Creating directories from list preserving whitespaces - bash

I have list of names in a file that I need to create directories from. The list looks like
Ada Lovelace
Jean Bartik
Leah Culver
I need the folders to be the exact same, preserving the whitespace(s). But with
awk '{print $0}' myfile | xargs mkdir
I create separate folders for each word
Ada
Lovelace
Jean
Bartik
Leah
Culver
Same happens with
awk '{print $1 " " $2}' myfile | xargs mkdir
Where is the error?

Using gnu xargs you can use -d option to set delimiter as \n only. This way you can avoid awk also.
xargs -d '\n' mkdir -p < file
If you don't have gnu xargs then you can use tr to convert all \n to \0 first:
tr '\n' '\0' < file | xargs -0 mkdir

#birgit:try: Completely based on your sample Input_file provided.
awk -vs1="\"" 'BEGIN{printf "mkdir ";}{printf("%s%s%s ",s1,$0,s1);} END{print ""}' Input_file | sh

awk '{ system ( sprintf( "mkdir \"%s\"", $0)) }' YourFile
# OR
awk '{ print"mkdir "\"" $0 "\"" | "/bin/sh" }' YourFile
# OR for 1 subshell
awk '{ Cmd = sprintf( "%s%smkdir \"%s\"", Cmd, (NR==1?"":"\n"), $0) } END { system ( Cmd ) }' YourFile
Last version is better due to creation of only 1 subshell.
If there are a huge amount of folder (shell parameter limitation), you could loop and create smaller command several times

Related

AWK -F with print all but last record

/Home/in/test_file.txt
echo /Home/in/test_file.txt | awk -F'/' '{ print $2,$3 }'
Gives the result as:
Home in
But I need /Home/in/ as the result .I have to get all except test_file.txt
How to achieve this?
$ echo '/Home/in/test_file.txt' | awk '{sub("/[^/]+$","")} 1'
/Home/in
$ echo '/Home/in/test_file.txt' | awk '{sub("[^/]+$","")} 1'
/Home/in/
$ echo '/Home/in/test_file.txt' | sed 's:/[^/]*$::'
/Home/in
$ echo '/Home/in/test_file.txt' | sed 's:[^/]*$::'
/Home/in/
$ dirname '/Home/in/test_file.txt'
/Home/in
Your attempt awk -F'/' '{ print $2,$3 }' didn't do what you wanted as -F'/' is telling awk to split the input into fields at every / and then print $2,$3 is telling awk to print the 2nd and 3rd fields separated by a blank char (the default value for OFS). You could do:
$ echo '/Home/in/test_file.txt' | awk 'BEGIN{FS=OFS="/"} { print "",$2,$3,"" }'
/Home/in/
to get the expected output but it'd be the wrong approach since it's removing the field you don't want AND removing the input separators AND then adding new output separators which happen to the have the same value as the input separators rather than simply removing the field you don't want like the other solutions above do.
echo /Home/in/test_file.txt | awk -F'/[^/]*$' '{ print $1 }'
..will print the everything but the trailing slash
There are several ways to achieve this:
Using dirname:
$ dirname /home/in/test_file.txt
/home/in
Using Shell substitution:
$ var="/home/in/test_file.txt"
$ echo "${var%/*}"
/home/in
Using sed: (See Ed Morton)
Using AWK:
$ echo "/home/in/test_file.txt" | awk -F'/' '{OFS=FS;$NF=""}1'
/home/in/
Remark: all these work since you can't have a filename with a forward slash (Is it possible to use "/" in a filename?)
Note: all but dirname will fail if you just have a single file_name without a path. While dirname foo will return ./ all others will return foo
awk behaves as it should.
When you define slash / as a separator, the fields in your expression become the content between the separators.
If you need the separator to be printed as well, you need to do it explicitly, like:
echo /Home/in/test_file.txt | awk -F'/' '{ printf "%s/%s/",$2,$3 }'
replace your last field with an empty string and
put the slash back in as the (builtin) Output Field Separator (OFS)
echo /Home/in/test_file.txt | awk -F'/' -vOFS='/' '{$NF="";print}

Writing an AWK instruction in a bash script

In a bash script, I need to do this:
cat<<EOF> ins.exe
grep 'pattern' file | awk '{print $2}' > results
EOF
The problem is that $2 is interpreted as a variable and the file ins.exe ends up containing
"grep 'pattern' file | awk '{print }' > results", without the $2.
I've tried using
echo "grep 'pattern' file | awk '{print $2}' > results" >> ins.exe
But it's the same problem.
How can I fix this?
Just escape the $:
cat<<EOF> ins.exe
awk '/pattern/ { print \$2 }' file > results
EOF
No need to pipe grep to awk, by the way.
With bash, you have another option as well, which is to use <<'EOF'. This means that no expansions will occur within the string.

How to awk -f everything after the last \

I have this where it could be one \ or multiple
C:\folder\file.log
C:\folder\folder\file.log
C:\folder\folder\folder\file.log
I want to get this
file.log
This works but its static with print $.
cat C:\folder\file.log | awk -F "\\" "{print $3}"
cat C:\folder\folder\file.log | awk -F "\\" "{print $4}
cat C:\folder\folder\folder\file.log | awk -F "\\" "{print $5}
How can i awk and always grab the data after the last \
You need the $NF special variable, which gives you the number of fields in your input.
echo C:\folder\file.log | awk -F "\\" "{print $NF}"
with grep:
grep -o '[^\\]*$' file
If you have awk, do you also have "basename"?
and as pointed out above, windows has similar capabilities built in.

Awk: Drop last record separator in one-liner

I have a simple command (part of a bash script) that I'm piping through awk but can't seem to suppress the final record separator without then piping to sed. (Yes, I have many choices and mine is sed.) Is there a simpler way without needing the last pipe?
dolls = $(egrep -o 'alpha|echo|november|sierra|victor|whiskey' /etc/passwd \
| uniq | awk '{IRS="\n"; ORS=","; print}'| sed s/,$//);
Without the sed, this produces output like echo,sierra,victor, and I'm just trying to drop the last comma.
You don't need awk, try:
egrep -o ....uniq|paste -d, -s
Here is another example:
kent$ echo "a
b
c"|paste -d, -s
a,b,c
Also I think your chained command could be simplified. awk could do all things in an one-liner.
Instead of egrep, uniq, awk, sed etc, all this can be done in one single awk command:
awk -F":" '!($1 in a){l=l $1 ","; a[$1]} END{sub(/,$/, "", l); print l}' /etc/password
Here is a small and quite straightforward one-liner in awk that suppresses the final record separator:
echo -e "alpha\necho\nnovember" | awk 'y {print s} {s=$0;y=1} END {ORS=""; print s}' ORS=","
Gives:
alpha,echo,november
So, your example becomes:
dolls = $(egrep -o 'alpha|echo|november|sierra|victor|whiskey' /etc/passwd | uniq | awk 'y {print s} {s=$0;y=1} END {ORS=""; print s}' ORS=",");
The benefit of using awk over paste or tr is that this also works with a multi-character ORS.
Since you tagged it bash here is one way of doing it:
#!/bin/bash
# Read the /etc/passwd file in to an array called names
while IFS=':' read -r name _; do
names+=("$name");
done < /etc/passwd
# Assign the content of the array to a variable
dolls=$( IFS=, ; echo "${names[*]}")
# Display the value of the variable
echo "$dolls"
echo "a
b
c" |
mawk 'NF-= _==$NF' FS='\n' OFS=, RS=
a,b,c

Bash: "xargs cat", adding newlines after each file

I'm using a few commands to cat a few files, like this:
cat somefile | grep example | awk -F '"' '{ print $2 }' | xargs cat
It nearly works, but my issue is that I'd like to add a newline after each file.
Can this be done in a one liner?
(surely I can create a new script or a function that does cat and then echo -n but I was wondering if this could be solved in another way)
cat somefile | grep example | awk -F '"' '{ print $2 }' | while read file; do cat $file; echo ""; done
Using GNU Parallel http://www.gnu.org/software/parallel/ it may be even faster (depending on your system):
cat somefile | grep example | awk -F '"' '{ print $2 }' | parallel "cat {}; echo"
awk -F '"' '/example/{ system("cat " $2 };printf "\n"}' somefile

Resources