CODEWITHSUNDEEP
bashshellawk
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Reputation: 2627

Right split using awk

I have a txt file filled with file sizes and full paths.

ex:

1 3788160 /nfs/extracted/UNENCRYPTED.bin.extracted/_0.extracted/vmlinux.img
2 5971656 /nfs/extracted/extracted/out_v2.1.3.zip.extracted/firmware.tar.extracted/_uImage.extracted/vmlinux.bin.30071
3 4202630 /nfs/extracted/v2.0.0.10/extracted/v2.0.0.10.zip.extracted/_install.img.extracted/package2/vmlinux.develop.avhdd.mars.nand.bin
4 117 /nfs/extracted/v1.0.0.14/extracted/1.0.0.14.zip.extracted

I'd like to print the second column and the filename (rsplit on "/") so that a[0] is the file name.

3788160 vmlinux.img
5971656 vmlinux.bin.30071
4202630 vmlinux.develop.avhdd.mars.nand.bin
117 1.0.0.14.zip.extracted

Is there an awk equilient to rsplit?

ex:

awk '{rsplit($2,a,"/"); print $1, a[0]}'

Upvotes: 3

Views: 992

Answers (4)

RavinderSingh13
RavinderSingh13

Reputation: 133600

Could you please try following, looks like this could be done with creating separate field delimiters tested and written with shown samples only. Written and tested in https://ideone.com/sk8zV1

awk -F'[ /]' '{print $2,$NF}' Input_file

Explanation: Simply making space and / as field delimiters and printing 2nd and last fields as per shown samples. Note NF denotes number of fields in current line so $NF will print last field's value here.

Upvotes: 1

markp-fuso
markp-fuso

Reputation: 34916

OP almost had it, just 2 smallish changes:

  • use the split() function (split() returns a count of the number of array entries which we'll catch and store in variable n)
  • access the last item of the array (instead of a[0])

Data set:

$ cat input.dat
1 3788160 /nfs/extracted/UNENCRYPTED.bin.extracted/_0.extracted/vmlinux.img
2 5971656 /nfs/extracted/extracted/out_v2.1.3.zip.extracted/firmware.tar.extracted/_uImage.extracted/vmlinux.bin.30071
3 4202630 /nfs/extracted/v2.0.0.10/extracted/v2.0.0.10.zip.extracted/_install.img.extracted/package2/vmlinux.develop.avhdd.mars.nand.bin
4 117 /nfs/extracted/v1.0.0.14/extracted/1.0.0.14.zip.extracted

One awk solution using the split() function:

$ awk '{ n=split($3,a,"/") ; print $2, a[n] }' input.dat
3788160 vmlinux.img
5971656 vmlinux.bin.30071
4202630 vmlinux.develop.avhdd.mars.nand.bin
117 1.0.0.14.zip.extracted

Upvotes: 2

anubhava
anubhava

Reputation: 785481

You may use this awk:

awk '{sub(/^.*\//, "", $3); print $2, $3}' file

3788160 vmlinux.img
5971656 vmlinux.bin.30071
4202630 vmlinux.develop.avhdd.mars.nand.bin
117 1.0.0.14.zip.extracted

Here:

  • sub(/^.*\//, "", $3): Removes everything upon start to last /
  • print $2, $3: Print 2nd and 3rd column

Upvotes: 3

Charles Duffy
Charles Duffy

Reputation: 295619

In native bash, no awk needed:

while read -r _ size name; do
  printf '%s\n' "$size ${name##*/}"
done

This throws away the first column by putting it into the variable $_; puts the second column into size, and puts the remaining content on each line into the variable name; then uses the parameter expansion ${name##*/} to strip everything up-to-and-including the last / from name on output.

Upvotes: 1

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