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bashsedawkgrepnawk

search (e.g. awk, grep, sed) for string, then look for X lines above and another string below

I need to be able to search for a string (lets use 4320101), print 20 lines above the string and print after this until it finds the string

For example:

Random text I do not want or blank line
16 Apr 2013 00:14:15
id="4320101"
</eventUpdate>
Random text I do not want or blank line

I just want the following result outputted to a file:

16 Apr 2013 00:14:15
id="4320101"
</eventUpdate>

There are multiple examples of these groups of text in a file that I want.

I tried using this below:

cat filename | grep "</eventUpdate>" -A 20 4320101 -B 100 > greptest.txt

But it only ever shows for 20 lines either side of the string.

Notes:
- the line number the text is on is inconsistent so I cannot go off these, hence why I am using -A 20.
- ideally I'd rather have it so when it searches after the string, it stops when it finds and then carries on searching.

Summary: find 4320101, output 20 lines above 4320101 (or one line of white space), and then output all lines below 4320101 up to

</eventUpdate>

Doing research I am unsure of how to get awk, nawk or sed to work in my favour to do this.

Solution

  • This might work for you (GNU sed):

    sed ':a;s/\n/&/20;tb;$!{N;ba};:b;/4320102/!D;:c;n;/<\/eventUpdate>/!bc' file

    EDIT:

    • :a;s/\n/&/20;tb;$!{N;ba}; this keeps a window of 20 lines in the pattern space (PS)
    • :b;/4320102!D; this moves the above window through the file until the pattern 4320102 is found.
    • :c;n;/<\/eventUpdate>/!bc the 20 line window is printed and any subsequent line until the pattern <\/eventUpdate> is found.