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?
Related
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Script parameters in Bash
(5 answers)
How do I parse command line arguments in Bash?
(40 answers)
Closed 2 years ago.
I made a .sh file with a program, the current input is as follows:
$ ./myprogram.sh
file.txt
How can I make it so the input is as follows instead:
$ ./myprogram.sh file.txt
Inside the shell script, you can refer to the arguments by their positions as $1, $2, etc. Note that the arguments start at 1, and $0 is the name of the executed script. $# contains the total number of arguments.
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:
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:
Capturing multiple line output into a Bash variable
(7 answers)
Closed 5 years ago.
I'm trying to capture a block of text into a variable, with newlines maintained, then echo it.
However, the newlines don't seemed to be maintained when I am either capturing the text or displaying it.
Any ideas regarding how I can accomplish this?
Example:
#!/bin/bash
read -d '' my_var <<"BLOCK"
this
is
a
test
BLOCK
echo $my_var
Output:
this is a test
Desired output:
this
is
a
test
You need to add " quotes around your variable.
echo "$my_var"