CODEWITHSUNDEEP
bashshell
mr_Sweet
mr_Sweet

Reputation: 53

how to pass string to bash command as a parameter

I have a string variable which contains a loop. 

loopVariable="for i in 1 2 3 4 5 do echo $i done"

I want to pass this variable to a bash command inside the shell script. But i am always getting an error 

bash $loopVariable

i've tried also 

bin/bash $loopVariable

But it also doesn't work. Bash treats the string giving me an error. But theoretically it execute it. I am not sure what am i doing wrong 

bash: for i in 1 2 3 4 5 do echo $i done: No such file or directory

I have also tried to use this approach using while loop. But getting the same error 

i=0 
loopValue="while [ $i -lt 5 ]; do make -j15 clean && make -j15 done"
bash -c @loopValue

when I use bash -c "@loopValue" i get following error

bash: -c: line 0: syntax error near unexpected token `done'

and when i use just use bash -c @loopValue 

[: -c: line 1: syntax error: unexpected end of file

Upvotes: 5

Views: 13915

Answers (3)

SLePort
SLePort

Reputation: 15471

You don't need to open a new process with bash -c. You can use a Bash function instead:

function loopVariable {
  for i in 1 2 3 4 5; do echo $i; done
}

loopVariable

Please note that since no suprocess is created, you don't need to export your variables to use them in a child process as with a bash -c. All of them are available in the script scope.

output:

1
2
3
4
5

Upvotes: 4

user000001
user000001

Reputation: 33387

You can add the -c option to read the command from an argument. The following should work:

$ loopVariable='for i in 1 2 3 4 5; do echo $i; done'
$ bash -c "$loopVariable"
1
2
3
4
5

from man bash:

  -c         If the -c option is present, then commands are read from  the
             first non-option argument command_string.  If there are argu‐
             ments after the command_string,  they  are  assigned  to  the
             positional parameters, starting with $0.

Another way is to use the standard input:

bash <<< "$loopVariable"

Regarding the updated command in the question, even if we correct the quoting issues, and the fact that the variable is not exported, you are still left with an infinite loop since $i never changes:

loopValue='while [ "$i" -lt 5 ]; do make -j15 clean && make -j15; done'
i=0 bash -c "$loopValue"

But it would almost always be better to use a function as in @Kenavoz' answer.

Upvotes: 10

Ipor Sircer
Ipor Sircer

Reputation: 3141

loopvariable='for i in 1 2 3 4 5; do echo $i; done'
bash -c "$loopvariable"

Upvotes: 2

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