get Nth line in file after parsing another file

Question

I have one of my large file as

foo:43:sdfasd:daasf bar:51:werrwr:asdfa qux:34:werdfs:asdfa foo:234:dfasdf:dasf qux:345:dsfasd:erwe ...............

here 1st column foo, bar and qux etc. are file names. and 2nd column 43,51, 34 etc. are line numbers. I want to print Nth line(specified by 2nd column) for each file(specified in 1st column). How can I automate above in unix shell. Actually above file is generated while compiling and I want to print warning line in code.

-Thanks,

kev · Accepted Answer · 2012-04-13 17:07:35Z

2

while IFS=: read name line rest
do
    head -n $line $name | tail -1
done < input.txt

edited Apr 13, 2012 at 17:07

answered Apr 13, 2012 at 14:46

kev

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1 Comment

glenn jackman Over a year ago

If you only want the line in question: head -n $line $name | tail -1

vyegorov · Accepted Answer · 2012-04-13 20:43:07Z

2

while IFS=: read file line message; do
    echo "$file:$line - $message:"
    sed -n "${line}p" "$file"
done <yourfilehere

edited Apr 13, 2012 at 20:43

vyegorov

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answered Apr 13, 2012 at 16:48

Mark Reed

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3 Comments

glenn jackman Over a year ago

for efficiency: sed -n "${line}{p;q}"

vyegorov Over a year ago

It is a good practice to save IFS before changing it and restore to it's original state after it is no longer needed. Like: OFS=$IFS; ...; IFS=$OFS

Mark Reed Over a year ago

vyegorov: the above only changes it for the duration of the read command. Even inside the body of the loop, it's back to its default value.

Teja · Accepted Answer · 2012-04-13 14:45:05Z

0

awk 'NR==4 {print}' yourfilename

or

cat yourfilename | awk 'NR==4 {print}'

The above one will work for 4th line in your file.You can change the number as per your requirement.

answered Apr 13, 2012 at 14:45

Teja

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2 Comments

glenn jackman Over a year ago

to pass the line number: awk -v n=$lineno 'NR==n {print;exit}' file

Teja Over a year ago

If the file is getting processed in the middle of another then we should use -v option.But if its an explicit process then we donot require -v option.

glenn jackman · Accepted Answer · 2012-04-13 17:45:02Z

0

Just in awk, but probably worse performance than answers by @kev or @MarkReed. However it does process each file just once. Requires GNU awk

gawk -F: '
    BEGIN {OFS=FS}
    { 
        files[$1] = 1
        lines[$1] = lines[$1] " " $2
        msgs[$1, $2] = $3 
    }
    END {
        for (file in files) {
            split(lines[file], l, " ")
            n = asort(l)
            count = 0
            for (i=1; i<=n; i++) {
                while (++count <= l[i])
                    getline line < file
                print file, l[i], msgs[file, l[i]]
                print line
            }
            close(file)
        }
    }
'

edited Apr 13, 2012 at 17:45

answered Apr 13, 2012 at 17:21

glenn jackman

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Comments

potong · Accepted Answer · 2012-04-13 18:36:00Z

0

This might work for you:

sed 's/^\([^,]*\),\([^,]*\).*/sed -n "\2p" \1/' file |
sort -k4,4 | 
sed ':a;$!N;s/^\(.*\)\(".*\)\n.*"\(.*\)\2/\1;\3\2/;ta;P;D' |
sh

edited Apr 13, 2012 at 18:36

answered Apr 13, 2012 at 18:22

potong

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Comments

user unknown · Accepted Answer · 2012-04-13 18:38:39Z

0

 sed -nr '3{s/^([^:]*):([^:]*):.*$/\1 \2/;p}' namesNnumbers.txt 
 qux 34

-n no output by default,
-r regular expressions (simplifies using the parens)
in line 3 do {...;p} (print in the end)
s ubstitute foobarbaz with foo bar

So to work with the values:

fnUln=$(sed -nr '3{s/^([^:]*):([^:]*):.*$/\1 \2/;p}' namesNnumbers.txt)

fn=$(echo ${fnUln/ */})
ln=$(echo ${fnUln/* /})
sed -n "${ln}p" "$fn"

edited Apr 13, 2012 at 18:38

answered Apr 13, 2012 at 17:21

user unknown

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2 Comments

glenn jackman Over a year ago

The question is asking how to get line 34 out of file "qux" (and similarly for all the other lines)

user unknown Over a year ago

@glennjackman: I know. With "qux" and "34" it is trivial: sed -n "${lineno}p" "$filename". You think I should include this in my answer?

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