使用 bash 将 XML 转换为竖线分隔的文件

Question

How can I remove the entry tag and convert this XML into a pipe-delimited file?

<entry><company>ABC</company><appname>XYZ</appname><appid>12345678</appid><updated>2014-04-29T20:58:00-07:00</updated><msgid>923605123</msgid><title>Crash</title><content type="text">Whenever you try to use the graph function.  I expect better from Schwab</content><version>4.1.3.6</version><rating>1</rating></entry>

Expected output format:

ABC|XYZ|12345678|2014-04-29T20:58:00-07:00|923605123|Crash|Whenever you try to use the graph function.  I expect better from Schwab|4.1.3.6|1|

Answer 1

Consider something akin to the following:

xmlstarlet sel -t -m '//entry' \
  -v ./company -o '|' \
  -v ./appname -o '|' \
  -v ./appid   -o '|' \
  -v ./content -n     \
  <test.xml

It would be possible to write a query which didn't call for spelling out each column in turn -- but writing it out is the better approach, as it ensures that column 3 in every line (in this case) always means appid, which otherwise isn't a guarantee that you have available.

Note that XMLStarlet, like many compliant parsers, requires a well-formed XML document -- meaning it the document being processed must have a single root element. If what you have is a file that contains a stream of documents (no root element in which the entries are contained), this can be faked; one ugly but functional way to do this follows: xmlstarlet ... < <(echo "<root>"; cat test.xml; echo "</root>") )

Answer 2

sed 's/<[^>]*>/|/g;s/||*/|/g' file1 > file2

Edited to remove ajacent "||" pairs

Answer 3

awk '$1 {printf s++ ? "|" $0 : $0}' RS='<[^>]+>'

• set Record Separator to a tag, example <entry>
• only print "lines" with a field, AKA dont print the tags
• if on the second "line" or more, print a | , otherwise just print the "line"

Result

ABC|XYZ|12345678|2014-04-29T20:58:00-07:00|923605123|Crash|Whenever you try to use the graph function.  I expect better from Schwab|4.1.3.6|1

Answer 4

With xidel :

xidel -s input.xml -e 'join(entry/*,"|")'