how to execute two commands stored in one bash variable

Question

Let's say there is one bash variable

run1="date"

I need to execute date by

${run1}

And it works, since it prints current time. But if I put two commands in the variable,

run2="date; echo foo"

I can't execute the commands stored in variable run2, since ${run2} complains

date;: command not found

favoretti · Accepted Answer · 2012-08-09 17:50:36Z

8

Try:

eval ${run2}

This should help.

answered Aug 9, 2012 at 17:50

favoretti

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2 Comments

it works... maybe i am inquisitive, but can you explain why it works for run1 but not for run2...

Because it is quoted, the semi-colon is part of the first word after quote removal, not a syntactic element to separate two commands.

twalberg · Accepted Answer · 2012-08-09 17:55:13Z

2

Try eval "${run2}". That will interpret the variable as a sequence of commands to be run.

answered Aug 9, 2012 at 17:55

twalberg

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