1

I have a text file like this:

first state
second state
third state

Getting the first word from every line isn't difficult, but the problem comes when adding the extra \n required to separate every word (selection) in dmenu, per its syntax:

echo -e "first\nsecond\nthird" | dmenu

I haven't been able to figure out how to add the separating \n. I've tried this:

state=$(awk '{for(i=1;i<=NF;i+=2)print $(i)'\n'}' text.txt)

But it doesn't work. I also tried this:

lol=$(grep -o "^\S*" states.txt | perl -ne 'print "$_"')

But same deal. Not sure what I'm doing wrong.

Tippitytopkek
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  • Could you please confirm first field one by one from each line to dmenu script? OR all first fields at a time you want to pass to dmenu? – RavinderSingh13 Jan 26 '21 at 10:58

4 Answers4

2

Your problem is in the AWK script. You need to identify each input line as a record. This way, you can control how each record in the output is separated via the ORS variable (output record separator). By default this separator is the newline, which should be good enough for your purpose.

Now to print the first word of every input record (each line in the input stream in this case), you just need to print the first field:

awk '{print $1}' textfile | dmenu

If you need the output to include the explicit \n string (not the control character), then you can just overwrite the ORS variable to fit your needs:

awk 'BEGIN{ORS="\\n"}{print $1}' textfile | dmenu
Jorge Bellon
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  • I tried the first method but for some reason dmenu needs them to be explicit. Your second example worked but only without escaping the newline! Not really sure why, but it's exactly what I needed. Thank you! – Tippitytopkek Jan 26 '21 at 20:58
2

This could be more easily done in while loop, could you please try following. This is simple, while is reading the file and during that its creating 2 variables 1st is named first and other is rest first contains first field which we are passing to dmenu later inside.

while read first rest
do
   dmenu "$first"
done < "Input_file"
RavinderSingh13
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0

Based on the text file example, the following should achieve what you require:

 awk '{ printf "%s\\n",$1 }' textfile | dmenu

Print the first space separated field of each line along with \n (\n needs to be escaped to stop it being interpreted by awk)

Raman Sailopal
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0

In your code

state=$(awk '{for(i=1;i<=NF;i+=2)print $(i)'\n'}' text.txt)

you attempted to use ' inside your awk code, however code is what between ' and first following ', therefore code is {for(i=1;i<=NF;i+=2)print $(i) and this does not work. You should use " for strings inside awk code.

If you want to merely get nth column cut will be enough in most cases, let states.txt content be

first state
second state
third state

then you can do:

cut -d ' ' -f 1 states.txt | dmenu

Explanation: treat space as delimiter (-d ' ') and get 1st column (-f 1)

(tested in cut (GNU coreutils) 8.30)

Daweo
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