how to pass string to bash command as a parameter

Question

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

Your while loop should be while [ $i -lt 5 ]; do make -j15 clean && make -j15; done, (missing ; before done), and your bash command: bash -c $loopValue, not bash - c @loopValue. — SLePort
– SLePort, Commented Nov 8, 2016 at 13:04
now its giving ` -c: line 1: syntax error: unexpected end of file` — mr_Sweet
– mr_Sweet, Commented Nov 8, 2016 at 13:09

user000001 · Accepted Answer · 2016-11-08 13:14:07Z

11

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.

edited Nov 8, 2016 at 13:14

answered Nov 8, 2016 at 12:33

user000001

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13 Comments

mr_Sweet Over a year ago

and what does -c do?

user000001 Over a year ago

It allows bash to read the command to be executed from the argument list, instead of the standard input.

mr_Sweet Over a year ago

can i use the same approach with while loop?

user000001 Over a year ago

@KvyatkovskySergey: Sure, why not? Any valid bash command (or list of commands) is accepted

mr_Sweet Over a year ago

i am assigning the loopVariable value to while [ 0 -lt 3 ] do make -j15 clean && make -j15 done but getting the same error

|

SLePort · Accepted Answer · 2016-11-08 14:37:17Z

4

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:

edited Nov 8, 2016 at 14:37

answered Nov 8, 2016 at 12:47

SLePort

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Comments

Ipor Sircer · Accepted Answer · 2016-11-08 12:33:54Z

2

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

answered Nov 8, 2016 at 12:33

Ipor Sircer

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