This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 6 months ago.
I want to grep a number of file in the current directory on linux.
E = $( grep -F "TOTEN" OUTCAR | tail -1 | awk '{printf "%12.9f \n",$5}')
When I run the script, it keeps telling me the command is not found.
I tried many ways to solve it by searching online but it did't work.
Related
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Why does a space in a variable assignment give an error in Bash? [duplicate]
(3 answers)
Closed 2 years ago.
I'm trying to store in a variable the temperature of the computer. I tried this but it doesn't work:
#!/bin/bash
temp = cat "/sys/class/thermal/thermal_zone0/temp"
echo "$temp"
i tried this too:
#!/bin/bash
temp = $(cat "/sys/class/thermal/thermal_zone0/temp")
echo "$temp"
but nothing works, it always says
./temp.sh: line 2: temp: command not found
Spaces are crucial! This works fine:
# NO space around `=`
temp=$(cat "/sys/class/thermal/thermal_zone0/temp")
echo "$temp"
This question already has answers here:
What is indirect expansion? What does ${!var*} mean?
(6 answers)
Dynamic variable names in Bash
(19 answers)
Is it possible to build variable names from other variables in bash? [duplicate]
(7 answers)
Closed 5 years ago.
I'm trying to execute below bash shell script, but not getting the expected output. Possible i'm doing something wrong or it's not the way of doing this.
#bin/bash
#set -x
path1_one=/home/dell/scripts
echo $path1_one
param_val=path1_one
param1=$( echo "$param_val" | awk -F '_' '{ print $0 }' )
#path2="$path1"
echo $param1
#echo $path2
Output:
/home/dell/scripts
path1_one
Expected Output:
/home/dell/scripts
/home/dell/scripts
Both variable value should be same,but don't know why param1 value is not reflecting with path1_one
You need to tell the script that you want to use the value of the variable path1, not the name path1.
Use:
path2="$path1"
This question already has answers here:
Length of string in bash
(11 answers)
Closed 6 years ago.
I use Cent OS 7, and have written a Bash script.
I tried to get the length of variable:
#!/bin/bash
URL_STRING="1";
VAR_LENGTH=${#URL_STRING}
echo $VAR_LENGTH;
But I get syntax error
Try this code
echo $URL_STRING| awk '{printf("%d",length($0))}' | read asd
This question already has answers here:
How do you grep a file and get the next 5 lines
(3 answers)
Closed 6 years ago.
I can only grep and display a particular line of a log file.
What if I want to display the next line together?
Many thanks in advance.
cat filelogname | grep keyword
example : cat /var/log/httpd/error.log |grep Hello
This question already has answers here:
Bash : extracting part of a string
(4 answers)
Closed 6 years ago.
Say I have a string 0.0.25, how do I delete the last part after dot (including it) to make it like 0.0? Note that last part can have variable number of digits.
In bash you can do the following:
echo "${var%.*}"
See the Shell Parameter Expansion section of the manual.
Using awk you could:
echo "0.0.25" | awk -F. '{print $1"."$2}'