grep или awk для извлечения xml из журнала на основе строки поиска

Prueba esto:

Max=`grep -c "^Printing" file.xml`

for count in `seq 1 $Max`
do
    sed -nr '/Printing/H;//,/ABC/G;s/\n(\n[^\n]*){'$count'}$//p'  file.xml | sed 's/Printing XML://' > $count.xml
done

Esto es lo que escribí, pero estoy seguro de que hay una forma más breve y más elegante de hacerlo.

searchstring=searchparam
filename=test.log
pattern1=ABC

linenums=($(grep -n "${searchstring}" ${filename} | awk -F":" '{print $1}'))
len=${#linenums[@]}

for (( i=0; i<${len}; i++ ));
do
  currentline=${linenums[$i]}
  relativeendlinearray=($(tail -n +${currentline} ${filename} | grep -n "${pattern1}" | awk -F":" '{print $1}'))
  actualendline=$(($currentline+${relativeendlinearray[0]}-1))

  index=$currentline
  while [ $index -ne 0 ]
        do
        found=`sed "${index}q;d" ${filename} | grep "${pattern1}"`
        if [ -n "$found" ]; then
            actualstartline=$index
            break;
        fi
        index=$[$index-1]
  done

  if [ -n "$found" ]; then
        echo ""
  else
        echo "Log break detected, content across multiple files"
  fi

  echo "Start Line" ${actualstartline}
  echo "Current Line" ${currentline}
  echo "End Line" ${actualendline}
  sed -n "${actualstartline},${actualendline}p" ${filename}
done