Executing a command stored in variable

Question

I would like to execute a series of commands that compile my program. But it does not seem to be working. This is my script

#!/bin/bash

task[0] = $(cd vssl/make/; make clean;)
compile[0] = $(make all)
error[0] = $(echo "We failed at vssl install")

task[1] = $(cd ../../web/make/; make clean;)
compile[1] = $(make install)
error[1] = $(echo "We failed at web install")

eval "$task[0]"
if (eval "$compile[0]"); then
    eval "$task[1]"
    if (eval "$compile[1]"); then
        echo "ALL DONE"
    else
        eval "$error[1]"
    fi
else
    eval "$error[0]"
fi

I have tried bunch of other methods, where I replace command declaration with this for example

task[0] = 'cd vssl/make/; make clean;'

This is just one error line, other lines say the same error But every time I get same errors saying

./install_everything.sh: line 4: task[0]: command not found

What exactly am I doing wrong here? I have tried simply substituting the actual command instead of variables like this

#!/bin/bash

cd vssl/make/
make clean
if(make all); then
    cd ../../web/make/
    make clean
    if(make install); then
        echo "ALL DONE"
    else
        echo "We failed at web install"
    fi
else
    echo "We failed at vssl install"
fi

And it worked perfectly. So I am quite stumped as to why my first method does not work.

codaddict · Accepted Answer · 2011-11-03 13:06:33Z

3

Remove the space:

task[0] = $(cd vssl/make/; make clean;)
       ^ ^

answered Nov 3, 2011 at 13:06

codaddict

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2 Comments

Quillion Over a year ago

Oh wow, thank you so much, that has worked. What is the difference between space and no space?

glenn jackman Over a year ago

@Quillion, it's the basic syntax of the shell. Commands and arguments are separated by space, so when the shell sees var = value, it's looking for the command var, passing 2 arguments to it. Variable assignment must be var=value. It hurts readability, but that's the way it is.

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