Consider the following script
#!/bin/bash
abc
Here I want the "abc" cmd to be aliased to echo "Hi"(say)
But I don't have control to modify the script.
Is there a way to make the cmd behavior change without touching the script?
$ cat ./someScript.sh
#!/bin/bash
abc
$ ./someScript.sh
./someScript.sh: line 2: abc: command not found
Define a function named abc, and export it.
$ abc() { echo Hi; }
$ export -f abc
Now:
$ ./someScript.sh
Hi
Related
How is Bash simple one line syntax to check and get the invalid or unrecognized command fed in shell or terminal prompt to be saved in a file? Illustration e.g.
$ run_bash_command FILE
$ foo
bash: foo: command not found
$ bar
bash: bar: command not found
$ baz
bash: baz: command not found
$ cat FILE
foo
bar
baz
With bash version >= 4.0 I suggest to add this function to ~/.bashrc and start a new session:
command_not_found_handle() {
echo "bash: $1: command not found" >&2;
echo "$#" >> /tmp/file.txt;
}
Look at this bash script (script.sh):
#!/bin/bash
echo "aaa"
echo "bbb"
...
echo $1
...
Now, i am trying to run this script this way:
./script.sh $(cat file1)
I have a problem:
the "cat file1" is run before script.sh. Bash is evaluating all arguments before running script.sh
I would like to run "cat file1" inside script.sh, on the "echo $1" line.
How can i do this ?
I have tried this:
./script.sh $(eval 'cat file1')
But it gives me the same result...
Thanks
I think, there is no way unless you can modify the script script.sh.
If you can modify script.sh, you can write your script like this:
#!/bin/bash
echo "$($1)"
Then call it in this way:
./script.sh "cat file1"
This means, you pass the command to be executed to the script and you echo the result of the executed command in the script.
But the above script is a little cumbersome. This one is simpler and would do the same:
#!/bin/bash
$1
Suppose I have the following shell program.
#!/bin/sh
FOO="foo"
echo $FOO | cat
I want to generate another shell program that does the same thing as this one, except that all shell variables have been substituted. For example,
#!/bin/sh
echo "foo" | cat
I know that I can get close if I run the above program using #!/bin/sh -x, but that output does not preserve redirections. Instead, I get
+ FOO=foo
+ echo foo
+ cat
foo
Any ideas?
The following shell:
$ cat eval.sh
echo "#!/bin/sh"
FOO="foo"
echo "echo $FOO | cat"
will write a shell:
$ sh eval.sh
#!/bin/sh
echo foo | cat
which does what you need.
I have the following test.sh script:
#!/bin/sh
echo "MY_VARIABLE=$MY_VARIABLE"
Well, if I execute the following:
export MY_VARIABLE=SOMEVALUE
/bin/bash test.sh
it prints:
MY_VARIABLE=
Why the MY_VARIABLE is not read in the test.sh script?
You can reproduce the context here using the following script:
touch test.sh
chmod a+x test.sh
echo "#!/bin/sh" >> test.sh
echo "echo "MY_VARIABLE=$MY_VARIABLE"" >> test.sh
export MY_VARIABLE=something
/bin/bash test.sh
In your script to create the context, the line
echo "echo "MY_VARIABLE=$MY_VARIABLE"" >> test.sh
creates the following line in test.sh:
echo MY_VARIABLE=
if MY_VARIABLE was unset before. The expansion of $MY_VARIABLE is done in the shell that prepares your context.
If you use single quotes
echo 'echo "MY_VARIABLE=$MY_VARIABLE"' >> test.sh
the script test.sh contains the correct line
echo "MY_VARIABLE=$MY_VARIABLE"
and prints MY_VARIABLE=something as expected.
Everything works well but if you want your parent process to keep environment update, you must source your script:
source test.sh
Otherwise, changes will only have effect during the execution of your script.
You can consider it the same as sourcing your ~/.bashrc file.
i have a bash script show below in a file called test.sh
#!/usr/bin/env bash
echo $1
echo "execution done"
when i execute this script using
Case-1
./test.sh "started"
started
execution done
showing properly
Case-2
If i execute with
bash test.sh "started"
i'm getting the out put as
started
execution done
But i would like to execute this using a cat or wget command with arguments
For example like.
Q1
cat test.sh |bash
Or using a command
Q2
wget -qO - "url contain bash" |bash
So in Q1 and Q2 how do i pass argument
Something simlar to this shown in this github
https://github.com/creationix/nvm
Please refer installation script
$ bash <(curl -Ls url_contains_bash_script) arg1 arg2
Explanation:
$ echo -e 'echo "$1"\necho "done"' >test.sh
$ cat test.sh
echo "$1"
echo "done"
$ bash <(cat test.sh) "hello"
hello
done
$ bash <(echo -e 'echo "$1"\necho "done"') "hello"
hello
done
You don't need to pipe to bash; bash runs as standard in your terminal.
If I have a script and I have to use cat, this is what I'll do:
cat script.sh > file.sh; chmod 755 file.sh; ./file.sh arg1 arg2 arg3
script.sh is the source script. You can replace that call with anything you want.
This has security implications though; just running an arbitrary code in your shell - especially with wget where the code comes from a remote location.