This question already has answers here:
Is there a list of 'if' switches anywhere?
(5 answers)
Closed 1 year ago.
What is the -n option in if statement in Linux?
Let's say there is a code:
if [ -n "$variable" ];then break;fi
What is it checking about the variable in if statement?
-n checks if the corresponding string variable (in your case $variable) is of non-zero length.
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:
Extract substring in Bash
(26 answers)
Closed 6 years ago.
How to retrieve the first 10 characters of a variable with Bash?
FOO="qwertzuiopasdfghjklyxcvbnm"
I need to get qwertzuiop.
If the variable is: FOO="qwertzuiopasdfghjklyxcvbnm"
then
echo ${FOO:0:10}
will give the first 10 characters.
Use the head command.
echo $FOO | head -c 10
=> qwertzuiop
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 $schema
But eventually in schema variable recorded only the first line from the result of bash execution.
How to make it use all lines?
This question already has answers here:
What does "-ne" mean in bash?
(3 answers)
Closed 3 years ago.
i am new to shell, I met a code "$# -ne 4". "$#" means the number of command-line arguments that were passed to the shell program, then what does "-ne" mean?
-ne in bash denotes 'not equal to'
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"