Linux shell scripting - Getting the sum of numbers within a String but also keeping the String with the sum number

Question

I have an xml file that has numbers within tags. Ex)

<REC_CNT>25000</REC_CNT>
<REC_CNT>25000</REC_CNT>
<REC_CNT>25000</REC_CNT>
<REC_CNT>11767</REC_CNT>

I need to get the sum of these numbers while at the same time the final output has to still have the tags around it.

<REC_CNT>86767</REC_CNT>

I've used awk already to remove the tags and add the numbers but that is not what I'm looking for. What I'm doing is a replace and if I only get the number then the replace will also replace every other number in that file that it is equal to, not just the REC_CNT... any input?

Shell is, quite simply the wrong tool for this. Could you explain why you don't want to consider alternative languages (such as Perl or Python)? — James Youngman
– James Youngman, Commented May 21, 2012 at 20:32
newcount=cat $pocfiles | grep '<REC_CNT>' | awk 'BEGIN {FS="<REC_CNT>"} {print substr($2,0)}' | awk 'BEGIN {FS="<"} {print substr($1,0)}' | awk '{SUM += $1} END {print SUM}' I ended up removing the tags, added the numbers, and reformatted my sed statement to only replace the first instance of the number: sed -i "0,/$oldcount/s//$newcount/" Thanks for the help! — user1408683
– user1408683, Commented May 22, 2012 at 15:34

Lars Kotthoff · Accepted Answer · 2012-05-21 20:37:12Z

3

awk -v FPAT='([A-Z_]+)|([0-9]+)' '{ sum += $2; fname = $1 } END { print "<" fname ">" sum "</" fname ">" }'

This assumes that your tags are uppercase letters or underscores. Adjust the regular expression as appropriate.

answered May 21, 2012 at 20:37

Lars Kotthoff

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1 Comment

user1408683 Over a year ago

thanks for the good effort in that reply.. what i ended up doing was just taking out the tags and then reformatting my sed to only replace the specific instance i want. Thanks again.

Mark Reed · Accepted Answer · 2012-05-21 21:09:05Z

0

You should use an actual XML processor instead of some regular-expression based text munging tool.

answered May 21, 2012 at 21:09

Mark Reed

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