Bash script dynamic file naming

teh_raab · 01-05-2009, 07:28 AM

I need to create a variable name on the fly then store information from a file to it e.g.

--------------------------------------------------
#Get Details from ini file
tmp=`cat "config.ini" | grep NumberOfLoadables=`

#Store info
arr_Load${x}_Type${j}=`echo ${tmp#*=}`
--------------------------------------------------

But, when i run the script i get an error like this..

arr_Load1_Type1=osver_db.xml: command not found

I think the problem is with how "arr_Load${x}_Type${j}" is being create! I need to have the two indexes in the variable name and dynamically.

Does anyone have any ideas?

Hko · 01-05-2009, 07:54 AM

What does a "config.ini" file look like?
How does you script generate "x" and "j"?
How do you expect/need the values of the array members to look like? (e.g. arr_Load1_Type1="osver_db.xml")

Anyways, (guessing this information) you could try this:

Code:

arr_Load${x}_Type${j}="${tmp#*=}"

colucix · 01-05-2009, 07:59 AM

You have to use eval to build the command line which does the assignment:

Code:

eval arr_Load${x}_Type${j}=$(echo ${tmp#*=})

in this way the shell makes the proper substitutions and then executes the resulting command. To retrieve the value of the variable, you have to use indirect reference in conjunction with eval:

Code:

eval echo \$arr_Load${x}_Type${j}

in this way the shell first substitutes ${x} and ${j} with their actual values, then substitutes the resulting variable name with its actual value.

teh_raab · 01-05-2009, 08:00 AM

Hi Hko,

Thanks for your quick reply.

I am already using arr_Load${x}_Type${j}=`echo ${tmp#*=}` and the script seems to only complain about the arr_Load${x}_Type${j} part.

As for x and j, they are the index of the nested loop in which the command is in!

And the config.ini contains bits like this

NumberOfLoadables=2
NumberOfFiles=13
NumberOfXMLs=4

But a bit too complex to explain properly.

teh_raab · 01-05-2009, 08:09 AM

Thanks Colucix, using the eval command seems to have worked.

Your saying to echo the content of the variable we have to use "eval echo \$arr_Load${x}_Type${j}"

could you explain why we cannot simply echo the result without the use of eval.

Thanks in advance.

colucix · 01-05-2009, 08:20 AM

Just try it out. If you use echo without indirect reference (that is without escaping the $ sign) the shell interprets it as a simple string concatenation:

Code:

echo $arr_Load${x}_Type${j}

the command above substitutes $arr_Load, ${x} and ${j} with their actual values, being null the value of $arr_Load.

If you escape the first $ sign without using eval, the shell interprets it as a literal $

Code:

echo \$arr_Load${x}_Type${j}

and the resulting output will be something like "$arr_Load1_Type1".

Finally the eval statement forces the shell to do two substitutions: the first for ${x} and ${j} protecting the first escaped $ sign; the second for the resulting $arr_Load1_Type1 that is the dynamic variable name you're looking for.

colucix · 01-05-2009, 08:27 AM

In addition, for a better understanding of what the shell is trying to do, you can run the script using bash -x. In this way you will get a detailed trace of all the commands executed by the shell. Very useful when doing some complex substitutions. For example, suppose you have a script called test.sh:

Code:

#!/bin/bash
x=1
j=1
echo Assign the dynamic variable
eval arr_Load${x}_Type${j}=$(echo ciao)
echo
echo Retrieve the value of the dynamic variable
eval echo \$arr_Load${x}_Type${j}
echo
echo Demonstrate the simple echo doing string concatenation
echo $arr_Load${x}_Type${j}
echo
echo Demonstrate the effect of the escaped \$ without eval
echo \$arr_Load${x}_Type${j}
echo

Running this script using the aforementioned method gives the following output

Code:

$ bash -x test.sh
+ x=1
+ j=1
+ echo Assign the dynamic variable
Assign the dynamic variable
++ echo ciao
+ eval arr_Load1_Type1=ciao
++ arr_Load1_Type1=ciao
+ echo

+ echo Retrieve the value of the dynamic variable
Retrieve the value of the dynamic variable
+ eval echo '$arr_Load1_Type1'
++ echo ciao
ciao
+ echo

+ echo Demonstrate the simple echo doing string concatenation
Demonstrate the simple echo doing string concatenation
+ echo 1_Type1
1_Type1
+ echo

+ echo Demonstrate the effect of the escaped '$' without eval
Demonstrate the effect of the escaped $ without eval
+ echo '$arr_Load1_Type1'
$arr_Load1_Type1
+ echo

teh_raab · 01-05-2009, 08:37 AM

Thanks, that explaination really helped