xmllint --xpath "concat((//identity)[$i]/name,', ',(//identity)[$i]/placeofbirth, ', ', (//identity)[$i]/photo)" - ; echo
done
结果:
Menten, Sint-Truiden, base64-string
Herbal, Paris, notavailable