In my project, I have two files.
The content userid is :
6534
4524
4522
6635
The content userpwinfo.txt is:
nsgg315_RJ:x:4520:100::/home-gg/users/nsgg315_RJ:/bin/bash
nsgg316_ZJY:x:4521:100::/home-gg/users/nsgg316_ZJY:/bin/bash
nsgg317_CPA:x:4522:100::/home-gg/users/nsgg317_CPA:/bin/bash
nsgg318_ZRL:x:4523:100::/home-gg/users/nsgg318_ZRL:/bin/bash
nsgg319_YYM:x:4524:100::/home-gg/users/nsgg319_YYM:/bin/bash
Now I want to print the username which id is in userid. I writed a bash shell like:
for i in $(cat userid)
do
#username=`awk -F: '{if($3=="$i") print $1}' /root/userpwinfo.txt`
#username=`awk -F: '$3=="$i" {print $1}' /root/userpwinfo.txt`
#username=`awk -F: '{if($3~/$i/) print $1}' /root/userpwinfo.txt`
username=`awk -F: '{if($3==$i) print $1}' /root/userpwinfo.txt`
echo $username
done
But unlucky, it shows nothing. The correct result should be:
nsgg319_YYM
nsgg317_CPA
I have tried in command line:
awk -F: '{if($3==4524) print $1}' /root/userpwinfo.txt
It is OK
Maybe if($3==$i) is wrong in shell, Who can help me?
Your $i is the shell variable, but it's inside the quotation mark ' so awk will try to interpret it instead of the shell.
Try this:
username=`awk -F: '{if($3=='$i') print $1}' /root/userpwinfo.txt`
Note that the $i is between ' marks, meaning it's outside of the block that will be interpreted by awk, meaning it should be interpreted by the shell.
Also note that if you have an empty line in the input file, your awk command would be if($3==) which is invalid and will yield an error.
I'd like to comment also that awk is meant to have a filter and an execution block. You shouldn't need to write an if inside a block, unless you want something unusual. Meaning, your command would be more appropriately written as:
username=`awk -F: '($3=='$i'){print $1}' /root/userpwinfo.txt`
Note that even this is not a very good solution, but you already have much to think about with only these changes. When you're more familiar with awk or getting more professional, come back and check the comments. ;)
If username is what you needed using the 2 files, you could try
$ cat userpwinfo.txt
nsgg315_RJ:x:4520:100::/home-gg/users/nsgg315_RJ:/bin/bash
nsgg316_ZJY:x:4521:100::/home-gg/users/nsgg316_ZJY:/bin/bash
nsgg317_CPA:x:4522:100::/home-gg/users/nsgg317_CPA:/bin/bash
nsgg318_ZRL:x:4523:100::/home-gg/users/nsgg318_ZRL:/bin/bash
nsgg319_YYM:x:4524:100::/home-gg/users/nsgg319_YYM:/bin/bash
$ cat userid.txt
6534
4524
4522
6635
$ awk -F":" ' { if( NR==FNR ) { a[$3]=$1; next } ; if(a[$1]) print a[$1] }' userpwinfo.txt userid.txt
nsgg319_YYM
nsgg317_CPA
Related
The below awk command (copied and pasted from stackoverflow) works fine from the command line but doesnt print anything when aliased
awk '/WORD/ {print $3}' log.log | awk 'BEGIN{c=0} length($0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'
alias getperc="awk '/WORD/ {print \$3}' log.log | awk 'BEGIN{c=0} length(\$0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'"
I am fairly new to using bash. What am I missing here?
Don't use aliases. They require an additional layer of quoting, which is troublesome (as here), and they prevent you from being able to usefully parameterize or add conditional logic to your code.
A simple transliteration to a function is:
getperc() { awk '/WORD/ {print $3}' log.log | awk 'BEGIN{c=0} length($0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'; }
A slightly more capable one, which will still use log.log by default, but which will also let you provide an alternate input file name (as in getperc alternate.log) or pipe to your function (as in cat alternate.log | getperc):
getperc() {
[[ -t 0 || $1 ]] || set -- - # use "-" (stdin) as input file if not a TTY
# ...this will let you pipe to your function.
awk '/WORD/ {print $3}' "${1:-log.log}" | awk 'BEGIN{c=0} length($0){a[c]=$0;c++}END{p5=(c/100*5); p5=p5%1?int(p5)+1:p5; print a[c-p5-1]}'
}
I think there is confusion by bash regarding $3 and $0 it thinks they are argument of the alias. you can verify this by
try this in bash
alias ech="echo {print \$3}"
it will print just
{print }
but now try
alias ech="echo {print \$\3}"
it will print what you expected
{print $3}
Let me know if this solves your problem
I have 88 folders, each of which contains the file "pair.'numbers'." (pair.3472, pair.7829 and so on). I need to treat the files with awk to extract the second column, but I need to save the numbers. If I try:
#!/bin/bash
for i in {1..88}; do
awk '{print $2}' ~/Documents/attempt.$i/pair* > ~/Results/pred.pair*
done
It doesn't save the numbers, but gives only one file: pred.pair*
Thanks for any tips.
You don't need a loop (and see https://unix.stackexchange.com/questions/169716/why-is-using-a-shell-loop-to-process-text-considered-bad-practice for why that's a Good Thing):
awk '
FNR==1 { close(out); out=FILENAME; sub(/\/Documents.*\//,"/Results/pred.",out) }
{ print $2 > out }
' ~/Documents/attempt.{1..88}/pair*
#!/bin/bash
for i in {1..88}; do
awk '{fname=FILENAME;sub(".*/", "", fname);print $2 > ("~/Results/pred."fname)}' ~/Documents/attempt.$i/pair*
done
Use AWK build in variable FILENAME. We need to get the basename fname from FILENAME. Then redirect $2 value to "~/Results/pred."fname
There are several ways to do it: awk has a FILENAME variable and you can redirect the output from within your awk script to a manipulated string which is based on FILENAME.
Or you can do it with bash
for i in {1..88}; do
to_be_processed_fname=$(ls ~/Documents/attempt.$i/pair*)
extension="${to_be_processed_fname/*./}"
awk '{print $2}' "${to_be_processed_fname}" > "$HOME/Results/pred.${extension}"
done
Now the above of course fails if you have more than one pair* files within the same directory. But I'm leaving that to you.
I'm trying to write a function that looks for a last name (from user input) in the second column and returns the line to the user. Here is what I've tried (and is not working). Most importantly, when I run this with the last name in place of $input_name at the command line, it works then. As you can see, the echo to confirm that the user input was read properly. What am I missing?
60 last_search () {
61 echo "What last name are you looking for?"
62 read input_name
63 echo "$input_name"
64 awk -F':' '$2 ~ /$input_name/{print $0}' temp_phonebook
65 }
Use -v name=value to pass a shell variable to awk
Shell variable won't be expanded in single quote
print $0 is default action so take it out.
You can use:
awk -F: -v input_name="$input_name" '$2 ~ input_name' temp_phonebook
Your awk command is in single ticks, so $input_name never gets expanded.
Try something like this:
awk -F':' '$2 ~ /'"$input_name"'/{print $0}' temp_phonebook
Alternately, supply the value in advance using -v name=value:
awk -F':' -v search="${input_name}" '$2 ~ search {print $0}' temp_phonebook
okay, I figured it out! I need single quotes around the variable '$input_name'
How do print my full name with the following?
awk -F: '($1==U){print $5}' U=$LOGNAME /etc/passwd
Per example, but with the echo command with some words on sides:
For example,
Hello Diogo Saraiva, happy to see you again
Where Diogo Saraiva is my full name which I have in Ubuntu records.
I tried some things, but I have not accomplished that script...
Another thing: Why, when I issue awk -F: '($1==U){print $5}' U=$LOGNAME /etc/passwd, is Diogo Saraiva,,, shown instead of Diogo Saraiva?
This happens too with:
grep $USER /etc/passwd | awk 'BEGIN { FS=":" } { print $5 }'
I need for my script to declare a variable, "like", with the command, and then echo that variable "like" various times along my script.
To output the string you want you can use this command.
awk -F: -v U="$LOGNAME" '$1==U{print "Hello " $5 ", happy to see you again"}' /etc/passwd
If awk -F: -v U="$LOGNAME" '$1==U{print $5}' /etc/passwd is outputting Diogo Saraiva,,, then that is what is in your /etc/passwd file for that field.
To save the output from a command in a variable just use:
var=$(command)
And remember to quote the variable when you use it:
echo "Hello $var, happy to see you again"
I have a line like:
one:two:three:four:five:six seven:eight
and I want to use awk to get $1 to be one and $2 to be two:three:four:five:six seven:eight
I know I can get it by doing sed before. That is to change the first occurrence of : with sed then awk it using the new delimiter.
However replacing the delimiter with a new one would not help me since I can not guarantee that the new delimiter will not already be somewhere in the text.
I want to know if there is an option to get awk to behave this way
So something like:
awk -F: '{print $1,$2}'
will print:
one two:three:four:five:six seven:eight
I will also want to do some manipulations on $1 and $2 so I don't want just to substitute the first occurrence of :.
Without any substitutions
echo "one:two:three:four:five" | awk -F: '{ st = index($0,":");print $1 " " substr($0,st+1)}'
The index command finds the first occurance of the ":" in the whole string, so in this case the variable st would be set to 4. I then use substr function to grab all the rest of the string from starting from position st+1, if no end number supplied it'll go to the end of the string. The output being
one two:three:four:five
If you want to do further processing you could always set the string to a variable for further processing.
rem = substr($0,st+1)
Note this was tested on Solaris AWK but I can't see any reason why this shouldn't work on other flavours.
Some like this?
echo "one:two:three:four:five:six" | awk '{sub(/:/," ")}1'
one two:three:four:five:six
This replaces the first : to space.
You can then later get it into $1, $2
echo "one:two:three:four:five:six" | awk '{sub(/:/," ")}1' | awk '{print $1,$2}'
one two:three:four:five:six
Or in same awk, so even with substitution, you get $1 and $2 the way you like
echo "one:two:three:four:five:six" | awk '{sub(/:/," ");$1=$1;print $1,$2}'
one two:three:four:five:six
EDIT:
Using a different separator you can get first one as filed $1 and rest in $2 like this:
echo "one:two:three:four:five:six seven:eight" | awk -F\| '{sub(/:/,"|");$1=$1;print "$1="$1 "\n$2="$2}'
$1=one
$2=two:three:four:five:six seven:eight
Unique separator
echo "one:two:three:four:five:six seven:eight" | awk -F"#;#." '{sub(/:/,"#;#.");$1=$1;print "$1="$1 "\n$2="$2}'
$1=one
$2=two:three:four:five:six seven:eight
The closest you can get with is with GNU awk's FPAT:
$ awk '{print $1}' FPAT='(^[^:]+)|(:.*)' file
one
$ awk '{print $2}' FPAT='(^[^:]+)|(:.*)' file
:two:three:four:five:six seven:eight
But $2 will include the leading delimiter but you could use substr to fix that:
$ awk '{print substr($2,2)}' FPAT='(^[^:]+)|(:.*)' file
two:three:four:five:six seven:eight
So putting it all together:
$ awk '{print $1, substr($2,2)}' FPAT='(^[^:]+)|(:.*)' file
one two:three:four:five:six seven:eight
Storing the results of the substr back in $2 will allow further processing on $2 without the leading delimiter:
$ awk '{$2=substr($2,2); print $1,$2}' FPAT='(^[^:]+)|(:.*)' file
one two:three:four:five:six seven:eight
A solution that should work with mawk 1.3.3:
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $1}' FS='\0'
one
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $2}' FS='\0'
two:three:four five:six:seven
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $1,$2}' FS='\0'
one two:three:four five:six:seven
Just throwing this on here as a solution I came up with where I wanted to split the first two columns on : but keep the rest of the line intact.
Comments inline.
echo "a:b:c:d::e" | \
awk '{
split($0,f,":"); # split $0 into array of fields `f`
sub(/^([^:]+:){2}/,"",$0); # remove first two "fields" from `$0`
print f[1],f[2],$0 # print first two elements of `f` and edited `$0`
}'
Returns:
a b c:d::e
In my input I didn't have to worry about the first two fields containing escaped :, if that was a requirement, this solution wouldn't work as expected.
Amended to match the original requirements:
echo "a:b:c:d::e" | \
awk '{
split($0,f,":");
sub(/^([^:]+:)/,"",$0);
print f[1],$0
}'
Returns:
a b:c:d::e