This question already has answers here:
Brace expansion with variable? [duplicate]
(6 answers)
how to use variables with brace expansion [duplicate]
(2 answers)
Closed 2 years ago.
I have the following in my bash file:
echo "Spawning $1 processes"
for i in {1..$1}
do
(go run loadgen.go $2 &)
echo "done."
done
However, I can only seem to get my go file to execute once. I know that they're started in the background, but each of my go files should append to the same log file (I can reproduce this by running my bash script multiple times). Am I doing something wrong to get this to iterate multiple times?
Related
This question already has answers here:
How do I use shell variables in Makefile actions?
(2 answers)
Using bash variables in Makefile
(3 answers)
Closed 5 months ago.
I have a simple makefile
SHELL := /bin/bash
test:
VAR='test' && echo ${VAR};
However, this does not exhibit the same behavior as typing this command into a normal shell, and it doesn't echo anything. Not sure what I am missing.
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Bash - variable variables [duplicate]
(4 answers)
Closed 2 years ago.
I recently started bash scripting and got stuck with a very basic usecase, searched stackoverflow/google but couldn't find a way to achieve what I am trying to do.
I have a script color.sh
#!/bin/bash
Apple="Red"
Orange="Orange"
Banana="Yello"
echo $$1
What I am trying to achieve is print the color of fruit and accept fruit from command line. The output I want is
./color.sh Apple -> Red, but what I get is some random number which I think is process Id.
This question already has answers here:
How do I parse command line arguments in Bash?
(40 answers)
How to get exact command line string from shell?
(2 answers)
Closed 3 years ago.
Suppose my script.sh could take a number of options and arguments. What is the best way to find out what the script was invoked with (form inside the script)?
For eg., someone called it with script.sh --foo_option bar_arg
Is there a way to echo that exact command they typed from inside the script?
I've tried echo !! which does not work inside a script.
This question already has answers here:
Why do you need to put #!/bin/bash at the beginning of a script file?
(10 answers)
What is the preferred Bash shebang ("#!")?
(6 answers)
Closed 4 years ago.
What is the significant of using #!/bin/bash in the starting of bash script? Can we write a bash script without #!/bin/bash ?
This line is called shebang. It’s a ‚magic‘ line telling the program loader (kernel) how to execute a script on unixoid systems.
Cf. https://en.m.wikipedia.org/wiki/Shebang_(Unix)
This question already has answers here:
Writing a Bash script without the shebang line
(2 answers)
Bash script execution with and without shebang in Linux and BSD
(2 answers)
Closed 5 years ago.
If after logging into my system I type: bash (to use bash subshell) and then try to run a bash script (e.g. example.sh), then does it matter if I do not put #!/bin/bash as the first line of the script or it is fine since I am already inside bash subshell?