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"
Related
This question already has answers here:
Assignment of variables with space after the (=) sign?
(4 answers)
Closed 2 years ago.
How to write command result into empty variable?
#!/bin/bash
today=''
$today = $(date)
echo $today
There shouldn't be a space around the =
On variable assignment, no need for the $
#!/bin/bash
today=''
today="$(date)"
echo "${today}"
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 2 years ago.
I ran into a strange problem that I do not understand. Why are multiple spaces not present in the output of the following command?
$ d='A B'
$ echo $d
A B
use in double quotes:
echo "$d"
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
When to wrap quotes around a shell variable?
(5 answers)
Closed 2 years ago.
I want to use:
schema=$(kubectl exec -n $namespace -it $podName -- bash -c "./spiral orm:schema")
echo $schema
But eventually in schema variable recorded only the first line from the result of bash execution.
How to make it use all lines?
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