Extract sub string from a variable ( using pattern match may be ) using bash unix?

Question

I have a value in a variable as :

partition_column='| PARTITIONED BY ( | | `part_col1` int, | | `part_col2` int) | | ROW FORMAT SERDE |'

I want to extract values between PARTITIONED BY and ROW FORMAT SERDE , in above case its part_col1 and part_col2

Desired output:
part_col1 part_col2

I have tried many commands nothing seems to be working :

result=$(echo $par_col | sed -nr '/`/p'|  cut -d '`' -f 2|xargs -n 1 echo -n "")

could you please correct above command , or suggest something else?

Please tag your question using the bash tag, if you are interested in only a bash solution. Further, your assignment statement partition_column=.... is incorrect (did you ever try to copy and paste it to the command line), since it would interpret part_col1 as the name of a command, and tries to run it. — user1934428
– user1934428, Commented Apr 22, 2020 at 7:10

user1934428 · Accepted Answer · 2020-04-22 07:15:25Z

1

Assuming that you have installed GNU cut, the following would work in bash:

 partition_column='| PARTITIONED BY ( | | `part_col1` int, | | `part_col2` int) | | ROW FORMAT SERDE |'
 result=$(cut -d '`' -f 2,4 --output-delimiter=' ' <<<"$partition_column")

This takes the advantage, that the delimiters are really not the keywords you posted, but backquote characters. The --output-delimiter is necessary, because otherwise the fields would also be separated by a backquote in the output.

answered Apr 22, 2020 at 7:15

user1934428

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KamilCuk · Accepted Answer · 2020-04-22 07:30:56Z

partition_column='| PARTITIONED BY ( | | `part_col1` int, | | `part_col2` int) | | ROW FORMAT SERDE |'
# extract everything between the patterns
<<<"$partition_column" sed 's/.*PARTITIONED BY\(.*\)ROW FORMAT SERDE.*/\1/' |
  # replace spaces for newlines
  tr ' ' '\n' |
  # filter only lines starting with \`
  grep '^`' |
  # remove the \`
  sed 's/`//g' |
  # join lines using a space
  paste -sd ' '

But it's possible with only sed, just use a global mask to first extract the chars within `, then replace them:

sed 's/.*PARTITIONED BY\(.*\)ROW FORMAT SERDE.*/\1/; s/[^`]*`\([^`]*\)`[^`]*/\1`/g; s/`/ /g; s/ $//;'

Or with sed you can loop with t until all the replacements are removed/shifted:

<<<"$partition_column" sed 's/.*PARTITIONED BY\(.*\)ROW FORMAT SERDE.*/\1/;
   # add a newline on the end
  s/$/\n/;
  :a;
     # find something within ` and move it behind the newline
     # remove everything in fron ` and after ` that is not a `
     s/[^`]*`\([^`]*\)`[^`]*\([^\n]*\n.*\)/\2 \1/;
  # loop until the `s` command above does something
  ta;
  # remove everything in front the newline and the space
  s/.*\n //;'

Pierre François · Accepted Answer · 2020-04-22 08:35:22Z

0

If you want a pure bash solution, try:

#!/bin/bash

partition_column='| PARTITIONED BY ( | | `part_col1` int, | | `part_col2` int) | | ROW FORMAT SERDE |'

left="${partition_column#*\`}"                  # remove everything until first `
target1="${left%%\`*}"                          # remove everything from first `
right="${partition_column%\`*}"                 # remove everything from last `
target2="${right##*\`}"                         # remove everything until last `

echo "$target1" "$target2"

edited Apr 22, 2020 at 8:35

answered Apr 22, 2020 at 8:27

Pierre François

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