Here's a MWE:
#!/bin/bash
INFILE=$1
echo `echo $INFILE | awk '{print(substr($0,8,3))}'`
PATH=${INFILE%/*}
echo `echo $INFILE | awk '{print(substr($0,8,3))}'`
exit
Apparently the first awkcommand runs fine, but in the second command bash doesn't recognize awkanymore!
This is what I get running it (assuming that f_mwe.sh is the name of the file):
$ ./f_mwe.sh /home/something/path/this_is_the_name.txt
ome
./f_mwe.sh: line 31: awk: command not found
$
I have tried defining /bin/sh and ksh at the beginning also but got the same results. I have no idea what's causing this.
Any help is appreciated.
You are overwriting the PATH variable and not appending to it I believe. You should append to the PATH variable.
Related
So I am trying to make a script that contains egrep and accepts a numeric variable
#!/bin/bash
var=$1
list="egrep "^.{$var}$ /usr/share/dict/words"
cat list
For example, if var is 5, I would like this script to print out every line with 5 characters. For some reason the script does not do that. Help would be greatly appreciated!
Your script doesn't work because there are several problems with these lines:
list="egrep "^.{$var}$ /usr/share/dict/words"
cat list
The first line isn't complete, it's missing a closing quote,
Even if you fixed it, you're assigning a literal string to list, not the output of a command,
RE and filename should be separated
cat doesn't print a variable's content, echo does that.
So:
#!/bin/bash
var="$1"
list="$(egrep '^.{'"$var"'}$' /usr/share/dict/words)"
echo "$list"
should work.
Or even better, you can use just an awk command:
awk 'length==5' /usr/share/dict/words
with $1 or any other variable:
awk -v n="$1" 'length==n' /usr/share/dict/words
#!/bin/bash
echo "Number of hosts entered are "$#
echo "Hostnames are "$#
for i in "$#"
do
echo "Logging in to the host "$i
pbsh root#$i '
ipaddr=`ip r | awk '{print $9}'`
if [ ipaddr = 172.*.*.* ]
then
echo "Script can not be run in this IP series"
exit
else
cd /var/tmp ; wget http://**********
fi'
done
After executing the above script it is throwing below error. The script is getting execute but not in the desired way.
awk: cmd. line:1: {print
awk: cmd. line:1: ^ unexpected newline or end of string
I am newbie to the scripting. Kindly correct me if anything wrong in the script.
In the listing which you posted, the opening single quote in the pbsh line is closed by single quote immediately followed by the awk command. You can escape the latter by prefixing it with a backslash.
If pbsh also accepts the command to be executed from stdin, an alternative would be to use a HERE document (see the bash man-page, section Here Documents).
UPDATE: Gordon Davisson is right in his comment. \' doesn't work either. pbsh insists on getting the command to be executed as a single argument, you could either fiddle around with the quotes, as he suggested, or put the whole input for pbsh into a separate file and use, i.e.,
pbsh root#$i $(<input_script.pbsh)
I know this seems fairly trivial. But I have no idea where I am going wrong. I have a shell script where I download a package based on the input argument and then extract the package name. This is how I do it:
wget $1
echo $1 | awk -F/ '{print $NF}'
I run it like this bash scrip.sh http://apache.claz.org/phoenix/apache-phoenix-4.10.0-HBase-1.2/bin/apache-phoenix-4.10.0-HBase-1.2-bin.tar.gz
I download the package and then the second line splits the input variable along the / delimiter I get apache-phoenix-4.10.0-HBase-1.2-bin.tar.gz. Now I want to assign the result of the second line to a variable. I change my script to a dir=$($1 | awk -F/ '{print $NF}') and add an echo $dir to the script to see the result. However I keep running into this error : line 2: http://apache.claz.org/phoenix/apache-phoenix-4.10.0-HBase-1.2/bin/apache-phoenix-4.10.0-HBase-1.2-bin.tar.gz: No such file or directory
I tried wrapping the command into `` but the problem persists. I am not cd-ing into any directory so I have no idea why this error keeps showing up.
I have to delete a line in a file from inside a shell script.
I am trying this:
linenumber=0
##CHeck If server IP exists
if grep -wq $serverip $FILE; then
echo "IP exists"
linenumber=$(awk -v serverip="$serverip" '$0 ~ serverip {print NR}' $FILE)
echo "$linenumber"
sed -e '${$linenumber}d' $FILE
fi
Basically I extract the line number and then want to delete it.
sed -e '1d' $FILE --> WOrks on CLI but inside script does not work
Why? How to get it working ?
This is simply a case of using the incorrect quotes around your sed command, so the variable isn't being used. Ignoring the fact that you're unnecessarily using 3 tools when 1 would suffice, the fix is this:
sed -e "${linenumber}d" "$FILE"
Perhaps your requirement is more complex than it appears but I would suggest changing your entire script to this:
awk -v serverip="$serverip" '!($0 ~ serverip)' "$FILE"
This prints every line that doesn't contain the shell variable $serverip. It is assumed that you have escaped any regex meta-characters present in the variable.
Alternatively (and more succinctly):
sed "/$serverip/d" "$FILE"
If you actually want the messages to be printed out (I assumed that they were for debugging), then that's easy enough to achieve:
awk -v serverip="$serverip" '$0 ~ serverip { print "IP exists"; print NR; next } 1' "$FILE"
If you're not familiar with the 1 at the end, it's just a common shorthand which causes awk to print each line (1 is always true and the default action is { print }).
I have the following line at the first line in my script file:
#!/bin/sh
So I'm using csh.(?)
I wanto assign the output of the following to an environment variable:
echo $MYUSR | awk '{print substr($0,4)}'
I try:
set $MYVAR = echo $MYUSR | awk '{print substr($0,4)}'
But it doesn't work,
How can I do it? I want to do it in a sh file.
Your script should look like
#!/bin/csh
set MYVAR = `echo $MYUSR | awk '{print substr($0,4)}'`
echo $MYVAR
I don't have a way to test this right now, let me now if it doesn't work.
If you've inherited the basis of your script from someone else, with the #!/bin/sh,
then you have to find out if /bin/sh is really the bourne shell, or if it is a link to /bin/bash
You can tell that by doing
ls -l /bin/sh /bin/bash
if you get back information on files where the size is exactly the same, the you're really using bash, but called as /bin/sh
So try these 2 solutions
MYVAR=$(echo $MYUSR | awk '{print substr($0,4)}')
echo $MYVAR
AND
MYVAR=``echo $MYUSR | awk '{print substr($0,4)}``
echo $MYVAR
# arg!! only one pair of enclosing back-ticks needed,
# can't find the secret escape codes to make this look exactly right.
in all cases (csh) included, the back-ticks AND the $( ... ) are known as command substitution.
What every output comes from running the command inside, is substituted into the command line AND then the whole command is executed.
I hope this helps.
if it's /bin/sh it's bourne shell or bash, and use back quotes to execute something and this to assign that...
MYVAR=`echo $MYUSR | awk ...`
That script first line indicates that it should be interpreted by the Bourne shell (sh), not csh. Change it to
#!/bin/csh
The first line of your code shows clearly you are not using a csh. You are using a plain sh environment/shell. You have 2 options:
Either change the first line to #!/bin/csh OR
Keeping first line unchanged, update the code for setting the variable.
MYVAR=`echo $MYUSR | awk '{print substr($0,4)}`
echo $MYVAR
Let me know, if you get any error.