How to return array with XPath?
I was wondering if XPath could return an array of value.
I have someting like this :
<Cities>
<City>Paris</City>
<City>Lyon</City>
<City>Marseille</City>
</Cities>
And I'd like to get an array 开发者_开发技巧of this form ['Paris', 'Lyon', 'Marseille'].
I am using the xpath UNIX utility.
Any idea ?
In addition to post-processing the output of:
xpath -q -e '//City/text()' inputfile
which is:
Paris
Lyon
Marseille
in any number of ways, including:
xpath -q -e '//City/text()' inputfile |
awk 'BEGIN {sq = "\047"; OFS = sq "," sq}
{a[$1] = NR}
END {
printf "[" sq;
for (i in a) {
printf "%s%s", d, i; d = OFS};
print sq "]"
}'
which gives the output:
['Lyon','Marseille','Paris']
you can use my modified version of xpath
. Since it's a simple Perl script, I modified it so you have more control over the formatting of the output. With my version, you can do:
xpath -q -b "[" -p "'" -i "," -s "'" -a "]"$'\n' -e '//City/text()'
to get the desired output. Here is the usage message:
Usage:
xpath [options] -e query [-e query...] [filename...]
If no filenames are given, supply XML on STDIN.
You must provide at least one query. Each supplementary
query is done in order, the previous query giving the
context of the next one.
Options:
-q quiet Only output the resulting PATH
-b before use before instead of nothing.
-p prefix use prefix instead of nothing.
-s suffix use suffix instead of linefeed.
-i infix use infix instead of nothing.
-a after use after instead of nothing.
You can download my version here.
This XPath expression:
/*/*/text()
selects exactly the three wanted text nodes.
You can use it with an array-like notation:
/*/*/text()[1]
selects the text-node
"Paris"
,
/*/*/text()[2]
selects the text-node
"Lyon"
,
/*/*/text()[3]
selects the text-node
"Marseille"
If you want just the string value of any of these text nodes, use (for example):
string(/*/*/text()[2])
The result of evaluating this XPath expression is the string
"Lyon"
i know that in .Net you can do it in xmlDocument
. xDoc.SelectNodes("//City");
it gives you a list of nodes and for each node you can do NODE.FirstChild.Value
to get the names.
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