I was basically trying to compare two files and as part of that I assigned the cksum of the file to a variable . But when I try to compare it, it did not work. I realized that when I tried to read the variable nothing gets printed out
The below commands worked just fine
s.joseph#VA-S-JOSEPH-900 /cygdrive/c/users/Anuprita
$ test=`cksum interface2 | awk -F" " '{ print $1 }'`
s.joseph#VA-S-JOSEPH-900 /cygdrive/c/users/Anuprita
$ echo "$test"
3021988741
But when these are part of a script and I try to echo $var, nothing gets printed
$ for i in `ls interface*`;
do chksum1=`cksum $i | awk -F" " '{ print "'$1'" }'`;
echo "$chksum1";
done
s.joseph#VA-S-JOSEPH-900 /cygdrive/c/users/Anuprita
$
I am using bash shell
Without assigning it to any variable, the output is as shown below
for i in interface*; do echo "interface=\"$i\""; cksum "$i"; done
interface="interface11"
4113442291 111 interface11
interface="interface17"
1275738681 111 interface17
interface="interface2"
3021988741 186 interface2
Looks like it is an issue only with bash on cygwin. The script seems to be working just fine on unix
for i in ls interface*; do chksum1=cksum $i | awk -F" " '{ print $1 }'; echo $i, $chksum1; done
interface1, 4294967295
interface2, 4294967295
Try this;
for i in ls interface*; do echo "interface=$i"; chksum1=$(cksum $i | awk -F" " '{ print "'$1'" }'); echo "$chksum1"; done
I like adding the echo statement to verify your getting what you think with the ls statement and the variable assignment should use $(cmd) or `cmd`
Cheers
What you have in your 2nd script:
print "'$1'"
is a completely different statement from what you have in your first one:
print $1
Think about it and ask yourself why you changed it and what it is you're trying to achieve. Also man awk and see g at http://cfajohnson.com/shell/cus-faq-2.html#Q24 for what print "'$1'" does.
Best I can tell without and provided sample input your script should be written:
for i in interface*; do chksum1=$(cksum "$i" | awk '{ print $1 }'); echo "$chksum1"; done
Related
I have a file that looks like:
>ref_frame=1
TPGIRYQYNVLPQGWKGSPAIFQSSMTKILEPFRKQNPDIVIYQYMDDLYVGSD
>ref_frame=2
HQGLDISTMCFHRDGKDHQQYSKVA*QKS*SLLENKIQT*LSINTWMICM*DLT
>ref_frame=3
TRD*ISVQCASTGMERITSNIPK*HDKNLRAF*KTKSRHSYLSIHG*FVCRI*
>test_3_2960_3_frame=1
TPGIRYQYNVLPQGWKGSPAIFQSSMTKILEPSRKQNPDIVIYQYMDDLYVGSD
I want to assign a bash variable so that echo $variable gives test_3_2960
The line/row that I want to assign the variable to will always be line 7. How can I accomplish this using bash?
so far I have:
variable=`cat file.txt | awk 'NR==7'`
echo $variable = >test_3_2960_3_frame=1
Using sed
$ variable=$(sed -En '7s/>(([^_]*_){2}[0-9]+).*/\1/p' input_file)
$ echo "$variable"
test_3_2960
No pipes needed here...
$: variable=$(awk -F'[>_]' 'NR==7{ OFS="_"; print $2, $3, $4; exit; }' file)
$: echo $variable
test_3_2960
-F is using either > or _ as field separators, so your data starts in field 2.
OFS="_" sets the Output Field Separator, but you could also just use "_" instead of commas.
exit keeps it from wasting time bothering to read beyond line 7.
If you wish to continue with awk
$ variable=$(awk 'NR==7' file.txt | awk -F "[>_]" '{print $2"_"$3"_"$4}')
$ echo $variable
test_3_2960
I have the following single line in my bash script:
echo "foo" | awk -F"=" '{char=system("echo $1 | cut -c1");}{print "this is the result: "$char;}' >> output.txt
I want to print the first letter of "foo" using awk, such that I would get:
this is the result: f
in my output file, but instead, I get:
this is the result: foo
What am i doing wrong?
Thanks
No, this is not the way system command works inside awk.
What's happening in OP's code:
You are giving a shell command in system which is good(for some cases) but there is a problem in this one that you should give it like system("echo " $0" | cut -c1") to get its first character AND you need NOT to have a variable etc to save its value and print it in awk.
You are trying to save its result to a variable but it will not have its value(system command's value) but its status. It doesn't work like shell style in awk in here.
So your variable named char will have 0 value(which is a success status from system command) and when you are printing $char it is printing whole line(because in awk: print $0 means print whole line).
You could do this in a single awk by doing:
echo "foo" | awk '{print substr($0,1,1)}'
OR with GNU awk specifically:
echo "foo" | awk 'BEGIN{FS=""} {print $1}'
you're not using much of awk, same can be done with printf
$ echo "foo" | xargs printf "this is the result: %.1s\n"
this is the result: f
or, directly
$ printf "this is the result: %.1s\n" foo
this is the result: f
I am trying to get the total disk usage of my machine. Below is the script code:
#!/bin/sh
totalUsage=0
diskUse(){
df -H | grep -vE '^Filesystem|cdrom' | awk '{ print $5 " " $1 }' | while read output;
do
diskUsage=$(echo $output | awk '{ print $1}' | cut -d'%' -f1 )
totalUsage=$((totalUsage+diskUsage))
done
}
diskUse
echo $totalUsage
While totalUsage is a global variable, I have tried to sum the individual disk usage to totalUsage in the line:
totalUsage=$((totalUsage+diskUsage))
An echo of totalUsage between do and done shows the correct value,
but when I try to echo it after my call diskUse, it stills prints a 0
Can you please help me, what is wrong here?
The variable totalUsage in a sub-shell doesn't change the value in the parent shell.
Since you tagged bash, you can use here string to modify your loop:
#!/bin/bash
totalUsage=0
diskUse(){
while read output;
do
diskUsage=$(echo $output | awk '{ print $1}' | cut -d'%' -f1 )
totalUsage=$((totalUsage+diskUsage))
done <<<"$(df -H | grep -vE '^Filesystem|cdrom' | awk '{ print $5 " " $1 }')"
}
diskUse
echo $totalUsage
I suggest to insert
shopt -s lastpipe
as new line after
#!/bin/bash
From man bash:
lastpipe: If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
This is what my script looks like essentially
......
rowNum=$(awk '{print NF}' temp)
i=1
while [ $i -lt $rowNum ]
do
echo "$rowNum"
echo "$i"
echo "$j"
awk -v text=$(awk -v numb=$i '{print $numb}' temp) -v num=$j 'BEGIN{FS=","} $1 ~ text {print $num}' > temp${i}
echo "testing flag"
i=$(expr $i + 1)
done
......
When I run it I get
101
1
3
And then it just hangs with "awk * script.sh text.txt" written on the tab of the terminal continuously so it's definately just hanging on the awk command but I can't figure out how to fix it.
Thank-you
Looks like you didn't supply input file for awk, so it's reading stdin.
I'm trying to pass a variable to nawk in a bash script, but it's not actually printing the $commentValue variable's contents. Everything works great, except the last part of the printf statement. Thanks!
echo -n "Service Name: "
read serviceName
echo -n "Comment: "
read commentValue
for check in $(grep "CURRENT SERVICE STATE" $nagiosLog |grep -w "$serviceName" | nawk -F": " '{print $2}' |sort -u ) ; do
echo $check | nawk -F";" -v now=$now '{ printf( "[%u]=ACKNOWLEDGE_SVC_PROBLEM;"$1";"$2";2;1;0;admin;$commentValue"\n", now)}' >> $nagiosCommand
done
$commentValue is inside an invocation to nawk, so it is considered as a variable in nawk, not a variable in bash. Since you do not have such a variable in nawk, you won't get anything there. You should first pass the variable "inside" nawk using the -v switch just like you did for the now variable; i.e.:
... | nawk -F";" -v now=$now -v "commentValue=$commentValue"
Note the quotes - they are required in case $commentValue contains whitespace.