Parsing String with Grep

Parsing string with grep

If the line prefix is always the same, simply use sed and replace the prefix with an empty string:

sed 's/\[INFO\] Total Time: //'

Assuming that the time is always the last thing in a line after a colon, use the following regex (replace each line with everything after the colon):

sed 's/^.*: \(.*\)$/\1/'

Shell scripting using grep to split a string

Using substitution with sed:

echo $myVAR | sed -E  's/(.*)#{3}(.*)/\1/'
>>> firstword

echo $myVAR | sed -E  's/(.*)#{3}(.*)/\2/'
>>> secondword

# saving to variables
myFIRST=$(echo $myVAR | sed -E  's/(.*)#{3}(.*)/\1/')

mySECOND=$(echo $myVAR | sed -E  's/(.*)#{3}(.*)/\2/')

parsing strings using either grep,awk or sed

sed version:

sed -e 's/\([^ ]*\).*Job\(ID\)\? \([0-9]\+\).*/\3 \1/g'

or with extended regex as pointed out by @spasic:

sed -E 's/^(\S+).*Job(ID)? ([0-9]+).*/\3 \1/'

Parsing unstructured text file with grep

Assuming you have more than one "Summarized attack:" in your input file this may be what you're looking for:

$ cat tst.awk
/^Summarized attack:/ {
    prt()
    atk = ($3 ~ /^4/ ? $3 : 0)
    cnt = 0
}
atk { cnt++ }
END {
    prt()
    print "TOTAL", tot
}

function prt() {
    if ( atk ) {
        cnt -= 2
        print atk, cnt
    }
    tot += cnt
}

.

$ awk -f tst.awk file

Grep / search for a string in a csv, then parse that line and do another grep

$ awk -v key=KEY1 -F, '$1==key{f=1} ($1!~/^ *$/)&&($1!=key){f=0} f{print $3}' file
TRACKINGNUMBER1-1
TRACKINGNUMBER1-2
TRACKINGNUMBER1-3
TRACKINGNUMBER1-4
TRACKINGNUMBER1-5

glennjackman helpfully points out that by using a "smarter" value for FS the internal logic can be simpler.

awk -v key=KEY1 -F' *,' '$1==key{f=1} $1 && $1!=key{f=0} f{print $3}' file

-v key=KEY1 assign the value KEY1 to the awk variable key

-F' *,' assign the value *, (which is a regular expression) to the awk FS variable (controls field splitting)

$1==key{f=1} if the first key of the line is equal to the value of the key variable (KEY1) then assign the value 1 to the variable f (find our first desired key line)

$1 && $1!=key{f=0} if the first field has a truth-y value (in awk a non-zero, non-empty string) and the value of the first field is not equal to the value of the key variable assign the value 0 to the variable f (find the end of our keyless lines)

f{print $3} if the variable f has a truth-y value (remember non-zero, non-empty string) then print the third field of the line

Parsing only first regex match in a line with several matches

How about using a ^ start anchor and restricting character set used:

grep -o '^[A-Za-z]1[A-Za-z]*1'

See this Bash demo or Regex Pattern at regex101

If you expect more digits or other characters in between, go with this

grep -oP '^[A-Za-z]1.*?[A-Za-z]1'

The lazy matching requires perl compatible mode. For not at line start, go with this

grep -oP '^.*?\K[A-Za-z]1.*?[A-Za-z]1'

\K resets beginning of the reported match and is a PCRE feature as well.

parsing results in bash

Without a sample file, it is hard to write a working example. But I see, both values are from the same text file and the same line. Therefore I would use awk to do it:

$ cat text
service1;some_other_text;2.3.4
service2;just_some_text;55.66

$awk -F ";" '{printf "%s:%s\n",  $1, $3}' test 
service1:2.3.4

For a JSON file, it would be easier if you can use jg (e.g. apt-get install jg):

$ cat test.json
[
  {
    "name": "service1",
    "customfield_10090": "1.2.3"
  },
  {
    "name": "service2",
    "customfield_10090": "23.3.2"
  }
]

$jq '.[] | .name + ":" + .customfield_10090' test.json | sed 's/"//g'
service1:1.2.3
service2:23.3.2

The sed is necessary to eliminate the quotes.

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