This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 6 years ago.
Following is the input from command line:
bash script.sh "input*" ";" "21" "yyyy-MM-dd"
and in my current directory i have two input files:
input1.txt
input2.txt
While i want to get value 'input*' from input '$1' i got 'input1.txt input2.txt' as input.
Is there any way to get value 'input*' from '$1'?
you can use "$1" where 1 is the number of first input parameter
#!/bin/bash
echo "$1"
but you have to pass the input*.txt like this input\*.txt
output
[shell] ➤ ./test7.sh input\*.txt
input*.txt
Regards
Claudio
Related
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:
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:
Capturing multiple line output into a Bash variable
(7 answers)
Closed 3 years ago.
I have a small script where I appended the output of linux mpstat to a log file.
#/bin/bash
CPU_USAGE=$(mpstat)
echo $CPU_USAGE >> temp.log
The problem is that the output of mpstat on the terminal is formatted properly in 3 lines like so
However, the output to the file is all in one line.
How do I format the output like the one on the terminal?
Just quote the variable so it is not seen as several different parameters to be printed one after the other:
echo "$CPU_USAGE" >> temp.log
You could just directly pipe the output to the file:
#!/bin/bash
mpstat >> temp.log
If you must store it in a variable, then quote it like:
#!/bin/bash
CPU_USAGE=$(mpstat)
echo "$CPU_USAGE" >> temp.log
Otherwise, bash will not interpret the newlines as part of the message to echo, but the whole output as a list of short strings to output.
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"