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I want to delete a specific line of a file in bash.

What I am currently doing is to get the line number and pass it so sed to delete this line:

awk '/qr/{ print NR; exit }' test | sed -i "${1}d" test

The awk part works well, but in this state, the sed part deletes all the content of the file (named test).

However, when I do it without the variable :

sed -i '1d' test

It works fine.

What am I doing wrong ?

Itération 122442
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  • sed does not read $1 from a pipeline (if it gets a filename arg, it won't read stdin at all). The $1 is expanded by the shell even before either of the processes in the pipeline are executed. So the sed expression is just "d", and it deleted every line. – Paul_Pedant May 25 '20 at 08:36
  • Ok, good to know. But do I have to store the return of the awk in a variable in order to use it with sed ? @Paul_Pedant – Itération 122442 May 25 '20 at 08:37
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    What speaks against doing the job completely in `awk`, as in `awk '/qr/{if (!i++) next}1' test` (*assuming you only want to remove the first occurence*)? If you need "inplace" function, GNU `awk` > 4.1.0 supports the `-i inplace` command-line option for that. – AdminBee May 25 '20 at 08:45
  • For non-huge files, I store the whole file in an array X[NR], mark lines for deletion in D[NR], and write the lot back to FILENAME in the END block. That means you can do stuff like mark multiple lines, paragraphs containing any number of specific words, etc. So a generic solution, but it's not for everybody. – Paul_Pedant May 25 '20 at 08:46
  • I didn't know about "inplace". My way does not use tmp files, and I can abort it cleanly if no change is needed, or I find a data error. I need to check what inplace does if I exit (1) somewhere. – Paul_Pedant May 25 '20 at 08:52
  • @AdminBee What prevents me is the knowladge about it ^^. Also, I need inplace and but have gnu awk 4.0.2. Thanks for the tip though. – Itération 122442 May 25 '20 at 08:53
  • Ok, that's bad luck ... ;) – AdminBee May 25 '20 at 08:54
  • @FlorianCastelain can't you install a more current gawk version (we're currently on version 5.1.0), you're missing some very useful functionality (see https://www.gnu.org/software/gawk/manual/gawk.html#Feature-History) and a few bug fixes. – Ed Morton May 25 '20 at 17:18

4 Answers4

4

${1} is replaced by the first argument in the current context; it doesn’t read anything from awk’s output in your example.

If you want to keep both awk and sed, one way to go about this is to store the output of awk in a variable:

line=$(awk '/qr/{ print NR; exit }' test); sed -i "${line}d" test

but that won’t work well if the file doesn’t have any line containing “qr”.

A better approach might be to use sed only:

sed -i "/qr/d" test

but that will delete all matching lines, not just the first one.

The AWK (or rather, gawk, the GNU implementation, starting with version 4.1.0) equivalent of the above is

gawk -i inplace '!/qr/' test

or, replacing only the first instance,

gawk -i inplace '/qr/ && !i { i++; next; } 1' test
Stephen Kitt
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    That first command would completely wipe out your input file if it didn't contain `qr`, you'd need to make it `line=$(awk '/qr/{ print NR; f=1; exit } END{exit !f}' test) && sed -i "${line}d" test` if you were really going to do something like that. – Ed Morton May 25 '20 at 16:54
  • Right, that’s why there’s a “but” ;-). Thanks for the working version! – Stephen Kitt May 25 '20 at 17:05
  • Yeah at first I didn't see the `but` and then once I did I thought `won’t work well` was kinda understating the impact so decided to comment :-). You're welcome! – Ed Morton May 25 '20 at 17:06
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Given this input file:

$ cat file
foo
qr
bar

If you were really going to use that approach then it'd be (without -i so you can see the effect of the sed):

$ awk '/qr/{print NR; exit}' file | xargs -n 1 -I {} sed '{}d' file
foo
bar

but of course that's a pointless pipe to xargs+sed and what you should do instead is just:

awk '!f && /qr/{f=1; next} 1' file > tmp && mv tmp file

or with a recent version of gawk that supports it:

awk -i inplace '!f && /qr/{f=1; next} 1' file
Ed Morton
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0

Assuming you only want to remove the first occurence of the pattern, you can use awk entirely:

awk '/qr/{if (!i++) next}1' test

This will print all lines (1) except for the first line matching /qr/ (where i is still zero, and hence the next command will be issued before the "print" action can become effective).

If you need the "inplace" editing function of sed -i, and you have GNU awk > 4.1.0, you can make use of the "inplace" extension:

awk -i inplace '/qr/{if (!i++) next}1' test
AdminBee
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0

If you are looking to just delete the first matching line then

sed  "0,/qr/{/qr/d;}" file

Just drop -inline in after testing

This only processes the addresses from the first line to the first match with qr 

0,/qr/

and the deletes any line in that address range that contains qr

{/qr/d;}
bu5hman
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